How to rename child element attribute and modify attribute value using xslt

Question

There is this certain element in my xml:

<para><link href="D52871.dita">abc</link>
</para>

I want the output to be

<para><link id="D52871">abc</link>
</para>

I have used identity transform in the beginning to copy everything. I tried this code snippet

<xsl:template match="link/@href">
  <xsl:attribute name="id">
    <xsl:value-of select="."/>
  </xsl:attribute>
</xsl:template>

but it's not working probably because I need to specify that link element is inside para. Have tried few approaches to include that but none worked so far.

What do you mean by "not working"? What does your "identity transform" look like? It works for me. Also, do you need to change the href value from D52871.dita to D52871? — Jim Garrison
– Jim Garrison, Commented Jul 7, 2016 at 18:58
Does you actual XML have namespace at all (Look for xlmns="..." somewhere in it)? — Tim C
– Tim C, Commented Jul 7, 2016 at 19:02

Jim Garrison · Accepted Answer · 2016-07-07 19:00:50Z

1

The following works fine for me

<xsl:stylesheet  xmlns="http://www.w3.org/TR/xhtml1/strict" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema"  version="1.0">
    <xsl:output method ="xml" indent="yes"/>

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

    <xsl:template match="link/@href">
        <xsl:attribute name="id">
            <xsl:value-of select="substring-before(.,'.')"/>
        </xsl:attribute>
    </xsl:template> 

</xsl:stylesheet>

Input:

<?xml version="1.0" encoding="UTF-8"?>
<para>
   <link href="D52871.dita">abc</link>
</para>

Output:

<?xml version="1.0" encoding="UTF-8"?>
<para>
   <link id="D52871">abc</link>
</para>

answered Jul 7, 2016 at 19:00

Jim Garrison

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