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Oracle® Database SQL Language Reference
11g Release 2 (11.2)

Part Number E10592-04
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REGEXP_SUBSTR

Syntax

Description of regexp_substr.gif follows
Description of the illustration regexp_substr.gif

Purpose

REGEXP_SUBSTR extends the functionality of the SUBSTR function by letting you search a string for a regular expression pattern. It is also similar to REGEXP_INSTR, but instead of returning the position of the substring, it returns the substring itself. This function is useful if you need the contents of a match string but not its position in the source string. The function returns the string as VARCHAR2 or CLOB data in the same character set as source_char.

This function complies with the POSIX regular expression standard and the Unicode Regular Expression Guidelines. For more information, refer to Appendix D, "Oracle Regular Expression Support".

Examples

The following example examines the string, looking for the first substring bounded by commas. Oracle Database searches for a comma followed by one or more occurrences of non-comma characters followed by a comma. Oracle returns the substring, including the leading and trailing commas.

SELECT
  REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA',
                ',[^,]+,') "REGEXPR_SUBSTR"
  FROM DUAL;

REGEXPR_SUBSTR
-----------------
, Redwood Shores,

The following example examines the string, looking for http:// followed by a substring of one or more alphanumeric characters and optionally, a period (.). Oracle searches for a minimum of three and a maximum of four occurrences of this substring between http:// and either a slash (/) or the end of the string.

SELECT
  REGEXP_SUBSTR('http://www.example.com/products',
                'http://([[:alnum:]]+\.?){3,4}/?') "REGEXP_SUBSTR"
  FROM DUAL;

REGEXP_SUBSTR
----------------------
http://www.example.com/

The next two examples use the subexpr argument to return a specific subexpression of pattern. The first statement returns the first subexpression in pattern:

SELECT REGEXP_SUBSTR('1234567890', '(123)(4(56)(78))', 1, 1, 'i', 1) 
"REGEXP_SUBSTR" FROM DUAL;

REGEXP_SUBSTR
-------------------
123

The next statement returns the fourth subexpression in pattern:

SELECT REGEXP_SUBSTR('1234567890', '(123)(4(56)(78))', 1, 1, 'i', 4) 
"REGEXP_SUBSTR" FROM DUAL;

REGEXP_SUBSTR
-------------------
78