CODEWITHSUNDEEP
xmlxslt-1.0
None
None

Reputation: 5670

Selecting specific named elements from XML using XSLT

I got an XML like this on va variable prdxml

  <category numberofproducts="0">
  <id>catering</id>
  <url>/products/kantine</url>
  <name>Kantine</name>
  <category numberofproducts="0">
    <id>madembal</id>
    <url>/products/mad-og-emballage</url>
    <name>Mad og emballage</name>
    <category numberofproducts="9">
      <id>stanniol</id>
      <url>/products/stanniol</url>
      <name>Stanniol</name>
      <category  numberofproducts="9">
        <id>stankd</id>
        <url>/products/stankd</url>
        <name>stankd</name>
      </category>
    </category>
    <category numberofproducts="5">
      <id>termo</id>
      <url>/products/termobakker</url>
      <name>Termobakker</name>
    </category>
  </category>
</category>

Now I want to get all the Id's present in this xml to a string.Tried usual for-each loop but failed,because of the structure of this xml. Now I thinks there may be a smarter way to achieve this ,that i never known.Can any one give me that light.

Upvotes: 1

Views: 204

Answers (1)

StuartLC
StuartLC

Reputation: 107357

If you mean flattening all id's irrespective of their nesting, level, then this should work with a ',' delimiter in between.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

  <xsl:template match="/">
    <xsl:apply-templates select="//id"/>
  </xsl:template>

  <xsl:template match="id">
    <xsl:value-of select="concat(./text(), ',')"/>
  </xsl:template>

</xsl:stylesheet>

Returns : catering,madembal,stanniol,stankd,termo,

If you need a more elegant mechanism for handling the last element delimiter, have a look at Dimitre's answer here

Update: To put the result into a variable:

<xsl:variable name="someVar">
  <xsl:apply-templates select="//id"/>
</xsl:variable>
Result: <xsl:value-of select="$someVar"/>

Upvotes: 1

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