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Re: sorting AND copying of XML via XSL

2003-09-09 10:48:39
Hi Marcel,
  Below is the complete XSL --

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0"
encoding="UTF-8" indent="yes"/>

<xsl:template match="/foo">
  <foo>
    <xsl:for-each select="bar">
     <xsl:sort select="@id" data-type="number"/>
        <bar id="{(_at_)id}">
          <xsl:for-each select="item">
           <xsl:sort select="@id" data-type="number"/>
                <item id="{(_at_)id}">
                  <xsl:value-of select="."/>
                </item>
          </xsl:for-each>
        </bar>
    </xsl:for-each>
  </foo>
</xsl:template>
</xsl:stylesheet>

Regards,
Mukul


--- marcel salathe <ms7141(_at_)gmx(_dot_)de> wrote:
Hi

I have a XSL question which might be easy to answer,
but I couldnt.

Suppose I have an XML document like this:

<foo>
  <bar id="2">
    <item id="6">hello</item>
    <item id="4">world</item>
  </bar>
  <bar id="1">
    <item id="7">how's</item>
    <item id="3">life</item>
  </bar>
</foo>

Now I want to sort the bar elements and then the
item elements according to
their Id's, and finally output 
the entire document sorted, e.g.

<foo>
  <bar id="1">
    <item id="3">life</item>
    <item id="7">how's</item>
  </bar>
  <bar id="2">
    <item id="4">world</item>
    <item id="6">hello</item>
  </bar>
</foo>

I could achieve sorting or copying, but I couldnt do
both together. I guess
it must be quite a simple 
XSLT. 

Any help very much appreciated.

Best regards,
marcel


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