Oracle PL/SQL Tutorial/Conversion Functions/NVL — различия между версиями

Текущая версия на 10:08, 26 мая 2010

Содержание

1 NVL() converts a null to a known value.
2 NVL Function is used if there are null values.
3 Use NVL() to convert date columns
4 Use nvl() to convert null value to 0

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().

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.

NVL Function is used if there are null values.

NVL function returns a substitute if a value is null.
NVL function takes two arguments.
The first argument is the field or attribute to check.
The second argument is the value that you want to replace the null value by.
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.
NVL does not change the actual data in the table.

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>

Use NVL() to convert date columns

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>

Use nvl() to convert null value to 0

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.

Oracle PL/SQL Tutorial/Conversion Functions/NVL — различия между версиями

Текущая версия на 10:08, 26 мая 2010

Содержание

NVL() converts a null to a known value.

NVL Function is used if there are null values.

Use NVL() to convert date columns

Use nvl() to convert null value to 0

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Версия 13:45, 26 мая 2010 (просмотреть исходный код) (обсуждение) ← Предыдущая правка	Текущая версия на 10:08, 26 мая 2010 (просмотреть исходный код) Admin (обсуждение \| вклад) м (1 версия)
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