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



<source lang="sql">
SQL> SELECT REPLACE ("This is a test","AAAA"," may be ")FROM dual;
REPLACE("THISI
--------------
This is a test
SQL></source>

== 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():



<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","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></source>

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



<source lang="sql">
SQL>
SQL>
SQL>
SQL> select replace ("beer bucket","beer") as replace_2
2 from dual;
REPLACE
-------
bucket
SQL></source>

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



<source lang="sql">
SQL>
SQL>
SQL>
SQL> select replace ("beer bucket","beer","milk") as replace_1
2 from dual;
REPLACE_1
-----------
milk bucket
SQL></source>

== 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.



<source lang="sql">
SQL> SELECT REPLACE ("This is a test"," is "," may be ") FROM dual;
REPLACE("THISISATE
------------------
This may be a test
SQL></source>

== translate vs replace==



<source lang="sql">
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></source>

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