Oracle PL/SQL/Select Query/Null Value Calculation — различия между версиями
Admin (обсуждение | вклад) м (1 версия) |
Admin (обсуждение | вклад) м (1 версия) |
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Текущая версия на 10:00, 26 мая 2010
Convert null number value as 0
SQL> create table emp
2 ( empno NUMBER(4) constraint E_PK primary key
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , sal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) default 10
11 ) ;
Table created.
SQL> insert into emp values(1,"Tom","N", "Coder", 13,date "1965-12-17", 800 , NULL, 20);
1 row created.
SQL> insert into emp values(2,"Jack","JAM", "Tester",6,date "1961-02-20", 1600, 300, 30);
1 row created.
SQL> insert into emp values(3,"Wil","TF" , "Tester",6,date "1962-02-22", 1250, 500, 30);
1 row created.
SQL> insert into emp values(4,"Jane","JM", "Designer", 9,date "1967-04-02", 2975, NULL, 20);
1 row created.
SQL> insert into emp values(5,"Mary","P", "Tester",6,date "1956-09-28", 1250, 1400, 30);
1 row created.
SQL> insert into emp values(6,"Black","R", "Designer", 9,date "1963-11-01", 2850, NULL, 30);
1 row created.
SQL> insert into emp values(7,"Chris","AB", "Designer", 9,date "1965-06-09", 2450, NULL, 10);
1 row created.
SQL> insert into emp values(8,"Smart","SCJ", "Coder", 4,date "1959-11-26", 3000, NULL, 20);
1 row created.
SQL> insert into emp values(9,"Peter","CC", "Designer",NULL,date "1952-11-17", 5000, NULL, 10);
1 row created.
SQL> insert into emp values(10,"Take","JJ", "Tester",6,date "1968-09-28", 1500, 0, 30);
1 row created.
SQL> insert into emp values(11,"Ana","AA", "Coder", 8,date "1966-12-30", 1100, NULL, 20);
1 row created.
SQL> insert into emp values(12,"Jane","R", "Manager", 6,date "1969-12-03", 800 , NULL, 30);
1 row created.
SQL> insert into emp values(13,"Fake","MG", "Coder", 4,date "1959-02-13", 3000, NULL, 20);
1 row created.
SQL> insert into emp values(14,"Mike","TJA","Manager", 7,date "1962-01-23", 1300, NULL, 10);
1 row created.
SQL>
SQL> select ename, sal, comm
2 , 12*sal+nvl(comm,0) as yearsal
3 from emp
4 where ename like "%T%";
ENAME SAL COMM YEARSAL
-------- ---------- ---------- ----------
Tom 800 9600
Take 1500 0 18000
SQL>
SQL>
SQL> drop table emp;
Table dropped.
NULL values on arithmetic operations
SQL>
SQL> -- create demo table
SQL> create table Employee(
2 ID VARCHAR2(4 BYTE) NOT NULL,
3 First_Name VARCHAR2(10 BYTE),
4 Last_Name VARCHAR2(10 BYTE),
5 Start_Date DATE,
6 End_Date DATE,
7 Salary Number(8,2),
8 City VARCHAR2(10 BYTE),
9 Description VARCHAR2(15 BYTE)
10 )
11 /
Table created.
SQL>
SQL> -- prepare data
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values ("01","Jason", "Martin", to_date("19960725","YYYYMMDD"), to_date("20060725","YYYYMMDD"), 1234.56, "Toronto", "Programmer")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("02","Alison", "Mathews", to_date("19760321","YYYYMMDD"), to_date("19860221","YYYYMMDD"), 2334.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("03","James", "Smith", to_date("19781212","YYYYMMDD"), to_date("19900315","YYYYMMDD"), 2334.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("04","Celia", "Rice", to_date("19821024","YYYYMMDD"), to_date("19990421","YYYYMMDD"), 2334.78, "Vancouver","Manager")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("05","Robert", "Black", to_date("19840115","YYYYMMDD"), to_date("19980808","YYYYMMDD"), 2334.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("06","Linda", "Green", to_date("19870730","YYYYMMDD"), to_date("19960104","YYYYMMDD"), 2334.78,"New York", "Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("07","David", "Larry", to_date("19901231","YYYYMMDD"), to_date("19980212","YYYYMMDD"), 2334.78,"New York", "Manager")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("08","James", "Cat", to_date("19960917","YYYYMMDD"), to_date("20020415","YYYYMMDD"), 2334.78,"Vancouver", "Tester")
3 /
1 row created.
SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
2 /
ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01 Jason Martin 25-JUL-96 25-JUL-06 1234.56 Toronto Programmer
02 Alison Mathews 21-MAR-76 21-FEB-86 2334.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 2334.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2334.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 2334.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 2334.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 2334.78 Vancouver Tester
8 rows selected.
SQL>
SQL>
SQL>
SQL>
SQL>
SQL> -- NULL values on arithmetic and concatenation operations
SQL>
SQL> SELECT ID, Salary + NULL AS "Result" FROM Employee;
ID Result
---- ----------
01
02
03
04
05
06
07
08
8 rows selected.
SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
SQL>
SQL>