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Lecture 5 SQL Continued

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Lecture 5 SQL Continued. SUBQUERIES. Queries inside query There are times when you need information from a table to answer query related to the same table or another table. Format: SQL>SELECT …. FROM WHERE colname1 condition(SELECT col2 . - PowerPoint PPT Presentation
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INSS 651 1 Lecture 5 SQL Continued
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Page 1: Lecture 5 SQL Continued

INSS 651 1

Lecture 5

SQL Continued

Page 2: Lecture 5 SQL Continued

INSS 651 2

SUBQUERIES.

Queries inside queryThere are times when you need information from a table to answer query related to the same table or another table

Page 3: Lecture 5 SQL Continued

INSS 651 3

Format:SQL>SELECT ….FROM WHERE colname1condition(SELECT col2 . . (SELECT . . ))

Page 4: Lecture 5 SQL Continued

INSS 651 4

List customers whose balance is greater than the average balance of all customers

Logic:

• To answer this query we need to know average balance of all customers

• We will have to put this in a sub query

Page 5: Lecture 5 SQL Continued

INSS 651 5

List customers whose balance are greater than the average balance of all customers

SQL> select *

from customer

where cust_balance >

(Select avg(cust_balance)

From customer);

Page 6: Lecture 5 SQL Continued

INSS 651 6

give the customer balance of customer whose order is 123Logic:

Before we can find balance of customer whose order is 123 we need to find the cust_numb first

Page 7: Lecture 5 SQL Continued

INSS 651 7

JOINING TABLES

when information needed is in more than one table, we need to join tables;

WHERE clause in the select SQL statement creates a join. Note some queries can also be answered using sub query

Page 8: Lecture 5 SQL Continued

INSS 651 8

Rules FOR joining

WHERE attribute1 condition attribute2

Ex: where employee.ssn=student.ssn

Value(s) from one table are matched value(s) from other tables all matching values are attached

allows joining of tables based on common attribute domains

without the WHERE clause it will produce a Cartesian product also

Page 9: Lecture 5 SQL Continued

INSS 651 9

Give the names of salesperson and their customers in maryland

SQL>Select cust_name, Sales_name from Customer C, salesperson Swherec.sales_numb= s. sales_numb’AndUpper(c.cust_st) =‘MD’;

C & S are aliases for tables Customer and Salesperson respectively

Page 10: Lecture 5 SQL Continued

INSS 651 10

BUILT-IN-FUNCTIONS

• AVG(value)• MAX(value)• MIN(value)• SUM(value)• STDDEV(value)• VARIANCE(value)• COUNT(value)• Etc…

Page 11: Lecture 5 SQL Continued

INSS 651 11

Nested functions

Select max (avg(grades)) etc..is allowed

Page 12: Lecture 5 SQL Continued

INSS 651 12

String functions

Several strings can be concatenated

Use string1||string 2

|| implies +

select custname|| ‘,’ || cust_street as address

From customer;

Page 13: Lecture 5 SQL Continued

INSS 651 13

Substrings

Returns substrings

Format:

Substr(stringvalue,m,n)

Where m is the starting value and n is the length of characters

Page 14: Lecture 5 SQL Continued

INSS 651 14

Assume orders have the format:”

Abc1234

cdf2345etc..

Get the first and last part of the order

Select substr (order_numb, 1,3),substr (order_numb,4,4)

From order;

Page 15: Lecture 5 SQL Continued

INSS 651 15

Product (p_code, P_price) table• P_CODE P_PRICE• -------- ----------• 11QER/31 109.99• 13-Q2/P2 14.99• 14-Q1/L3 17.49• 1546-QQ2 39.95• 1558-QW1 43.99• 2232/QTY 109.92• 2232/QWE 99.87• 2238/QPD 38.95• 23109-HB 9.95• 23114-AA 14.4• 54778-2T 4.99

• P_CODE P_PRICE• -------- ----------• 89-WRE-Q 256.99• PVC23DRT 5.87• SM-18277 6.99• SW-23116 8.45• WR3/TT3 119.95

• 16 rows selected.

Page 16: Lecture 5 SQL Continued

INSS 651 16

Select three characters of price• SQL> select substr(p_price,1,3) from product;

• SUB• ---• 109• 14.• 17.• 39.• 43.• 109• 99.• 38.• 9.9• 14.• 4.9

• SUB• ---• 256• 5.8• 6.9• 8.4• 119

• 16 rows selected.

Page 17: Lecture 5 SQL Continued

INSS 651 17

Padding charactersRpad (string, length,’set’)Lpad (string,length,’set’)

1* select rpad (p_code,15,'.') as rightpad,p_price from product• SQL> /

• RIGHTPAD P_PRICE• --------------- ----------• 11QER/31....... 109.99• 13-Q2/P2....... 14.99• 14-Q1/L3....... 17.49• 1546-QQ2....... 39.95• 1558-QW1....... 43.99• 2232/QTY....... 109.92• 2232/QWE....... 99.87• 2238/QPD....... 38.95• 23109-HB....... 9.95• 23114-AA....... 14.4• 54778-2T....... 4.99

• RIGHTPAD P_PRICE• --------------- ----------• 89-WRE-Q....... 256.99• PVC23DRT....... 5.87• SM-18277....... 6.99• SW-23116....... 8.45• WR3/TT3 ....... 119.95

• 16 rows selected.

Page 18: Lecture 5 SQL Continued

INSS 651 18

Length of string

formatlength (string)

Returns length of the string

Page 19: Lecture 5 SQL Continued

INSS 651 19

SQL> select P_code, length(P_code) from product;

• P_CODE LENGTH(P_CODE)• -------- --------------• 11QER/31 8• 13-Q2/P2 8• 14-Q1/L3 8• 1546-QQ2 8• 1558-QW1 8• 2232/QTY 8• 2232/QWE 8• 2238/QPD 8• 23109-HB 8• 23114-AA 8• 54778-2T 8

• P_CODE LENGTH(P_CODE)• -------- --------------• 89-WRE-Q 8• PVC23DRT 8• SM-18277 8• SW-23116 8• WR3/TT3 8

• 16 rows selected.

Page 20: Lecture 5 SQL Continued

INSS 651 20

Trimming data

LTrim, Rtrim..remove unwanted characters

Format:

RTRIM (string, ‘set’)

Ltrim (string, ‘set’)

Set is the collection of characters you want to trim

Page 21: Lecture 5 SQL Continued

INSS 651 21

SQL> select P_code, length(rtrim (P_code,' '))

from product;

• P_CODE LENGTH(RTRIM(P_CODE,''))• -------- ------------------------• 11QER/31 8• 13-Q2/P2 8• 14-Q1/L3 8• 1546-QQ2 8• 1558-QW1 8• 2232/QTY 8• 2232/QWE 8• 2238/QPD 8• 23109-HB 8• 23114-AA 8• 54778-2T 8

• P_CODE LENGTH(RTRIM(P_CODE,''))• -------- ------------------------• 89-WRE-Q 8• PVC23DRT 8• SM-18277 8• SW-23116 8• WR3/TT3 7

• 16 rows selected.

Page 22: Lecture 5 SQL Continued

INSS 651 22

Remove . From price• SQL> select ltrim(p_price,'.'), p_code from product;

• LTRIM(P_PRICE,'.') P_CODE• ---------------------------------------- --------• 109.99 11QER/31• 14.99 13-Q2/P2• 17.49 14-Q1/L3• 39.95 1546-QQ2• 43.99 1558-QW1• 109.92 2232/QTY• 99.87 2232/QWE• 38.95 2238/QPD• 9.95 23109-HB• 14.4 23114-AA• 4.99 54778-2T

• LTRIM(P_PRICE,'.') P_CODE• ---------------------------------------- --------• 256.99 89-WRE-Q• 5.87 PVC23DRT• 6.99 SM-18277• 8.45 SW-23116• 119.95 WR3/TT3

• 16 rows selected.

Page 23: Lecture 5 SQL Continued

INSS 651 23

INSTR function

Allows searching for a string of characters, gives the position of the string but does Not cut off anything

Format:

Instr(string, start,occurrence)Start is the start of the stringOccurrence is the position of occurrence

that you want to search

Page 24: Lecture 5 SQL Continued

INSS 651 24

Search for first “/” in p_code

SQL> select p_code, instr(p_code,'/') from product;;

• P_CODE INSTR(P_CODE,'/')• -------- -----------------• 11QER/31 6• 13-Q2/P2 6• 14-Q1/L3 6• 1546-QQ2 0• 1558-QW1 0• 2232/QTY 5• 2232/QWE 5• 2238/QPD 5• 23109-HB 0• 23114-AA 0• 54778-2T 0

• P_CODE INSTR(P_CODE,'/')• -------- -----------------• 89-WRE-Q 0• PVC23DRT 0• SM-18277 0• SW-23116 0• WR3/TT3 4

• 16 rows selected.

Page 25: Lecture 5 SQL Continued

INSS 651 25

Separate P_code in two parts: before _ and after _ for names that contain -

SQL> select p_code, substr(P_code, 1, instr(p_code,'/')) part1from product where p_code like '%/%';• 2 3• P_CODE PART1• -------- --------• 11QER/31 11QER/• 13-Q2/P2 13-Q2/• 14-Q1/L3 14-Q1/• 2232/QTY 2232/• 2232/QWE 2232/• 2238/QPD 2238/• WR3/TT3 WR3/

• 7 rows selected.

Page 26: Lecture 5 SQL Continued

INSS 651 26

SQL> select p_code, substr(P_code, 1, instr(p_code,'/')-1) part1from productwhere p_code like '%/%'; 2 3• P_CODE PART1• -------- --------• 11QER/31 11QER• 13-Q2/P2 13-Q2• 14-Q1/L3 14-Q1• 2232/QTY 2232• 2232/QWE 2232• 2238/QPD 2238• WR3/TT3 WR3

• 7 rows selected.

Page 27: Lecture 5 SQL Continued

INSS 651 27

Get the right partSQL> select p_code, substr(P_code, instr(p_code,'/')) part2from productwhere p_code like '%/%'; 2 3

• P_CODE PART2• -------- --------• 11QER/31 /31• 13-Q2/P2 /P2• 14-Q1/L3 /L3• 2232/QTY /QTY• 2232/QWE /QWE• 2238/QPD /QPD• WR3/TT3 /TT3

• 7 rows selected.

Page 28: Lecture 5 SQL Continued

INSS 651 28

SQL> select p_code, substr(P_code, instr(p_code,'/')+1) part2from productwhere p_code like '%/%';2 3• P_CODE PART2• -------- --------• 11QER/31 31• 13-Q2/P2 P2• 14-Q1/L3 L3• 2232/QTY QTY• 2232/QWE QWE• 2238/QPD QPD• WR3/TT3 TT3

• 7 rows selected.

Page 29: Lecture 5 SQL Continued

INSS 651 29

SQL> select p_code, substr(P_code, 1, instr(p_code,'/')-1) part1, substr(P_code, instr(p_code,'/')+1) part2from productwhere p_code like '%/%'; 2 3

• P_CODE PART1 PART2• -------- -------- --------• 11QER/31 11QER 31• 13-Q2/P2 13-Q2 P2• 14-Q1/L3 14-Q1 L3• 2232/QTY 2232 QTY• 2232/QWE 2232 QWE• 2238/QPD 2238 QPD• WR3/TT3 WR3 TT3

• 7 rows selected.

Page 30: Lecture 5 SQL Continued

INSS 651 30

Remove the period from price

Page 31: Lecture 5 SQL Continued

INSS 651 31

SQL> select (substr(P_price, 1, instr(p_price,'.')-1)|| substr (P_price, instr(p_price,'.')+1)) as pricefrom product; 2 3

• PRICE• --------------------------------------------------------------------------------• 10999• 1499• 1749• 3995• 4399• 10992• 9987• 3895• 995• 144• 499

• PRICE• --------------------------------------------------------------------------------• 25699• 587• 699• 845• 11995

• 16 rows selected.

Page 32: Lecture 5 SQL Continued

INSS 651 32

COMMON ERRORS

Page 33: Lecture 5 SQL Continued

INSS 651 33

INTEGRITY CONSTRAINT VIOLATION

SQL> select * from trial1;

SN SCITY--- --------------------111 baltimore

SQL> insert into trial2 values (234,222);insert into trial2 values (234,222)• *ERROR at line 1:ORA-02291: integrity constraint

(AGGARWAL.SYS_C0026818) violated - parent key not found

Page 34: Lecture 5 SQL Continued

INSS 651 34

TOO MANY VALUES

• SQL> l• 1 select * from invoice where cust_code in• 2* (select inv-num, inv_date from invoice)• SQL> /• (select inv-num, inv_date from invoice)• *• ERROR at line 2:• ORA-00913: too many values

Page 35: Lecture 5 SQL Continued

INSS 651 35

LEASE TABLE

• SQL> desc lease;• Name Null? Type• ----------------------------------------- --------

----------------------------• L_NO NOT NULL CHAR(6)• P_NO CHAR(5)• RENTER_NO CHAR(4)• RENT NUMBER(5)• PAYMENT CHAR(5)• START_DATE DATE• FIN_DATE DATE

Page 36: Lecture 5 SQL Continued

INSS 651 36

MISMATCH COMPARISON

SQL> /

• select * from lease where rent in (select payment from lease)

*

ERROR at line 1:

ORA-01722: invalid number

Page 37: Lecture 5 SQL Continued

INSS 651 37

UNIQUE CONSTRAINT VIOLATION

SQL> select * from trial1;SN SCITY--- --------------------111 baltimoreSQL> insert into trial1 values (111,'mass');insert into trial1 values*ERROR at line 1:ORA-00001: unique constraint

(AGGARWAL.SYS_C0026745) violated

Page 38: Lecture 5 SQL Continued

INSS 651 38

NOT a Single_Group Function

• select distinct l_no, sum(rent)

• *

• ERROR at line 1:

• ORA-00937: not a single-group group function


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