How do I transform an XML element name into an attribute value using XSLT?

Question

Given the XML snippet:

<transactions>
  <tran id="1">
    <E8>
      <datestamp>2012-05-17T15:16:57Z</datestamp>
    </E8>
  </tran>
</transactions>

How do I transform the element <E8> into <event type="E8"> using XSLT?

Edit: Expected output:

<transactions>
  <tran id="1">
    <event type="E8">
      <datestamp>2012-05-17T15:16:57Z</datestamp>
    </event>
  </tran>
</transactions>

Dimitre Novatchev · Accepted Answer · 2013-02-20 02:49:00Z

3

This transformation:

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

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

 <xsl:template match="tran/E8">
  <event type="E8">
   <xsl:apply-templates/>
  </event>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document:

<transactions>
  <tran id="1">
    <E8>
      <datestamp>2012-05-17T15:16:57Z</datestamp>
    </E8>
  </tran>
</transactions>

produces the wanted, correct result:

<transactions>
   <tran id="1">
      <event type="E8">
         <datestamp>2012-05-17T15:16:57Z</datestamp>
      </event>
   </tran>
</transactions>

answered Feb 20, 2013 at 2:49

Dimitre Novatchev

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Comments

Kirill Polishchuk · Accepted Answer · 2013-02-20 01:37:31Z

2

Use:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

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

  <xsl:template match="tran/*">
    <event type="{name()}">
      <xsl:value-of select="."/>

    </event>
  </xsl:template>

</xsl:stylesheet>

Output:

<transactions>
  <tran id="1">
    <event type="E8">2012-05-17T15:16:57Z</event>
  </tran>
</transactions>

answered Feb 20, 2013 at 1:37

Kirill Polishchuk

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4 Comments

Jordan Allan Over a year ago

Thanks, that answers my question, but it's not quite the output I was expecting (see edit). I still need to keep all child elements of the E8 element (there can be more than one). Sorry for not clarifying.

Jordan Allan Over a year ago

I figured it out. Used <xsl:copy-of select="@*|node()" /> instead of <xsl:value-of select="."/>

Dimitre Novatchev Over a year ago

@JordanDedels, actually, using xsl:copy-of isn't the best solution -- if in the future you'd need to transform these children, then you'll have to remove this instruction.

Jordan Allan Over a year ago

@DimitreNovatchev, That's a good point. I'm new to XLST, so thanks for the heads up. I'll use the <xsl:apply-templates/> as shown in your answer.

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