Returns an integer representing the starting position of the first occurrence of a pattern in a string.
SyntaxPATINDEX( '%pattern%', string-expression )
Parameterspattern The pattern to be searched for. If the leading percent wildcard is omitted, the PATINDEX function returns one (1) if the pattern occurs at the beginning of the string, and zero if not.
The pattern uses the same wildcards as the LIKE comparison. These are as follows:
| Wildcard | Matches |
|---|---|
| _ (underscore) | Any one character |
| % (percent) | Any string of zero or more characters |
| [] | Any single character in the specified range or set |
| [^] | Any single character not in the specified range or set |
string-expression The string to be searched for the pattern.
ReturnsINT
RemarksThe PATINDEX function returns the starting position of the first occurrence of the pattern. If the pattern is not found, it returns zero (0).
See also Standards and compatibilitySQL/2008 Vendor extension.
ExampleThe following statement returns the value 2.
SELECT PATINDEX( '%hoco%', 'chocolate' ); |
The following statement returns the value 11.
SELECT PATINDEX( '%4_5_', '0a1A 2a3A 4a5A' ); |
The following statement returns 14 which is the first non-alphanumeric character in the string expression. Note that the pattern
'%[^a-z0-9]%' can be used instead of '%[^a-zA-Z0-9]%' if the database is case insensitive.
SELECT PATINDEX( '%[^a-zA-Z0-9]%', 'SQLAnywhere12 has many new features' ); |
To get the first alphanumeric word in a string, you can use something like the following:
SELECT LEFT( @string, PATINDEX( '%[^a-zA-Z0-9]%', @string ) ); |
 |
Discuss this page in DocCommentXchange.
|
Copyright © 2010, iAnywhere Solutions, Inc. - SQL Anywhere 12.0.0 |