Question

I’m in a process converting html to XML, and all works fine but I’m struggling a bit on converting my original path structure to a new one.

Below example explains what I want to achieve :

What I have now (simplified example) :

<links>
<link>../../../folder1/folder2/folder3/folder4/somefile.png</link>
<link>../../../foldera/folderb/folder/some_other_file.gif</link>
<link>folder_x/yet_another_file.jpg</link>
</links>

What I like to get :

<links>
<link>somefile.png</link>
<link>some_other_file.gif</link>
<link>yet_another_file.jpg</link>
</links>

Or in other words, what’s the simplest approach using xslt (2) to get the string after the final “/” in my text

Thanks for any advice on this !

Answer 1

Simply do e.g.

<xsl:template match="link">
  <xsl:copy>
    <xsl:value-of select="tokenize(., '/')[last()]"/>
  </xsl:copy>
</xsl:template>

plus the identity transformation of course e.g.

<xsl:template match="@* | node()">
  <xsl:copy>
    <xsl:apply-templates select="@* , node()"/>
  </xsl:copy>
</xsl:template>