Question

I want help in getting a value from one xml into another using xsl, i have the following xml:

<metadata>
  <idinfo>Node Text</idinfo>
</metadata>

and i have a general xsl to display the nodes like:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" 
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
  xmlns:dmd="http://www.digitalmeasures.com/schema/data-metadata"
  xmlns:dm="http://www.digitalmeasures.com/schema/data">
    <xsl:output method="html" encoding="utf-8"/>
    <xsl:template match="*[(child::*)]">
        <fieldset>
            <legend><xsl:value-of select="local-name()"/></legend>
            <xsl:apply-templates/>
        </fieldset>
    </xsl:template>
    <xsl:template match="*[not(child::*)]">
        <strong><xsl:value-of select="local-name()"/></strong>
            <i><xsl:apply-templates/></i>
    </xsl:template>
</xsl:stylesheet>

And the output will be:

<fieldset>
<legend>metadata</legend>
  <strong>idinfo</strong><i>Node Text</i>
</fieldset>

Now what I want is replacing node names by another xml, so i want to replace idinfo by Identification Information

The xml which I ant to use is as follows:

<?xml version="1.0" encoding="UTF-8"?>
<labels>
<element name="idinfo">
<label>Identification Information</label>
</element>
</label>

I hope that my question is clear.

Answer 1

first, setup your reference table like:

<xsl:variable name="labels_info" select="document('another.xml')/labels"/>

And then use it like:

<xsl:value-of select="$labels_info/element[@name=local-name(current())]/label"/>

in case it doesn’t work (and I heard reports on that), use a variable:

<xsl:variable name="current_name" select="local-name"/> 
<xsl:value-of select="$labels_info/element[@name=$current_name]/label"/>