Oracle PL/SQL Tutorial/Character String Functions/REPLACE

If the look for string is not present, then the replacing does not occur

SQL> SELECT REPLACE ("This is a test","AAAA"," may be ")FROM dual;
REPLACE("THISI
--------------
This is a test
SQL>

REPLACE(x, search_string, replace_string) searches x for search_string and replace it with replace_string.

REPLACE() doesn"t modify the actual row in the database.

REPLACE() only deals with rows in the result set.

The following example retrieves the first_name column from the employee table and replaces the string J with P using REPLACE():

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",     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   James      Cat        17-SEP-96 15-APR-02    1232.78 Vancouver  Tester
8 rows selected.
SQL>
SQL>
SQL>
SQL> SELECT REPLACE(first_name, "J", "P") FROM employee;
REPLACE(FI
----------
Pason
Alison
Pames
Celia
Robert
Linda
David
Pames
8 rows selected.
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
  2  /
Table dropped.
SQL>

select replace ("beer bucket","beer") as replace_2

SQL>
SQL>
SQL>
SQL> select replace  ("beer bucket","beer") as replace_2
  2  from   dual;
REPLACE
-------
 bucket
SQL>

select replace ("beer bucket","beer","milk") as replace_1

SQL>
SQL>
SQL>
SQL> select replace  ("beer bucket","beer","milk") as replace_1
  2  from   dual;
REPLACE_1
-----------
milk bucket
SQL>

The REPLACE Function

The REPLACE function has the following general syntax:

REPLACE (string, look for, replace with)

All three arguments are necessary.

The "look for" string will be replaced with the "replace with" string every time it occurs.

SQL> SELECT REPLACE ("This is a test"," is "," may be ") FROM dual;
REPLACE("THISISATE
------------------
This may be a test
SQL>

translate vs replace

SQL>
SQL> select translate("beer bucket","beer","milk") as translate
  2  ,      replace  ("beer bucket","beer","milk") as replace_1
  3  ,      replace  ("beer bucket","beer")        as replace_2
  4  from   dual;
TRANSLATE   REPLACE_1   REPLACE
----------- ----------- -------
miik muckit milk bucket  bucket
SQL>