Re: [xsl] Is there a way in XSLT/XPath to find the portion of a string that matches a pattern?

["Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

On 11/03/2025 23:12, Roger L Costello costello@xxxxxxxxx wrote:
> Liam gave this wicked cool way to return the portion of a string that
matches a regex pattern:
> replace($input, '^.*(\d+).*$', '$1')
>
>
>
> However, I did some testing, and it's not returning the desired results:
>
>
>
> <xsl:variable name="TEXT" select="'The person put 12 dollars into the
jar'"/>
> <xsl:variable name="INTEGER" select="'[0-9]+'"/>
>
> <xsl:value-of select="replace($TEXT,'^.*(' || $INTEGER || ').*$', '$1')"/>
>
>
>
> Result: 2  <-- should be 12
>
>
>
> Another test:
>
>
>
> <xsl:variable name="TEXT" select="'SATSET'"/>
>
> <xsl:variable name="VOWELS" select="'A|E|I|O|U'"/>
>
> <xsl:value-of select="replace($TEXT,'^.*(' || $VOWELS || ').*$', '$1')"/>
>
>
>
> Result: E  <-- should be A
>
>
>
> What am I doing wrong?

I guess you need non-greedy matching for the .* with .*? e.g.
replace($TEXT,'^.*?(' || $INTEGER || ').*$', '$1')