в 13:45, 26 мая 2010

2010-05-26T13:45:46Z

Admin: 1 версия

2010-05-26T10:08:16Z

1 версия

Новая страница

== NVL() converts a null to a known value.==

NVL(x, value) returns value if x is null; otherwise, x is returned.
The following example selects the id and first_name from the employee table.
Null values for the first_name are converted to the string "Unknown first_name" by NVL().



<source lang="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"), 6661.78, "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"), 6544.78, "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"), 2344.78, "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"), 4322.78,"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"), 7897.78,"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",NULL, "Cat", to_date("19960917","YYYYMMDD"), to_date("20020415","YYYYMMDD"), 1232.78,"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 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL>
SQL>
SQL> SELECT id, NVL(first_name, "Unknown first name") FROM employee;
ID NVL(FIRST_NAME,"UN
---- ------------------
01 Jason
02 Alison
03 James
04 Celia
05 Robert
06 Linda
07 David
08 Unknown first name
8 rows selected.
SQL>
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.</source>

== NVL Function is used if there are null values.==

<OL><LI>NVL function returns a substitute if a value is null.</LI><LI>NVL function takes two arguments.</LI><LI>The first argument is the field or attribute to check.</LI><LI>The second argument is the value that you want to replace the null value by.</LI><LI>For example, "NVL(value, 10)": we are looking for null values in the "value" column, and would like to replace the null value in the "value" column by 10.</LI><LI>NVL does not change the actual data in the table.</LI></OL>



<source lang="sql">
SQL>
SQL> -- create demo table
SQL> create table myTable(
2 id NUMBER(2),
3 value NUMBER(6,2)
4 )
5 /
Table created.
SQL>
SQL> -- prepare data
SQL> insert into myTable(ID, value)values (1,9)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (2,2.11)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (3,3.44)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (4,NULL)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (5,10)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (6,3)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (7,-5.88)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (8,123.45)
2 /
1 row created.
SQL> insert into myTable(ID, value)values (9,98.23)
2 /
1 row created.
SQL>
SQL> select * from myTable
2 /
ID VALUE
---------- ----------
1 9
2 2.11
3 3.44
4
5 10
6 3
7 -5.88
8 123.45
9 98.23
9 rows selected.
SQL>
SQL> SELECT id, NVL(value, 10) From myTable
2 /
ID NVL(VALUE,10)
---------- -------------
1 9
2 2.11
3 3.44
4 10
5 10
6 3
7 -5.88
8 123.45
9 98.23
9 rows selected.
SQL>
SQL> -- clean the table
SQL> drop table myTable
2 /
Table dropped.
SQL></source>

== Use NVL() to convert date columns==



<source lang="sql">
SQL>
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"), 6661.78, "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"), 6544.78, "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"), 2344.78, "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"), 4322.78,"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"), 7897.78,"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", NULL, to_date("20020415","YYYYMMDD"), 1232.78,"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 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL>
SQL> SELECT id, first_name, last_name, NVL(start_date, "01-JAN-2000") FROM employee;
ID FIRST_NAME LAST_NAME NVL(START
---- ---------- ---------- ---------
01 Jason Martin 25-JUL-96
02 Alison Mathews 21-MAR-76
03 James Smith 12-DEC-78
04 Celia Rice 24-OCT-82
05 Robert Black 15-JAN-84
06 Linda Green 30-JUL-87
07 David Larry 31-DEC-90
08 James Cat 01-JAN-00
8 rows selected.
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL></source>

== Use nvl() to convert null value to 0==



<source lang="sql">
SQL>
SQL> create table employees(
2 empno NUMBER(4)
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , msal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) ) ;
Table created.
SQL>
SQL>
SQL> insert into employees values(1,"Jason", "N", "TRAINER", 2, date "1965-12-18", 800 , NULL, 10);
1 row created.
SQL> insert into employees values(2,"Jerry", "J", "SALESREP",3, date "1966-11-19", 1600, 300, 10);
1 row created.
SQL> insert into employees values(3,"Jord", "T" , "SALESREP",4, date "1967-10-21", 1700, 500, 20);
1 row created.
SQL> insert into employees values(4,"Mary", "J", "MANAGER", 5, date "1968-09-22", 1800, NULL, 20);
1 row created.
SQL> insert into employees values(5,"Joe", "P", "SALESREP",6, date "1969-08-23", 1900, 1400, 30);
1 row created.
SQL> insert into employees values(6,"Black", "R", "MANAGER", 7, date "1970-07-24", 2000, NULL, 30);
1 row created.
SQL> insert into employees values(7,"Red", "A", "MANAGER", 8, date "1971-06-25", 2100, NULL, 40);
1 row created.
SQL> insert into employees values(8,"White", "S", "TRAINER", 9, date "1972-05-26", 2200, NULL, 40);
1 row created.
SQL> insert into employees values(9,"Yellow", "C", "DIRECTOR",10, date "1973-04-27", 2300, NULL, 20);
1 row created.
SQL> insert into employees values(10,"Pink", "J", "SALESREP",null,date "1974-03-28", 2400, 0, 30);
1 row created.
SQL>
SQL>
SQL>
SQL> select ename, msal, comm, 12* msal+nvl(comm,0) as yearsal
2 from employees;
ENAME MSAL COMM YEARSAL
-------- ---------- ---------- ----------
Jason 800 9600
Jerry 1600 300 19500
Jord 1700 500 20900
Mary 1800 21600
Joe 1900 1400 24200
Black 2000 24000
Red 2100 25200
White 2200 26400
Yellow 2300 27600
Pink 2400 0 28800
10 rows selected.
SQL>
SQL>
SQL> drop table employees;
Table dropped.</source>

← Предыдущая	Версия 13:45, 26 мая 2010
(нет различий)

Oracle PL/SQL Tutorial/Conversion Functions/NVL - История изменений

в 13:45, 26 мая 2010

Admin: 1 версия