t15 beyond sql rahimi

7/31/2019 T15 Beyond SQL Rahimi

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Outline

Data Warehousing concepts

On-Line Analytical Processing (OLAP) Using SQL to perform analytic functions

Oracles advanced analytic functions

Acknowledgement:

Some of the material in this presentation are based on thebook, Databases and Transaction Processing, fromAddison Wesley.

Copyright 2011 Saeed Rahimi 2


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Data Warehouse Concepts OLTP vs. OLAP

On Line Transaction Processing (OLTP)

Short burst transactions Frequent modifications

Updates Inserts Deletes

Normalized (3NF, BCNF or 4NF) Transactions access only a small fraction of the database

On-Line Analytic Processing (OLAP) Main use is in decision support, business analysis

Complex, aggregate and time based queries

Almost never updates (bulk update or load) Queries access a large portion of the database Queries usually take longer to run Normalized (1NF, 2NF)



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Data Warehouse Concepts OLTP vs. OLAP example

OLTP query: Update the balance of an account to show a deposit Insert into the item table to show addition of a new

inventory

Transaction: Check the balance of checking and savings

Transfer $500 from savings to checking

OLAP query:

How many skis did we sell in the Northeast and Midwestregions of the US during the last quarter of the last fiveyears



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Data Warehouse Concepts OLAP: Traditional vs. Newer Applications

Traditional

Uses data the enterprise gathers in its usual activities,perhaps in its OLTP system

Queries are ad hoc, perhaps designed and carried out-

Newer Applications

Gather data (information) actively

Buy if have to Mine the information to find out better ways of serving the

customer and/or selling more products

Hire professional to do so



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Data Warehouse Concepts Example: Traditional vs. Newer Applications

Traditional

How many skis were sold in all Northeast warehouses inthe years 2004 and 2005?

Newer Prepare a profile of the skiers for the residents of the

Northeast region

Customize our advertising and marketing to actively sell

products types these residents would want The newer approach requires Data Mining

Finding of nuggets of gold in the vast see of informationcollected in the warehouse



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Data Warehouse Concepts

BitmapJoin

MaterializedMaterializedViews

WarehouseSql mergeMultitable

Insert

Externaltables

Data Warehouse Life Cycle

OperationalSystems

StoreTransform UsersExtract PerformanceAnalyzeLoad



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Data Warehouse Concepts Data Mining

Data Mining is the art of knowledge discovery

Knowledge is used to better the business

Data mining vs. OLAP OLAP: What percentage of people who make over

$50,000 defaulted on their mortgage in the year 2010?

Data Mining: How can information about salary, networth, and other historical data be used to predict whowill default on their mortgage?



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Data Warehouse Concepts Data warehouse as a database

OLAP applications are based on a table called, fact table

For example, a supermarket application might be based onthe fact table Sales

Sales Market_Id, Product_Id, Time_Id, Sales_Amt

The table is viewed as multidimensional

The first three columns are the dimensions representing

supermarkets, products and time intervals

The fourth column, the Sales_Amt, is a function of theother three



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Data Warehouse Concepts A Cube

The fact table can be viewed as a three-dimensional cube

Each entry in this three-dimensional view represents aspecific sales amount for a given market, product and for aspecific time period



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Data Warehouse Concepts Dimension Tables

The dimensions of the fact table can be furtherdescribed with dimension tables

Sales (Market_id, Product_Id, Time_Id, Sales_Amt)

Dimension Tables

Market (Market_Id, City, State, Region)

Product (Product_Id, Name, Category, Price)

Time (Time_Id, Week, Month, Quarter)



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Data Warehouse Concepts Star Schema

Time

ProductMarket Sales



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Data Warehouse Concepts Schema of the Sales data warehouse

Time

Time_Id

Week

Month

Quarter

A2

A10

A10

A8

Identifier_1

Sales Data

Warehouse

Market Product

Time

Market

Market_Id

City

State

Region

A2

A20

A20

A20

Identifier_1

Product

Product_Id

Name

Category

Price

A2

A20

A20

N6,2

Identifier_1

Sales

Sales_Amt N8,2



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Data Warehouse Concepts Warehouses dimension tables

Marketcreate table MARKET (

MARKET_ID CHAR(2) not null,CITY CHAR(20),

STATE CHAR(20),

REGION CHAR(20),

constraint PK_MARKET primary key (MARKET_ID)

)

ro uctcreate table PRODUCT (

PRODUCT_ID CHAR(2) not null,

NAME CHAR(20),

CATEGORY CHAR(20),

PRICE NUMBER(6,2),

constraint PK_PRODUCT primary key (PRODUCT_ID)

)

Timecreate table TIME (

TIME_ID CHAR(2) not null,

WEEK CHAR(10),

MONTH CHAR(10),

QUARTER CHAR(8),

constraint PK_TIME primary key (TIME_ID)

)Copyright 2011 Saeed Rahimi 15


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Warehouses fact tablecreate table SALES (

MARKET_ID CHAR(2) not null,

PRODUCT_ID CHAR(2) not null,

TIME_ID CHAR(2) not null,

SALES_AMT NUMBER(8,2),

constraint PK_SALES primary key (MARKET_ID, PRODUCT_ID, TIME_ID),

constraint FK_SALES_MARKET_MARKET foreign key (MARKET_ID)

references MARKET (MARKET_ID),

constraint FK_SALES_PRODUCT_PRODUCT foreign key (PRODUCT_ID)

references PRODUCT (PRODUCT_ID),

constraint FK_SALES_TIME_TIME foreign key (TIME_ID)

references TIME (TIME_ID)

)



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Data Warehouse Concepts Star Schema of the Warehouse

Time Dimension Table

Time

Time_Id

Week

Month

Quarter

CHAR(2)

CHAR(10)

CHAR(10)

CHAR(8)

Market Dimension Table

Sales Fact Table

Product Dimension Table

Market

Market_Id

City

State

Region

CHAR(2)

CHAR(20)

CHAR(20)

CHAR(20)

Product

Product_Id

Name

Category

Price

CHAR(2)

CHAR(20)

CHAR(20)

NUMBER(6,2)

Sales

Market_Id

Product_Id

Time_Id

Sales_Amt

CHAR(2)

CHAR(2)

CHAR(2)

NUMBER(8,2)



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SQL> select * from sales;

MI PI TI SALES_AMT

-- -- -- ----------

M1 P1 T1 1000

M1 P2 T1 2000

M1 P3 T1 1500

M1 P4 T1 2500

M2 P1 T1 500

SQL> select * from Market;

MI CITY STATE REGION

-- -------------------- -------------------- ----------

M1 Stony Brook New York East

M2 Newark New Jersey East

M3 Oakland California West

SQL> select * from Product;

The warehouse tables

M2 P3 T1 0

M2 P4 T1 3333M3 P1 T1 5000

M3 P2 T1 8000

M3 P3 T1 10

M3 P4 T1 3300

M1 P1 T2 1001

M1 P2 T2 2001

M1 P3 T2 1501

M1 P4 T2 2501

M2 P1 T2 501

M2 P2 T2 801

...

...

...

36 rows selected.

PI NAME CATEGORY PRICE

-- -------------------- -------------------- ----------P1 Beer Drink 1.98

P2 Diapers Soft Goods 2.98

P3 Cold Cuts Meat 3.98

P4 Soda Drink 1.25

SQL> select * from Time;

TI WEEK MONTH QUARTER

-- ---------- ---------- --------

T1 Wk-1 January FirstT2 Wk-24 June Second

T3 Wk-52 December Fourth



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Data Warehouse Concepts Constellation (snow flake) schema

A data warehouse may use more than one fact

table These fact tables may share the same dimension

therefore forming a schema that looks like a snow

flake.

Time

ProductMarket Sales

WarehouseInventory



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An Introduction to

OLAP Basic O erations

using SQL


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OLAP Operations

Aggregation OLAP queries usually total(aggregate) information in the fact table

,

product, in each market, for each quarter, weuse the following:


SELECT S.Market_Id, S.Product_Id, time_ID, SUM (S.Sales_Amt) AS Total_Sale

FROM Sales SGROUP BY S.Market_Id, S.Product_Id, time_ID

order by time_id;

Aggregation


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Data Warehouse ConceptsSELECT S.Market_Id, S.Product_Id, time_ID, SUM (S.Sales_Amt) AS Total_SaleFROM Sales S

GROUP BY S.Market_Id, S.Product_Id, time_ID

order by time_id;

MA PR TI TOTAL_SALE

-- -- -- ----------

M1 P1 T1 1000

M2 P1 T1 500

M3 P1 T1 5000

M1 P2 T1 2000

M2 P2 T1 800


M3 P2 T1 8000

M1 P3 T1 1500

M2 P3 T1 0

M3 P3 T1 10

M1 P4 T1 2500

M2 P4 T1 3333

M3 P4 T1 3300

M1 P1 T2 1001

M2 P1 T2 501M3 P1 T2 5001

M1 P2 T2 2001

M2 P2 T2 801

M3 P2 T2 8001

M1 P3 T2 1501

Not all rows are shown


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Data Warehouse Concepts OLAP Operations The query on previous page returns a three dimensional view of the

results (Cube)

We can collapse the time dimension and show sales for each product in

each market. This is a two-dimensional view of the same result.

SELECT S.Market_Id, S.Product_Id, SUM (S.Sales_Amt) AS Total_Sale

FROM Sales S

GROUP BY S.Market_Id, S.Product_Id;

MI PI TOTAL_SALES

-- -- -----------

M1 P1 3003

M1 P2 6003

M1 P3 4503

M1 P4 7503

M2 P1 1503

M2 P2 2403

M2 P3 3

M2 P4 7000

M3 P1 15003

M3 P2 24003

M3 P3 33

M3 P4 9903



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Data Warehouse Concepts OLAP Operations

The same result can be viewed as follows.

This is a pivoted view of the same results.

Total Sales M1 M2 M3

Product_Id

Mar et_I

P1 3003 1503 15003

P2 6003 2402 24003

P3 4503 3 33

P4 7503 7000 9903



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Data Warehouse Concepts Pivoted view of the same result

select Product_id,

sum(case when Market_ID = 'M1'

then Sales_Amt

else NULL end)as M1,


then Sales_Amt

else NULL end)as M2,



then Sales_Amt

else NULL end)as M3

FROM Sales

GROUP BY Product_Id;

PR M1 M2 M3

-- ---------- ---------- ----------P1 3003 1503 15003

P2 6003 2403 24003

P3 4503 3 33

P4 7503 7000 9903


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Data Warehouse Concepts OLAP Operations We can now get the product sales for all markets in all quarters as:

SELECT S.Market_Id, SUM (S.Sales_Amt) AS Total_Sale

FROM Sales S

GROUP BY S.Market_Id;

MA TOTAL_SALE

-- ----------

M1 21012

M2 10909

And, finally get the total sales over all products for all markets for alltime periods as:

M3 48942


SELECT SUM (S.Sales_Amt) AS Total_Sale

FROM Sales S;

TOTAL_SALE

----------

80863


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Drilling Down Some dimension tables represent a hierarchy

For example:

Market dimension has: City State Region

Time dimension has: Week Month Quarter

When we execute queries that move down a hierarchy (e.g., fromaggregation over regions to aggregation over states) we are drillingdown.

We are adding more columns of the dimension to the query

To be able to drill down, we must have access to more specific

information.



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Dimensions do not always form a hierarchy

Some dimensions may have a lattice

For example, time dimension can be represented as a lattice

Weeks are not contained in months

We can roll up days into weeks or months, but we can only rollup weeks into quarters



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Drilling Down Example:

The first query aggregates total sales for products in each region

The second query drills down to state level.

SELECT S.Product_Id,M.Region, SUM (S.Sales_Amt)

FROM Sales S, Market MWHERE M.Market_Id = S.Market_Id

GROUP BY M.Region, S.Product_Id;

SELECT S.Product_Id,M.State, SUM (S.Sales_Amt)

FROM Sales S, Market M

WHERE M.Market_Id = S.Market_Id

GROUP BY M.State, S.Product_Id;



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OLAP Operations

Rolling Up When we execute queries that move upthe hierarchy (e.g., from states to regions) we arerolling up

e can ro up n e erarc y or use e resu s oprevious queries from lower aggregates



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Rolling Up:

The following query creates a table containing the total sales for each stateas:

CREATE TABLE State_Sales AS

SELECT S.Product_Id, M.State, SUM (S.Sales_Amt)Sales_Amt

FROM Sales S, Market M

WHERE M.Market_Id = S.Market_Id

. , . _

Table created.

select * from state_sales;

PR STATE SALES_AMT

-- -------------------- ----------

P1 California 15003

P2 California 24003

P3 California 33

P4 California 9903

P1 New Jersey 1503P2 New Jersey 2403

P3 New Jersey 3

P4 New Jersey 7000

P1 New York 3003

P2 New York 6003

P3 New York 4503

P4 New York 7503

12 rows selected.



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Rolling Up: Example

Then we can use the following to roll up the total sales for each region as

SELECT T.Product_Id, R.Region, SUM (T.Sales_Amt)

FROM State_Sales T,

(SELECT DISTINCT M.Region, M.State FROM Market M) R

WHERE R.State = T.State

GROUP BY R.Region, T.Product_Id;

PR REGION SUM(T.SALES_AMT)

-- -------------------- ----------------

P1 East 4506

P2 East 8406

P3 East 4506

P4 East 14503

P1 West 15003P2 West 24003

P3 West 33

P4 West 9903

8 rows selected.



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Pivot pivoting is changing the orientation of the cube.

Dimensions that we are pivoting on are used in the GROUP BYclause aggregation (SUM) is used on the remaining attributes


PR QUARTER SUM(SALES_AMT)

-- -------- --------------

P3 Fourth 1516

P1 First 6500

P2 First 10800

P2 Second 10803

P2 Fourth 10806

P1 Second 6503

P3 Second 1513

P1 Fourth 6506

P3 First 1510

P4 First 9133

P4 Second 9136

P4 Fourth 6137

PR Q1 Q2 Q4

-- ---------- ---------- ----------

P1 6500 6503 6506

P2 10800 10803 10806

P3 1510 1513 1516

P4 9133 9136 6137


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OLAP Operations

Product sales per quarter for all regions.SELECT S.Product_Id, T.Quarter, SUM (Sales_Amt)

FROM Sales S, Time T

WHERE T.Time_Id = S.Time_Id

GROUP BY S.Product_Id, T.Quarter

ORDER BY S.Product_Id, T.Quarter;


-- -------- --------------

P1 First 6500

P1 Fourth 6506

P1 Second 6503

P2 First 10800

P2 Fourth 10806

P2 Second 10803

P3 First 1510

P3 Fourth 1516

P3 Second 1513

P4 First 9133

P4 Fourth 6137

P4 Second 9136



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Data Warehouse Concepts OLAP Operations Pivoted results so that we can see the sales for each quarter

over all products.

Note: T3 is Q4 and not Q3 in our time table

SQL> select S.Product_id,

2 sum(case when S.Time_id = 'T1'

3 then Sales_Amt


4 else NULL end) as Q1,

5 sum(case when S.Time_id = 'T2'6 then Sales_Amt

7 else NULL end) as Q2,

8 sum(case when S.Time_id = 'T3'

9 then Sales_Amt

10 else NULL end) as Q4

11 FROM Sales S

12 GROUP BY S.Product_Id;

PR Q1 Q2 Q4

-- ---------- ---------- ----------

P1 6500 6503 6506

P2 10800 10803 10806

P3 1510 1513 1516

P4 9133 9136 6137


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Data Warehouse Concepts Pivot

Oracle 11g has a pivot operation that can also be used

Unpivot does exactly the opposite of the pivot

SQL> select * from (

2 select Product_ID, Market_ID, Sales_Amt

3 from Sales )


4 pivot

5 (6 Sum(Sales_Amt)

7 for Market_ID in ('M1','M2','M3')

8 )

9 order by Product_ID;

PR 'M1' 'M2' 'M3'-- ---------- ---------- ----------

P1 3003 1503 15003

P2 6003 2403 24003

P3 4503 3 33

P4 7503 7000 9903


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Can you write a pivot statement that generates the report onpage 35?



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What if you do not know the exact number of values forMarket_ID ini the statement on page 36?



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Data Warehouse Concepts Pivot Oracle has the capability to deal with any number in the IN

construct of the pivot statement and generate an XML report.

SET LONG 99999

select * from (

select Product_ID, Market_ID, Sales_Amt

See answer on next page


_ _ _

from Sales )

pivot xml(

Sum(Sales_Amt)

for Market_ID in (any)

)

order by Product_ID;


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Data Warehouse ConceptsPR--MARKET_ID_XML

--------------------------------------------------------------------------------

P1

M13003M21503M315003

P2

M1


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OLAP Operations

Slice - A slice is a subset of a cube corresponding toa single value for one or more members of thedimensions not in the subset

value, we are performing a slice Slicing Sales Cube in the time dimension:

total sales of each product in Wk-1

SELECT S.Product_Id, SUM (Sales_Amt)


WHERE T.Time_Id = S.Time_Id AND T.Week = Wk-1

GROUP BY S. Product_Id;



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OLAP OperationsDice The dice operation is a slice on more than two

dimensions of a cube (or more than two consecutiveslices).

When we use a GROUP BY clause in a uer to

specify part of a hierarchy, we are partitioning themulti-dimensional cube into sub-cubes. Therefore, weare dicing the cube

Example:

SELECT S.Product_Id, T.Quarter, SUM (Sales_Amt)FROM Sales S, Time T


GROUP BY T.Quarter, S.Product_Id;



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OLAP Operations Dice Dicing Sales in the time dimension: total sales for each product in each quarter.

SELECT S.Product_Id, T.Quarter, SUM (Sales_Amt)



GROUP BY T.Quarter, S.Product_Id

ORDER BY Product_ID;



-- -------- --------------P1 First 6500

P1 Fourth 6506

P1 Second 6503

P2 First 10800

P2 Fourth 10806

P2 Second 10803P3 First 1510

P3 Fourth 1516

P3 Second 1513

P4 First 9133

P4 Fourth 6137

P4 Second 9136


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OLAP Basic Operations

usin Oracles Anal tic Functions


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OLAP Operations

OLAP queries use the GROUP BY clause ofSQL to get the answer Standard options for GROUP BY are limited

It is not easy to formulate all OLAP needs in SQL92

SQL 1999 has extended SQL with additionalaggregate functions to support OLAP needs

Oracle 11g supports these extensions We will examine these functions next



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OLAP Operations

The Cube Operator Suppose we want to obtain a tabular view of the

information that contains:

Total sales of each product for each market Total sales of each market for each product

And the grand total of all sales for all market for allproducts!

What we are after is depicted on the next page.


D t W h C t


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OLAP Operations

Sales application in the form of a spreadsheet

Market_Id

Sum(Sales_Amt) M1 M2 M3 Total

P1 3003 1503 15003 19509

P2 6003 2402 24003 32408

Product_Id P3 4503 3 33 4539

P4 7503 7000 9903 24406

Total 21012 10908 48942 80862


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OLAP Operations

To create this sheet using the standard SQL operations, we need to usethe following three queries:

-- One query to calculate the entries, without the totals

--

SELECT S.Market_Id, S.Product_Id, SUM (S.Sales_Amt)

FROM Sales S

. ar e _ , . ro uc _ ;

MA PR SUM(S.SALES_AMT)-- -- ----------------

M1 P1 3003

M1 P2 6003

M1 P3 4503

M1 P4 7503

M2 P1 1503

M2 P2 2403M2 P3 3

M2 P4 7000

M3 P1 15003

M3 P2 24003

M3 P3 33

M3 P4 9903



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OLAP Operations-- One to calculate the row totals

--

SELECT S.Product_Id, SUM (Sales_Amt)FROM Sales S

GROUP BY S.Product_Id;

PR SUM(SALES_AMT)

-- --------------

P1 19509

P2 32409

P3 4539

P4 24406

-- And one to calculate the column totals

--

SELECT S.Market_Id, SUM (Sales_Amt)

FROM Sales S

GROUP BY S.Market_Id;

MA SUM(SALES_AMT)

-- --------------

M1 21012

M2 10909

M3 48942




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Question:

Can we use some of the queries we used before togenerate the same results?

Give it a try (hint see slide on page 35)




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SQL> select C.Product_id,

2 sum(case when Market_ID = 'M1'

3 then c.Sales_Amt

4 else NULL end)as M1,

5 sum(case when Market_ID = 'M2'6 then c.Sales_Amt



9 then c.Sales_Amt


e se en as ,

11 sum(c.sales_amt) as Total

12 FROM (SELECT S.Market_Id, S.Product_Id, SUM (S.Sales_Amt)as sales_amt

13 FROM Sales S

14 GROUP BY S.Market_Id, S.Product_Id) C

15 GROUP BY C.Product_Id;



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SQL> select C.Product_id,


3 then c.Sales_Amt


5 sum(case when Market_ID = 'M2'6 then c.Sales_Amt



9 then c.Sales_Amt


e se en as ,

11 sum(c.sales_amt) as Total

12 FROM (SELECT S.Market_Id, S.Product_Id, SUM (S.Sales_Amt)as sales_amt

13 FROM Sales S

14 GROUP BY S.Market_Id, S.Product_Id) C

15 GROUP BY C.Product_Id;

PR M1 M2 M3 TOTAL

-- ---------- ---------- ---------- ----------

P4 7503 7000 9903 24406

P1 3003 1503 15003 19509

P2 6003 2403 24003 32409

P3 4503 3 33 4539


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OLAP Operations Using three queries is wasteful

The first query does much of the work of the other two

If we could save that result of the first query and thenaggregate over Market_Id and Product_Id, we could computethe other queries more efficiently

The Cube operator in SQL 1999 has been designed to helpwith these types of requirements in OLAP

The CUBE function is used in the GROUP BY clause as

GROUP BY CUBE(v1, v2, , vn)

This is equivalent to a collection of GROUP BYs, one foreach value of v



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Example using the CUBE-- Doing the three queries in one using the CUBE operator

--

SELECT S.Market_Id, S.Product_Id, SUM (S.Sales_Amt)FROM Sales S

GROUP BY CUBE (S.Market_Id, S.Product_Id);

MARK PROD SUM(S.SALES_AMT)

---- ---- ----------------

NULL NULL 80863

Questions: What do NULLs represent?

How many of them are there?

Why?

NULL P1 19509

NULL P2 32409

NULL P3 4539

NULL P4 24406

M1 NULL 21012

M1 P1 3003

M1 P2 6003

M1 P3 4503

M1 P4 7503

M2 NULL 10909

M2 P1 1503

M2 P2 2403

M2 P3 3

M2 P4 7000

M3 NULL 48942

M3 P1 15003

M3 P2 24003

M3 P3 33

M3 P4 9903



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OLAP Operations

The ROLLUP Operator ROLLUP is similar to CUBE except that instead ofaggregating all subsets of the arguments, it creates subsetsmovin from ri ht to left

ROLLUP is also supported in SQL1999 ROLLUP does exactly what is sounds

It first finds the fine-grained aggregations of the dimensions,

Then, it uses them to calculate coarse-grained aggregations,

and Uses these aggregations to find the grand total


h


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OLAP Operations ROLLUP Example:


FROM Sales S

GROUP BY ROLLUP(S.Market_Id, S.Product_Id)

GROUP BY S.Market_Id, S.Product_Id

Then with the next level of granularity aggregating the product sales foreach MarketsGROUP BY S.Market_Id

And finally, using the total sales of all products in each market it figuresout the grand total which corresponds to an empty GROUP BY clause


D W h C


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OLAP Operations

Example of ROLLUP-- The ROLLUP operator--


FROM Sales S

GROUP BY ROLLUP (S.Market_Id, S. Product_Id);

MARK PROD SUM(S.SALES_AMT)

---- ---- ----------------

M1 P1 3003M1 P2 6003

M1 P3 4503

M1 P4 7503

M1 NULL 21012

M2 P1 1503

M2 P2 2403

M2 P3 3

M2 P4 7000M2 NULL 10909

M3 P1 15003

M3 P2 24003

M3 P3 33

M3 P4 9903

M3 NULL 48942

NULL NULL 80863


D t W h C t


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What does the following Rollup generate?

SELECT S.Market_Id, S.Product_Id, S.Time_ID, SUM (S.Sales_Amt)FROM Sales S

GROUP BY ROLLUP (S.Market_Id, S. Product_Id, s.Time_ID);


D t W h C t


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MA PR TI SUM(S.SALES_AMT)

-- -- -- ----------------

M1 P1 T1 1000

M1 P1 T2 1001M1 P1 T3 1002

M1 P1 3003

M1 P2 T1 2000

M1 P2 T2 2001


M1 P2 T3 2002

M1 P2 6003M1 P3 T1 1500

M1 P3 T2 1501

M1 P3 T3 1502

M1 P3 4503

M1 P4 T1 2500

M1 P4 T2 2501

M1 P4 T3 2502

M1 P4 7503

M1 21012

And so on

D t W h C t


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OLAP Operations ROLLUP Vs. CUBE

By contrast, the same query with CUBE

first aggregates with the finest granularity

GROUP BY S.Market Id S.Product Id_ _

then with the next level of granularity (both subsets)

GROUP BY S.Market_Id

GROUP BY S.Product_Id

then the grand total with

GROUP BY


Data Warehousing


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Data Warehousing

Oracles Advanced OLAP Operations



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Advanced OLAP Operations

For this portion of the presentation, we will

use a general purpose practice database This database has the following schema

Copyright (c) 2011 Saeed Rahimi 62



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Department

Department_name

Department Payroll_numberSocial_security_numberFk_department

Last_nameFirst_nameStreetCityStatePhone

Employee

Payroll_number

Security_option

Sectab

Wge_maint

M1 1 M

Fk_payroll_numberPurchase_dateOpticianCostCheck_number

Glasses

Fk_payroll_numberPurchase_dateTool_nameTool_costPayroll_deductPaymentLast_paymentFirst_payment_dat

Emp_tools

senseCurrent_positionEmployment_dateBirth_dateWagesGender

Tax_rateBottom_wageTop_wage

ax_ra eFk_payroll_number

Fk_department_numberClassificationClassification_dateOld_wagesNew_wages

1M

MM

1 1




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ROLLUP revisited

ROLLUP is used with the GROUP BY option

GROUP BY ROLLUP (expr1, expr2)

To compute, Oracle will first group by data by expr2

,

different values of expr1 It rolls up these aggregates to figure out sub-totals for eachvalue of expr1

And it adds up these sub-totals into a grand total




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GROUP BY vs. ROLLUPSQL> -- Group by - one expression

SQL> --

SQL> select department, SUM(wages)

SQL> -- Rollup - one expression

SQL> --

SQL> select department, SUM(wages)


2 from department, employee

3 where department = fk_department

4 group by department

5 order by 1;

DEPA SUM(WAGES)

---- ----------

INT 65000

POL 87700

WEL 52000



4 group by rollup(department)

5 order by 1;

DEPA SUM(WAGES)

---- ----------

INT 65000

POL 87700

WEL 52000

204700



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GROUP BY vs. ROLLUPSQL> -- Rollup - two expressions

SQL> --

SQL> select gender, department, SUM(wages)

SQL> select gender, department, SUM(wages)


3 where de artment = fk de artment2 from department, employee


4 group by rollup(department, gender)5 order by 1,2;

G DEPA SUM(WAGES)

- ---- ----------

F POL 9800

F WEL 7000

M INT 65000M POL 77900

M WEL 45000

INT 65000

POL 87700

WEL 52000

204700

_

4 group by department, gender

5 order by 1,2;

G DEPA SUM(WAGES)

- ---- ----------

F POL 9800

F WEL 7000

M INT 65000

M POL 77900

M WEL 45000




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ROLLUP revisited

Example: figuring out sub-totals for each gender andthen grand total for the department

SQL> select gender, department, sum(wages)


3 where department = fk_department4 group by rollup (gender, department)

5 order by 1,2;

GENDER DEPARTMENT SUM(WAGES)

------ ---------- ----------

F POL 9800

F WEL 7000F 16800

M INT 65000

M POL 77900

M WEL 45000

M 187900

204700

Aggregate value

Rollup total female wages

Rollup total male wages

Rollup total wages




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Partial ROLLUP

If you place GROUP BY expressions outside theROLLUP option, Oracle will:

Aggregate values based on these expressions outside

the GROUP BY Calculates ROLLUP or subtotals on the expressions

within the ROLLUP parameter list

Computes a ROLLUP values for each unique occurrence

of the expressions outside the ROLLUP Does NOT figure out the grand total




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Partial ROLLUP example:

Figuring out a partial rollup of wages per gender within adepartment


2 from de artment em lo ee

Rolled-up values for the

POL department

Aggregate or grouped

Value for the POL

department


4 group by rollup(gender), department

5 order by 1,2;


------ ---------- ----------

F POL 9800

F WEL 7000

M INT 65000

M POL 77900M WEL 45000

INT 65000

POL 87700

WEL 52000




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ROLLUP vs. Partial ROLLUPROLLUP Partial ROLLUPgroup by rollup (gender), departmentgroup by rollup (gender, department)


------ ---------- ----------

F POL 9800

F WEL 7000

F 16800

M INT 65000

M POL 77900

M WEL 45000

M 187900

204700


------ ---------- ----------

F POL 9800

F WEL 7000

M INT 65000

M POL 77900

M WEL 45000

INT 65000

POL 87700

WEL 52000

What does this ROLLUP generate?

select gender Gender, department, sum(wages)

from department, employee

where department = fk_department

group by rollup(department, gender)

order by 1,2;Copyright (c) 2011 Saeed Rahimi 70



71/127



ROLLUP vs. Partial ROLLUPROLLUP Partial ROLLUPgroup by rollup (gender), departmentWhat does this ROLLUP generate?

select gender Gender, department, sum(wages)


------ ---------- ----------

F POL 9800

F WEL 7000

M INT 65000

M POL 77900

M WEL 45000

INT 65000

POL 87700

WEL 52000

rom epartment, emp oyee


group by rollup(department, gender)order by 1,2;


G DEPA SUM(WAGES)

- ---- ----------

F POL 9800

F WEL 7000

M INT 65000

M POL 77900M WEL 45000

INT 65000

POL 87700

WEL 52000

204700



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CUBE operator revisited

Unlike ROLLUP, CUBE will figure out sub-totals for allexpressions inside the CUBE and then rolls them up




4 group by cube (gender, department)

5 order by 1,2;


------ ---------- ----------

F POL 9800

F WEL 7000

F 16800

M INT 65000

M POL 77900M WEL 45000

M 187900

INT 65000

POL 87700

WEL 52000

204700

11 rows selected.Copyright (c) 2011 Saeed Rahimi 72



73/127

Data Warehouse Concepts Advanced OLAP Operations

Partial CUBE operator

Similar to partial ROLLUP, partial CUBE will calculate rollup

values for each unique occurrence of expression(s) outsidethe cube




4 group by cube (gender), department5 order by 1,2;


------ ---------- ----------

F POL 9800

F WEL 7000

M INT 65000M POL 77900

M WEL 45000

INT 65000

POL 87700

WEL 52000

8 rows selected.

Why is this partial cube exactly

the same as the partial Rollup?




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p


CUBE vs. Partial CUBEPartial CUBE

group by cube (gender), department

CUBE

group by cube (gender, department)


------ ---------- ----------

F POL 9800

F WEL 7000

M INT 65000

M POL 77900

M WEL 45000

INT 65000

POL 87700

WEL 52000


------ ---------- ----------

F POL 9800

F WEL 7000

F 16800

M INT 65000

M POL 77900

M WEL 45000

M 187900

INT 65000

POL 87700

WEL 52000

204700




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p


CUBE vs. ROLLUPCUBE

group by cube (gender, department)

ROLLUP

group by rollup (gender, department)

------ ---------- ----------

F POL 9800

F WEL 7000

F 16800

M INT 65000

M POL 77900

M WEL 45000

M 187900

INT 65000

POL 87700

WEL 52000

204700

------ ---------- ----------

F POL 9800F WEL 7000

F 16800

M INT 65000

M POL 77900

M WEL 45000

M 187900

204700




76/127

p


Next pages shows the side by side views of

the results for CUBE, ROLLUP, PartialCUBE and Partial ROLLUP

ny n eres ng o serva ons




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CUBE Partial CUBE

Department

Sum(Wages) INT POL WEL Total

F 0 9800 7000 16800

Gender M 65000 77900 45000 187900

Department


F 0 9800 7000

Gender M 65000 77900 45000

Tota l 65000 87 700 52 00 0 2 04 700

ROLLUP Partial ROLLUP

Total 65 000 87 700 5200 0

Department


F 0 9800 7000 16800

Gender M 65000 77900 45000 187900

Total 204700

Department


F 0 9800 7000

Gender M 65000 77900 45000

Total 65 000 87 700 5200 0



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p

Advanced OLAP Operations GROUPING Function

What is the problem with CUBE and ROLLUPfunction?

cu y n en y ng e rows a are su - o a

One way is to find the rows that contain NULL values Expressions that are sub-totaled will have a value in the

column that determines the ROLLUP

The other expressions will contain null values

This works well if the database does not have any nullvalues in it otherwise, it will be confusing

GROUPING function can help with this.




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GROUPING Function Usage: GROUPING (expression)

More than one GROUPING function calls areallowed in one SQL statement

GROUPING will return a 1 if the row is a sub-totalrow for the expression

GROUPING will return a 0 if the row is NOT a sub-total row for the expression




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GROUPING Function

Example:SQL> select gender, department, sum(wages),

2 grouping(gender) as gdr,

3 grouping(department) as dpt



6 group by rollup(gender, department)7 order by 1,2;

G DEPA SUM(WAGES) GDR DPT

- ---- ---------- ---------- ----------

F POL 9800 0 0

F WEL 7000 0 0

F 16800 0 1

M INT 65000 0 0

M POL 77900 0 0

M WEL 45000 0 0

M 187900 0 1

204700 1 1




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Making the report more readable Use of DECODE Function

column value

It acts as a complex IFTHENELSE or a CASEstatement




82/127


Making the report more readable Example: Assume the following query results

SQL> select empno, deptno

2 from emp

3 order by deptno;

EMPNO DEPTNO

---------- ----------

7782 10

7839 10

7934 10

7369 20

7876 20

7902 20

7788 20

7566 207499 30

7698 30

7654 30

7900 30

7844 30

7521 30




83/127


Making the report more readable

Now assume we want to print Ten for 10, Twenty for 20 and thirty for

30 in deptno column Decode can do this very easily

SQL> select empno, decode (deptno,

2 10, 'Ten',

3 20, 'Twent ',

4 30, 'Thirty', 'OTHER')

5 from emp

6 order by deptno;

EMPNO DECODE

---------- ------

7782 Ten

7839 Ten

7934 Ten

7369 Twenty

7876 Twenty

7902 Twenty7788 Twenty

7566 Twenty

7499 Thirty

7698 Thirty

7654 Thirty

7900 Thirty

7844 Thirty

7521 Thirty




84/127


Use of DECODE function

Example:SQL> select decode(grouping(gender),1, 'Total Wages', gender) gender,

2 decode(grouping(department),1, 'Total Per Gender', department) department,



6 group by rollup(gender, department)7 order by 1;


----------- ---------------- ----------

F POL 9800

F WEL 7000

F Total Per Gender 16800

M INT 65000

M POL 77900

M WEL 45000

M Total Per Gender 187900

Total Wages Total Per Gender 204700




85/127

How do we generate the following report?

DEPARTMENT GENDER SUM(WAGES)

----------- -------------------- ----------

INT Total Per Department 65000

INT M 65000

POL M 77900

POL F 9800

POL Total Per Department 87700

WEL F 7000

WEL M 45000

WEL Total Per Department 52000

Total Wages Total Per Department 204700




86/127

How do we generate the following report?SQL> select decode(grouping(department),1, 'Total Wages', department) department,

2 decode(grouping(gender),1, 'Total Per Department', gender) gender,3 sum(wages)



6 group by rollup(department, gender);

DEPARTMENT GENDER SUM(WAGES)

----------- -------------------- ----------

INT Total Per Department 65000

INT M 65000

POL M 77900

POL F 9800

POL Total Per Department 87700

WEL F 7000

WEL M 45000

WEL Total Per Department 52000

Total Wages Total Per Department 204700




87/127


Use of DECODE function

Example: To suppress the extra Fs and MsSQL>break on gender;

SQL> select decode(grouping(gender),1, 'Total Wages', gender) gender,

' ', , , ,

3 sum(wages)


5 where department = fk_department6 group by rollup(gender, department)

7 order by 1;


----------- ---------------- ----------

F POL 9800

WEL 7000

Total Per Gender 16800M INT 65000

POL 77900

WEL 45000

Total Per Gender 187900

Total Wages Total Per Gender 204700




88/127

In the previous example, for the row thatindicates the Total Wages, we should NOT

print the Total Per Gender How do we do that?


----------- ---------------- ----------

F POL 9800

WEL 7426.3

Total Per Gender 17226.3

M INT 65000

POL 77900

WEL 47740.5


Total Wages 207866.8

Do not print anything here!


Data Warehouse ConceptsGENDER DEPARTMENT SUM(WAGES)


89/127


----------- ---------------- ----------

F POL 9800

WEL 7426.3


M INT 65000

POL 77900

WEL 47740.5

Tota Per Gen er .

Total Wages 207866.8Do not print anything here!


break on gender;

select decode(grouping(gender),1, 'Total Wages', gender) gender,

decode(grouping(department),1,

decode(grouping(gender),0,'Total Per Gender',' '), department)department,

sum(wages)

from department, employee

where department = fk_departmentgroup by rollup(gender, department)

order by 1;



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The RANK Function Oracle 11g providesseveral functions for ranking rows returnedfrom a SELECT statement

The functions can calculate rankings,percentiles and n-tiles

These functions are performed after the select

statement returns the rows and prior toprinting the results




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Advanced OLAP Operations The RANK Function Syntax

Rank() over (

[partition by expression, expression]Order by expression[collate clause] [asc | desc]

nu s rs nu s as

Only Rank and Order by are mandatory clauses That is because in order to rank rows, the result set must be

sorted The expression in the order by clause is used for ranking Default sort order is ascending

By default, NULL values are considered the largest Can change where NULL values will appear using nulls first or

nulls last




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Advanced OLAP Operations The RANK function Example:

Departments CEN and TRF do not have employees sum of thewages is null

Nulls by default are printed first

ere are emp oyees w ou sa ary n e a a ase as we

SQL> select department DPT, sum(wages),2 rank() over (order by sum(wages) desc)

3 as rank_all


5 where department = fk_department(+) -- Outer join department

6 group by department;

DPT SUM(WAGES) RANK_ALL

---- ---------- ----------

CEN 1

TRF 1

POL 87700 3

INT 65000 4

WEL 52000 5


Rank does not

advance for equal

values



93/127

S L> select de artment DPT sum wa es

Advanced OLAP Operations The RANK function Example continued

NULLs last

2 rank() over (order by sum(wages) desc nulls last )as rank_all


4 where department = fk_department(+) -- Outer join department5 group by department;

DPT SUM(WAGES) RANK_ALL

---- ---------- ----------

POL 87700 1

INT 65000 2

WEL 52000 3CEN 4

TRF 4




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Advanced OLAP Operations The RANK function Example continued

We will use the function nvl to provide values for where there is

none Syntax nvl(col_name, val)

If column does not have a value, then val is used instead

In this example, we provide the value of 90,000 for null values

of salary

Value of 4 is missing

SQL> select department, sum(nvl(wages,90000)) total_wages,

2 rank() over (order by sum(nvl(wages,90000)) desc)

3 as rank_all


5 where department = fk_department(+)


DEPA TOTAL_WAGES RANK_ALL

---- ----------- ----------

INT 155000 1

WEL 142000 2

CEN 90000 3

TRF 90000 3

POL 87700 5




95/127

Advanced OLAP Operations The following does not look right since on page 93 the INT department

had total wages of 65500. Now it says, 155000 what is going on?

SQL> select department, sum(nvl(wages,90000)) total_wages,

2 rank() over (order by sum(nvl(wages,90000)) desc)

_




DEPA TOTAL_WAGES RANK_ALL

---- ----------- ----------

INT 155000 1

WEL 142000 2

CEN 90000 3

TRF 90000 3POL 87700 5




96/127

Advanced OLAP Operations The following does not look right since on page 93 the INT department

had total wages of 65500. Now it says, 155000 what is going on?SQL> select payroll_number, wages

2 from employee

= ' '_

PAYROLL_NUMBER WAGES

-------------- ----------25 9500

46 9500

36 14000

33 13000

29

28 11000

22 8000

7 rows selected.


One employee without salary.

NVL replaces this with 90000




97/127

Advanced OLAP Operations The DENSE_RANK Function

Dense RANK function does the same thing as the RANK function

except that it does not count the number of equal ranks. It makes sure that all ranks are used

Here is the same example as before using the dense RANK

SQL> select department, sum(nvl(wages, 90000)) total_wages,

2 dense_rank() over (order by sum(nvl(wages,90000)) desc) as rank_dense




DEPA TOTAL_WAGES RANK_DENSE

---- ----------- ----------INT 155000 1

WEL 142000 2

CEN 90000 3

TRF 90000 3

POL 87700 4

Value of 4 is NOT missing




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Top-N and Bottom-N queries Rank functions rank the rows of the result set - -

portion of the ranked rows from the top or thebottom

There are two steps required to do this:

Create an inline view to develop the data and the

rankings Use the RANK expression in the where clause to identify

the number of Top and Bottom ranked records




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The Top-N rows example: Display the top three salary earning people in thecom an if salar is missin re lace it with 0

SQL> select last_name, first_name, wages, emp_wage_rank

2 from (select last_name, first_name, wages,

3 rank() over(order by nvl(wages,0) desc) as emp_wage_rank4 from employee)

5 where emp_wage_rank


100/127


How do get the three lowest paid employeesin the organization?LAST_NAME FIRST_NAME WAGES EMP_WAGE_RANK

--------------- --------------- ---------- -------------

EISENHOWER DWIGHT 1

ROOSEVELT ELEANOR 1

ANTHONY SUSANNE 7000 2

JOHNSON ANDREW 7500 3




101/127


How do get the three lowest paid employeesin the organization?LAST_NAME FIRST_NAME WAGES EMP_WAGE_RANK

--------------- --------------- ---------- -------------

EISENHOWER DWIGHT 1

ROOSEVELT ELEANOR 1

ANTHONY SUSANNE 7000 2

JOHNSON ANDREW 7500 3


select last_name, first_name, wages, emp_wage_rank

from (select last_name, first_name, wages,

dense_rank() over(order by nvl(wages,0) asc) as emp_wage_rank

from employee)where emp_wage_rank


102/127

result set SQL> select last_name as LNAME, wages,2 row_number() over(order by wages) as Row_number

3 from employee;

LNAME WAGES ROW_NUMBER

--------------- ---------- ----------

ANTHONY 7000 1

JOHNSON 7500 2

ROOSEVELT 8000 3

TAFT 8500 4


COOLIDGE 9500 6

MILLER 9500 7

DWORCZAK 9800 8HOOVER 10000 9

ROOSEVELT 10400 10

TRUMAN 11000 11

KENNEDY 11500 12

JOHNSON 12000 13

NIXON 12500 14

FORD 13000 15

CARTER 13000 16REAGAN 13500 17

BUSH 14000 18

CLINTON 15000 19

EISENHOWER 20

ROOSEVELT 21

21 rows selected.



103/127

Advanced OLAP Operations Top-N and RANK functions allow selecting top n rows

in a set of records that the select statement returns Consider the following:

We need to print the top two paid employees within each

This requires ranking of employees based on their salaries

within each department The PARTITION function can achieve this!

NOTE: This partition function is NOT the same as physically

partitioning tables for performance purposes This is a logical partitioning (temporary) that results inmemory and is lost after the query executes




104/127


The use of PARTITION clause in conjunction with the

RANK FunctionSQL> select department_name, last_name, first_name, wages, emp_wage_rank

2 from (select department_name, last_name, first_name, wages,

_ , _ _

4 from department, employee where department = fk_department)

5 where emp_wage_rank


105/127

Advanced OLAP OperationsWindowing Oracle has the functionality that allows

you to calculate values based on a period of time(called a window)

The functions in this class can be used to com utemoving, cumulative and centered aggregates

They include moving averages, moving sums,moving min/max, cumulative sum, and LAG/LEAD

These functions create a value that is based onvalues that precede or follow the record

The windowing functions can be used in the SELECTand ORDER BY clauses




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The syntax{Sum | Avg |Max | Min | Count | Stddev | Variance| First_value |

Last_value}

*

Over ({partiton by [,]]

Order by [collate clause>][asc |desc] [nulls first |nulls last] [, ]

Rows | range {{unbounded preceding | preceding} between {unbounded preceding |


107/127


Windowing functions clauses

Over Tell Oracle that the function will operate over a query result set.

Partition by Determines how the data will be segmented for analysis

Order By Determines how the data will be sorted within the partition. Options are ASC (default),, .

Rows | Range These keywords determine the windows used for the calculation. The rows keyword isused to specify the window as a set of rows. Range sets the window as a logical offset.

This function cannot be used unless the order by clause is used

Between .. AND Determines the starting point and end point of the window. Omitting the betweenkeyword and specifying only one end point will cause Oracle to consider the endpoint asthe starting point. The current row will then consist of the current row.

UnboundedPreceding

Sets the first row of the partition as the starting point of the window

UnboundedFollowing

Sets the last row of the partition as the endpoint of the window

Current Row Sets the current row as the starting point or as the end point of the window



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Advanced OLAP Operations Windowing function: cumulative aggregate function

example find cumulative cost of tools purchased within POL departmentSQL> select department, last_name, first_name, tool_cost, sum(tool_cost)2 over (order by purchase_date rows unbounded preceding) balance

3 from department, employee, emp_tools


5 and payroll_number = fk_payroll_number

6 and department = 'POL';

DEPA LAST_NAME FIRST_NAME TOOL_COST BALANCE

---- --------------- --------------- ---------- ----------

POL JOHNSON ANDREW 5.95 5.95

POL JOHNSON ANDREW 10.75 16.7

POL WILSON WOODROW 4.95 21.65

POL WILSON WOODROW 100 121.65


POL ROOSEVELT FRANKLIN 12 145.65


POL NIXON RICHARD 12.75 166.4



Data Warehouse Concepts Advanced OLAP Operations Windowing function: cumulative aggregate function


109/127

example find cumulative cost of tools purchased within each department

SQL> select department, last_name, first_name, tool_cost, sum(tool_cost)

2 over (partition by department order by purchase_date rows unbounded preceding) balance



5 and payroll_number = fk_payroll_number;

DEPA LAST_NAME FIRST_NAME TOOL_COST BALANCE

---- --------------- --------------- ---------- ----------

INT ROOSEVELT THEODORE 34 34

INT ROOSEVELT THEODORE 290 324

INT COOLIDGE CALVIN 25 349

INT COOLIDGE CALVIN 10 359

INT EISENHOWER DWIGHT 25 384

INT EISENHOWER DWIGHT 200 584INT EISENHOWER DWIGHT 150 734

INT FORD GERALD 12 746



INT BUSH GEORGE 2.75 748.75



INT MILLER KEVIN 100 892.2

INT MILLER KEVIN 0 892.2

POL JOHNSON ANDREW 5.95 5.95POL JOHNSON ANDREW 10.75 16.7

POL WILSON WOODROW 4.95 21.65











110/127

p Windowing Moving Averages

Moving averages can be computed when the

window is changed over time A moving average can be computed if several

function

A range interval is needed to identify the number ofvalues used

A time unit is needed. Common time units are year,month and day

The preceding and/or following keywords are neededto indicate which records in the ordered set will beused





111/127

Moving average function example

This example computes the moving average tools cost for INTdepartment in the past 20 years

SQL> select department as DPT, to_char(purchase_date, 'YYYY-DD-MON'), tool_cost,

2 avg(tool_cost) over (order by purchase_date range interval '20' year preceding)as Average


0 records in the

previous 20 years

1 records in the

previous 20 years

2 records in the

previous 20 years

1 records in the

previous 20 years


4 where department = fk_department and payroll_number = fk_payroll_number and department = 'INT'

5 order by purchase_date;

DPT TO_CHAR(PUR TOOL_COST AVERAGE

---- ----------- ---------- ----------

INT 1903-01-FEB 34 34

INT 1905-10-MAR 290 162

INT 1922-01-OCT 25 116.333333

INT 1923-01-FEB 10 89.75

INT 1953-01-MAR 25 25

INT 1953-31-MAR 200 125

INT 1953-31-MAR 150 125

INT 1974-01-JAN 12 12INT 1974-10-AUG 0 6

INT 1977-23-MAR 0 4

INT 1988-23-SEP 2.75 3.6875

INT 1988-10-NOV 35.95 10.14

INT 1989-23-FEB 7.5 9.7

INT 2001-08-APR 100 36.55

INT 2001-23-MAY 0 29.24




112/127

Moving average function example

This example computes the total salary for employees in a 40

year period around the hire date of the current employeeSQL> select department as DPT, first_name, Last_name,

2 to_char(employment_date, 'YYYY-DD-MON') hire_date, wages,

3 SUM(wages) OVER (ORDER BY employment_date RANGE BETWEEN

4 INTERVAL '20' YEAR PRECEDING AND INTERVAL '20' YEAR FOLLOWING) ctrd_sum



7 and department = 'INT' order by 4;

DPT FIRST_NAME LAST_NAME HIRE_DATE WAGES CTRD_SUM

---- --------------- --------------- ----------- ---------- ----------

INT THEODORE ROOSEVELT 1902-20-NOV 8000 17500

INT CALVIN COOLIDGE 1921-07-AUG 9500 17500

INT HAROLD TRUMAN 1945-15-APR 11000 11000

INT DWIGHT EISENHOWER 1953-20-MAR 11000INT GERALD FORD 1973-20-MAY 13000 27000

INT GEORGE BUSH 1988-05-JAN 14000 36500

INT KEVIN MILLER 2000-12-OCT 9500 23500

7 rows selected.

Copyright (c) 2011 Saeed Rahimi

112


113/127

Questions




114/127


Question 1: Print the cost of tools per classification

(position) within gender. Subtotal the costs for eachgender GENDER CURRENT_POSITIO Tool Cost------ --------------- ----------

F ADMINISTRATOR

F SALESPERSON 2 88.85

F SYSTEM ANALYST 61.95

F 150.8

M CLERK 1 20

M CLERK 2 46.2M CONTROLLER 324

M COUNSELER 2

M GUARD 4 375

M JANITOR 35

M LABORER 2 12

M LABORER 3

M MAINT. MAN 2 24

M MAINT. MAN 3 116.95

M PRESIDENT 28.7M PROGRAMMER 1

M SALESPERSON 1 16.7

M TREASURER 18.5

M TREASURER CLERK

M VICE PRESIDENT 123

M 1140.05

1290.85

22 rows selected.




115/127


Question 1: Print the cost of tools per classification

(position) within gender. Subtotal the costs for eachgender

select gender, current_position,sum(tool_cost) "Tool Cost"

from employee, emp_tools

where payroll_number = fk_payroll_number(+)group by rollup (gender, current_position)

order by 1,2;




116/127


Question 2: Determine the two employees in each

department who had the largest cost of eye glasses

' '_ , _ _ _ _ _

---------- -------------------------------- ------------- -------------------------

INT BUSH, GEORGE 1

INT EISENHOWER, DWIGHT 15 2POL CLINTON, WILLIAM 1

POL KENNEDY, JOHN 1

POL DWORCZAK, ALICE 1

WEL HOOVER, HERBERT 1

WEL ANTHONY, SUSANNE 120 2

7 rows selected.




117/127


Question 2: Determine the two employees in each

department who had the largest cost of eye glasses

*

from (select department, last_name||', '||first_name, sum(cost) eyeglass_cost,

rank() over (partition by department

order by sum(cost) asc nulls first) Lowest_eyeglass_cost_rankfrom department, employee, glasses


and payroll_number = fk_payroll_number(+)

group by department, last_name, first_name)

where lowest_eyeglass_cost_rank


118/127


Question 3:Create a checkbook style

cumulative cost of eye glassesPURCHASE_ LAST_NAME||','||FIRST_NAME EYEGLASS_COST BALANCE

--------- -------------------------------- ------------- ----------

12-MAR-03 ROOSEVELT, THEODORE 123 123

06-MAY-04 ROOSEVELT, THEODORE 145 268

08-NOV-10 TAFT, WILLIAM 145 413

01-JAN-17 WILSON, WOODROW 123 536

15-NOV-23 COOLIDGE, CALVIN 175 71103-JUN-33 ROOSEVELT, FRANKLIN 129 840

01-JUL-33 ROOSEVELT, ELEANOR 134 974

20-JUL-35 ROOSEVELT, ELEANOR 143 1117

12-AUG-40 ANTHONY, SUSANNE 120 1237

12-OCT-47 TRUMAN, HAROLD 110 1347

31-MAR-53 EISENHOWER, DWIGHT 15 1362

31-JAN-64 JOHNSON, LYNDON 170 1532

31-MAY-67 JOHNSON, ANDREW 165 1697

23-JUN-70 NIXON, RICHARD 123 182001-FEB-74 FORD, GERALD 145 1965

08-SEP-77 CARTER, JIMMY 164 2129

12-AUG-79 CARTER, JIMMY 175 2304

23-OCT-83 REAGAN, RONALD 165 2469

21-DEC-00 MILLER, KEVIN 165 2634

19 rows selected.




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Question 3:Create a checkbook style

cumulative cost of eye glasses

select purchase_date, last_name||', '||first_name,

cost eyeglass_cost,

sum(cost) over (order by purchase_date

rows unbounded preceding) balance

from employee, glasses

where payroll_number = fk_payroll_number;




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Question 4: Write a SQL statement that

counts the number of eyeglasses within oneof the four cost classes: less than $100,$100 to $125, $126 to $150 and Above $150

cost cat COUNT

------------- ----------

100 to 125 5126 to 150 6

Above 150 7

Less than 100 1




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Question 4: Write a SQL statement that counts the number ofeyeglasses within one of the four cost classes: less than $100,

$100 to $125, $126 to $150 and Above $150

select (case when cost = 100 and cost 126 and cost 150 then 'Above 150' end)

"cost cat", count(*) as count

from glasses

group by (case when cost = 100 and cost 126 and cost 150 then 'Above 150' end);


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Questions?

Contact information

[email protected] 962 5514

Copyright (c) 2011 Saeed Rahimi122

SQL Statement with Group By


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The LAG and LEAD Functions

These functions are useful for computing thedifference between values in different rows

The LAG and LEAD functions return the value of a

preceding or following row to the current row Syntax:

Lag (expression, record offset)

Lead (expression, record offset) For example, offset 1 refers to the row immediately

before the current row (for Lag) and the rowimmediately after the current row (for Lead)

Data Warehouse Concepts Advanced OLAP Operations

The use of Lag and Lead

Example: find the wage salary difference of each employee and the one


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immediately hired before and after the employee

SQL> select last_name, wages,

2 lead(wages, 1) over(order by employment_date) - wages as "Lead Diff",

3 lag(wages, 1) over(order by employment_date) - wages as "Lag Diff"4 from employee;

LAST_NAME WAGES Lead Diff Lag Diff

--------------- ---------- ---------- ----------

ROOSEVELT 8000 500

TAFT 8500 500 -500

WILSON 9000 500 -500

COOLIDGE 9500 500 -500

HOOVER 10000 -500

ROOSEVELT

ROOSEVELT 10400 -3400

ANTHONY 7000 4000 3400

TRUMAN 11000 -4000

EISENHOWER

KENNEDY 11500 500

JOHNSON 12000 -4500 -500

JOHNSON 7500 5000 4500

NIXON 12500 500 -5000

FORD 13000 0 -500

CARTER 13000 500 0

REAGAN 13500 500 -500

BUSH 14000 1000 -500

CLINTON 15000 -5200 -1000

DWORCZAK 9800 -300 5200

MILLER 9500 300




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p

Question: Join table employee with itself (self join) to

find out the name of the person immediately hiredafter each employee for department WEL

e are oo ng or s s

LAST_NAME Hire Date Next emp--------------- ---------- -----------

TAFT 1908-06-01 HOOVER

HOOVER 1928-04-06 ROOSEVELT

ROOSEVELT 1932-03-20 ANTHONY

ANTHONY 1940-03-30 CARTERCARTER 1976-07-10 REAGAN

REAGAN 1980-03-03

6 rows selected.




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d a ced O Ope at o s

Question:

Does the use of self join provide the right answer?

If not why?

How do we get the right answer?

Use of Lag function

Repeat the work but this time use the Lag (orLead) functions to get the right answer

_ _

Prev emp

-------------- --------------- -------------------------

20 ANTHONY 19

ROOSEVELT

35 REAGAN 34

CARTER

t15 beyond sql rahimi

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