RDBMS- Day 4 - WordPress.com · 1 RDBMS- Day 4 •Grouped results •Sub queries •Relational...

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RDBMS- Day 4

•Grouped results•Sub queries•Relational algebra•Use of EXISTS and NOT EXISTS

In today’s session we will discuss about the concept of subqueries,

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Grouped results

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ER/CORP/CRS/DB07/003

Version No: 2.03Copyright © 2004,

Infosys Technologies Ltd

SQL - Using GROUP BY

• Related rows can be grouped together by GROUP BY clause by specifying a column as a grouping column.

• GROUP BY is associated with an aggregate function• Example: For each part supplied get the part number and the total

shipment quantity

SELECTSELECT PNO, SUM(QTY)

FROMFROM SP

GROUP BY GROUP BY PNO

In the output table all the rows with an identical value in the grouping column will be grouped together.

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SELECTSELECT SNO, PNO, SUM(QTY)

FROMFROM SP


SELECTSELECT SNO, PNO, SUM(QTY)

FROMFROM SP

GROUP BY GROUP BY SNO, PNO

Retrieval using GROUP BY

Get SNO, PNO, total qty for each part supplied

Why the first form is considered wrong is that, for each part , multiple suppliers would have supplied the part in different combinations. For e.g say for a part p1, s1 might have supplied twice, s2 once , s3 thrice etc. so we want the results to be grouped based on part number and within that by supplier number and take the count.

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Get PNO for parts which have more than two shipments

SELECTSELECT PNO,COUNT(*) FROMFROM SP


HAVINGHAVING COUNT(*)>2

Retrieval using HAVING

• Used to specify condition on group

Get supplier numbers who have at least two shipments.

SELECTSELECT SNO , COUNT(*) FROMFROM SP

GROUP BY GROUP BY SNO

HAVINGHAVING COUNT(*)>=2

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Can you identify any error…?

SELECTSELECT SNO , COUNT(*)

FROMFROM SP

GROUP BY GROUP BY SNO

HAVINGHAVING PNO=‘p2’;

Ans: The Having condition has to be based on some column that appears in the select list

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Independent subqueries

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Independent sub-queries

• Inner query is independent of outer query.

• Inner query is executed first and the results are stored.

• Outer query then runs on the stored results.

These are queries where there are two parts to the query. We need to collect one type of information based on which other set of information has to be retrieved from the table.

For e.g :

Select all sales reps who have a higher quota than sales rep 101.

We need to analyze this query and understand how to break it into sub problems

1. First we nee dto find out what is the quota of seles rep 101

2. Based on this info, we need to select sales reps who have a higher quota than this value

3. So, the inner query will find the quota of sales rep 101 and the outer query will extract sakes reps exceeding this quota value. The solution would look like:

SELECT Rep

FROM SalesReps

WHERE Quota >

SELECT Quota

FROM SalesReps

WHERE Empl_Num = 101;

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Get supplier names for all suppliers who supply pa rt P2

SELECTSELECT SNAME

FROMFROM S

WHEREWHERE SNO IN

(SELECTSELECT SNO

FROMFROM SP

WHERE WHERE PNO =‘P2’)

Retrieval using SUB QUERIES

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Get SNO for suppliers who are located in the same city as S1

SELECTSELECT SNO

FROM FROM SWHEREWHERE CITY =

(SELECTSELECT CITY

FROMFROM S

WHERE WHERE SNO=‘S1’)


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Get SNO for suppliers who supply at least one par t supplied by S2

SELECT SELECT SNO

FROMFROM SP

WHEREWHERE PNO IN

(SELECTSELECT PNO

FROMFROM SP

WHEREWHERE SNO=‘S2’)


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SELECTSELECT SNO

FROMFROM S

WHEREWHERE STATUS <

(SELECTSELECT STATUS

FROMFROM S

WHEREWHERE SNO=‘S1’)

Get SNO who have status less than the status of ‘S1 ’


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Get supplier numbers for suppliers with status les s than the current maximum in the supplier table

SELECTSELECT SNO

FROMFROM S

WHEREWHERE STATUS <

(SELECT MAXSELECT MAX(STATUS)

FROMFROM S)


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Get SNO for suppliers who do not supply any part supplied by S2

SELECTSELECT SNO FROMFROM SP

WHEREWHERE PNO NOT INNOT IN

(SELECT PNO FROM SP

WHERE SNO=‘S2’))


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For each part supplied, get the part number, maximu m quantity, and minimum quantity supplied for that part

SELECTSELECT SP.P#,

MAX (SP.QTY) ASAS MXQ,

MIN (SP.QTY) ASAS MNQ

FROMFROM SP

GROUP BY SP.P#;GROUP BY SP.P#;

Write a query for the following ...

Let us take few more examples and see how to solve:

1. Who are the customers whose rep is 102 and have at least one order more than $1000?

Select cust_name from customers

where cust_rep=102 and cust_num in

(select cust from orders

where amount > 1000)

2. Which product has the maximum total order amount?

Select product from orders group by product having sum(amount) =(select max(sum(amount)) from orders group by product);

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Correlated Sub Queries

• You can refer to the table in the FROM clause of the outer query in the inner query using Correlated sub-queries.

• The inner query is executed separately for each row of the outer query.

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Get PNO for all parts supplied by more than on e supplier

SELECTSELECT PNO

FROMFROM SP XX

WHEREWHERE PNO IN

(SELECTSELECT PNO

FROMFROM SP YY

WHEREWHERE YY.SNO<>XX.SNO)

Correlated Sub Queries

In this query, clearly there is a case of comparison of each row with some set of rows of the table to determine whether the given rows qualifieto go into the result. So, the typical way to solve is to select each row and check it with some other rows selected based on a criteria iteratively. Typically, the same table will be involved in both the outer query and the inner queryTo solve these type of correlated queries, we typically consider the same table as two copies each with a different name . This enables comparing the table with itself. To illustrate this, consider the above query. Let’s take a sample snapshot:SNO PNO JNO QtyS1 P1 J1 75S2 P3 J1 40S1 P1 J2 30S4 P3 J1 55Here, both p1 and p3 have been supplied twice. But p1 has been supplied by the same supplier and p3 by two different suppliers.So , when we consider the same table as two copies X and Y and compare, the processing would go something like this…X YSNO PNO JNO Qty SNO

PNO JNO QtyS1 P1 J1 75 S1 P1

J1 75S2 P3 J1 40 S2 P3

J1 40S1 P1 J2 30 S1 P1

J2 30S4 P3 J1 55 S4 P3

J1 55The inner query will be executed completely once for every row of the outer query. So, the result returned by the inner query would be Iteration 1 :Is p1 in (p3, p3) - not qualified -> not selectedIteration 2 :Is p3 in (p1, p1, p3) -> qualified -> selectedIteration 3:Is p1 in (p3,p3) -> not qualified -> not selectedIteration 4:Is p3 in (p1,p1,p3)-> qualified -> selectedSo the result will contain two rows displaying PnoP3P3To avoid this we may use in the outer query select distinct rather than select

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SELECTSELECT DISTINCT SNO

FROMFROM SPJ X

WHEREWHERE PNO=‘P1’

ANDAND QTY> (SELECT AVGSELECT AVG(QTY)

FROMFROM SPJ YWHEREWHERE PNO=‘P1’

ANDAND X.JNO=Y.JNO)

Get SNO for suppliers supplying some project wit h P1 in a quantity greater than the average qty of P1 supplied to that project

Correlated Sub Queries ...

Take a sample snapshot of the shipment tables as follows:

SNO PNO JNO Qty

S1 P1 J1 75

S2 P1 J1 40

S3 P2 J2 30

S4 P1 J1 55

Here, s1, s2 , s4 are supplying project j1 with part p1. The average quantity of p1 supplied to j1 is 56. 7. out of this, s1 supplies individually a quantity 75 of part p1 to the project j1.

So according to the query, s1 is the supplier supplying some project (here it is j1) with p1 in a quantity(75) which is greater than the average quantity of p1 supplied to that project(56.7). So the result should be s1.

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Exercise

• List names of sales reps who have a higher target than their managers

Select s.name from salesreps s wheres.quota > (select m.quota from selesreps m

where m.emplnum=s.manager);

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Exercise

• Who are the customers whose rep is 102 and have at least one order more than $1000?

Select cust_name from customers

Where cust_rep=102 and cust_num in

(select cust from orders where amount > 1000)

To solve this problem, we split the query as follows:

We have to find the list of customers having an order amount above 1000

Then we have to select customers whose rep is 102 and whose custno is in the list generated by the inner query.

Here there are two clear cut components which are independent so, this will be solved as an independent subquery.

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Relational algebra operations

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SET operations

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Get a list of all the parts cities and supplier ci ties

SELECTSELECT CITY

FROMFROM P

UNIONUNION

SELECTSELECT CITY

FROMFROM S

Retrieval using UNION

Supplier Parts

The results of two independent SELECT statements can be worked with using the SET operation –UNION. By default, UNION returns only distinct values. Union is like an “OR” operation. If the tupleoccurs in relation 1 or relation 2, it is selected. Set theoretic notation indicates union as indicated in the slide

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Get a list of all common parts and supplier cities

SELECTSELECT CITY

FROMFROM P

INTERSECTINTERSECT

SELECTSELECT CITY

FROMFROM S

Retrieval using INTERSECT

An intersection is an AND operation. It retrieves those tuples which are present in both relation A and B

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Minus

• Get the list of cities specific to suppliers which are not the cities of any parts

SELECT city FROM supplier MINUS

SELECT city FROM parts

This is the difference operation. It retrieves tuples which are present in relation 1 but not in relation 2.

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Other RA operations

• Restriction• Projection• Join

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Restriction

• Restricts the rows that can be chosen from a relation using a where clause• Takes a horizontal subset of values from the original relation• E.g select * from employee where salary>10000;

This will retrieve only those rows of the table which satisfy the condition in the where clause

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Projection

• Projection is projecting a set of attributes of a relation so that rows of values corresponding to those colums will figure in the output

• For e.g • select empid, name, salary from employee;

• This takes a vertical subset of the relation

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Join

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JOIN

• Inner join• Equi join• Outer join

– Left-outer join– Right-outer join

• Self join

In relational databses, data is spread over multiple tables. Sometimes we may want dta from two or more tables. A join is an operation which combines results from two or more tables.

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Inner Joins

• Common type of join• Combines records from two tables with matching values on a column.

In this type of join, rows from two different tables are selected based on a matching column. Which rows are selected is restricted by a where clause. For example,

Customers Orders

Custid Orderid

custname custid

cust_city amount

Here, custid of customer table and custid of order table correspond.

Note :

pl note that there need not be any primary key fore ign key relationship between these two columns which would be used in a join operation

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Get all combinations of supplier and part informati on such that the supplier and part are co-located.

SELECTSELECT S.*, P.*

FROMFROM S, P

WHEREWHERE S.CITY=P.CITY

Retrieval from Multiple tables-Equi join

Here the where clause is based on the equality condition “=“. Hence it is called equi join

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Get SNO,PNO combinations where the part’s city f ollows the supplier’s alphabetically

SELECTSELECT SNO, PNO

FROMFROM S, P

WHEREWHERE S.CITY<P.CITY

Retrieval from Multiple tables- Non Equi join

Here the where clause is based on a non quality condition (<). ?Hence, it is called non-equi join

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Get SNO and PNO for co-located suppliers and par ts omitting suppliers with status<20

SELECTSELECT P.PNO, S.SNO

FROMFROM P, S

WHEREWHERE S.CITY = P.CITY

AND S.STATUS > = 20

Retrieval from Multiple tablesRetrieval from Multiple tables

There may be multiple conditions in the where clause apart from the join condition

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Outer join

• Retrieve all rows that match the WHERE clause and those that have a NULL

The inner join takes into account only those non NULL rows from the tables involved. If you want the result to include even those rows having a NULL for a particular row in the selected column, then go for an outer join. The syntax for representing this is slightly different in each rdbms product. What follows in the next slide is the oracle style.

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Left/Right-Outer join

• Left outer joins include all records from the first (left) of two tables, A=B +

Right outer joins include all records from the second (right) of two tables,

A += B

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Example of left-join

SELECT S.SNO, SP.QTYFROM S, SPWHERE S.SNO = SP.SNO (+);

All unmatched rows of S are also selected

List all SNO with QTY supplied or SNO which have no t yet supplied any QTY

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Example of right outer join

• Select the orders and coresponding customernames. Select all customer names, even if there are no orders placed by any.

select o.order_num, c.cust_name from orders o, customers c where o.cust (+) = c.cust_num;

The (+) symbol is next to the column which needs to be expanded to include null values also. In the example above, there may be some customers who have not made any orders, so if we select their names from the customers table(the second table based on int position in the query), the corresponding order detail would be null. Eevne then such values have to be selected . That’s what is indicated. A typical output would look like:

ORDER_NUM CUST_NAME

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

5 radha

first corp

jcp inc.

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Get all pairs of SNO who are co-located

SELECTSELECT FIRST.SNO, SECOND.SNO

FROMFROM S FIRST,

S SECOND

WHEREWHERE FIRST.CITY=SECOND.CITY

Self join-Joining a table with itself

When you wish to join a table with itself based on some criteria, use the concept of synonyms. Treat the table as two different tables by giving synonyms

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Cartesian Product (a) Cross Join

• A join without any conditions is called a Cartesian product

• It will be a collection of all possible combinations of rows from the tables involved

• Practically of little use

• Just included for conceptual completeness

• E.g select s.sno, sp.sno from supplier s , shipment sp;

Let us assume a snapshot of the supplier table and shipment table are as follows:Supplier Shipment

SNO Name status SNO PNO qtys1 Ram 20 s3 p1 56s2 John 45 s4 p2 100s3 Rafi 32 s3 p3 45s4 Inder 34 s1 p2 34The cartesian product would be like:SNO SNOs1 s3s2 s3s3 s3s4 s3s1 s4s2 s4s3 s4s4 s4s1 s3s2 s3s3 s3s4 s3s1 s1s2 s1s3 s1s4 s1which is hardly of any use

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EXISTS Vs NOT EXISTS

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Get PNO’s supplied in the same QTY by at least one different supplier

SELECTSELECT PNO

FROMFROM SP A

WHERE EXISTSWHERE EXISTS(SELECT SELECT *

FROMFROM SP BWHEREWHERE A.PNO=B.PNO

ANDAND A.QTY=B.QTY ANDANDA.SNO<>B.SNO)

Retrieval using EXISTS

Exists check for the existence of a situation/condition.

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Get all part names from parts table which have been shipped

SELECTSELECT PNAME

FROMFROM P

WHERE EXISTSWHERE EXISTS

(SELECTSELECT *

FROMFROM SP

WHEREWHERE SP.PNO=P.PNO)

Retrieval using EXISTS

Let us take another example:Find the deptno of those departments having employees who can do some work done by employee in depertment d1.

The solution would look like:Select deptno

from department where exists(

select * from employee x

where x.dept= ‘d1’ and exists(select *

from employee y

y.job=x.job and y.dept=department.deptno)

)

and deptno <> ‘d1’;

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Get the list of all prospective suppliers, i. e.,su ppliers for whom no shipments exist yet.

SELECTSELECT SNAME

FROMFROM S

WHERE NOT EXISTSWHERE NOT EXISTS

(SELECTSELECT *

FROMFROM SP

WHEREWHERE SP.SNO=S.SNO)

Retrieval using NOT EXISTS

A not exists checks for the opposite condition than an Exists. It checks for the non-existence of a condition.

Not exists can be used to arrive at a result which can’t otherwise be arrived at using exists.

The logic is as follows:For all x there exists an y such that f(x,y) is true is the same asThere does not exist and x for which there does not exists a y so that f(x,y)is true

Taking a slight variant of the previous example:Find the deptno of those departments having employees who can do all work done by employees in depertment d1.

Select deptnofrom department

where not exists(select *

from employee x

where x.dept= ‘d1’ and not exists(

select * from employee y

y.job=x.job and y.dept=department.deptno)

)and deptno <> ‘d1’

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Exercise

• List cities that have received shipments from every customer.

SELECT A.CITY_NAME

FROM CITY AS A

WHERE NOT EXISTS(

SELECT *

FROM CUSTOMER B

WHERE NOT EXISTS(

SELECT *

FROM SHIPMENT C

WHERE A.CITY_NAME=C.DESTINATION

AND B.CUST_ID=C.CUST_ID));

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Summary• The result of a query can be grouped based on a grouping column

• While checking for conditions after grouping by a column , Having is used instead of where

• Grouped queries help look at data category wise

• When the query consists of more than one component, it is implemented in the form of a nested query depending on the nature of the query.

• Sub queries help split a problem involving different levels of data

• Relational algebra operations like union, intersect, difference, restriction, projection and join help us get different combinations of data from more than one table

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Thank You!

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