Re: [xsl] efficiently extracting a capture group from analyze-string()

["Imsieke, Gerrit, le-tex gerrit.imsieke@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Why not use the classic XPath 2.0 replace function?

replace($file, '^.+/my-(w+)\.xml$', '$1')

On 04.06.2022 17:30, Chris Papademetrious 
christopher.papademetrious@xxxxxxxxxxxx wrote:
> Hi everyone!
>
> Given some filename string like
>
>  B  <xsl:variable name="file" select="'path/my-sometype.xml'"/>
>
> Ibd like to extract the bmy-(\w+)b part of the filename using 
> analyze-string(). So far I have
>
>  B  <xsl:variable name="my-file-type" select="data(analyze-string($file, 
> 'my-(\w+)')//fn:group[1])"/>
>
> which works by extracting the <group> descendant from the 
> <analyze-string-result> tree, then converting it to text. Is there a 
> more concise way to do this?
>
> I had to define
>
>  B  xmlns:fn=http://www.w3.org/2005/xpath-functions 
> <http://www.w3.org/2005/xpath-functions>
>
> in my stylesheet for fn:group to match, and Ibm not sure if therebs a 
> better way to do that too.
>
> Thanks!
>
>   * Chris