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XSLT - apply a template to call-template result

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xslt

Is it possible to apply a template to the result of a call-template?

For example, the xml and 2 xslt below.

index.xml:

<?xml version="1.0" encoding="UTF-8"?>
<root>
  <book>
    <title>Ethics</title>
  </book>
  <book>
    <title>Beyond Good and Evil</title>
  </book>
</root>

index2.xsl

<xsl:stylesheet version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:template match="/" name="temp2">
    <foo>
    <xsl:for-each select="//book">
      <bar><xsl:value-of select="."/></bar>
    </xsl:for-each>
    </foo>
  </xsl:template>
</xsl:stylesheet>

and index.xsl is calling index2.xsl

<xsl:stylesheet version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
<xsl:include href="index2.xsl" />
<!-- apply a template to the <xsl:call-template name="temp2" /> result -->
</xsl:stylesheet>

Is there any way to apply a template to the result of <xsl:call-template name="temp2" /> in index.xsl?

Thanks in advance.

like image 775
Eric Avatar asked Dec 17 '11 21:12

Eric


1 Answers

Is there any way to apply a template to the result of <xsl:call-template name="temp2" /> in index.xsl?

Yes, this is called multi-pass processing. The only peculiarity is that in XSLT 1.0 you must apply the implementation-dependent xxx:node-set() extension function to any intermediate result, which in general will be of the infamous RTF (Result Tree Fragment) type and thus will need to be converted to a regular tree.

Here is a complete example of multi-pass processing (I am using xsl:apply-templates and not xsl:call-template, but you can freely modify this example to use named templates and xsl:call-template instead):

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:ext="http://exslt.org/common" exclude-result-prefixes="ext">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/*">
  <xsl:variable name="vrtfPass1">
   <xsl:apply-templates select="num"/>
  </xsl:variable>

  <xsl:apply-templates mode="pass2"
    select="ext:node-set($vrtfPass1)/*"/>
 </xsl:template>

 <xsl:template match="num">
  <num><xsl:value-of select="2* ."/></num>
 </xsl:template>

 <xsl:template match="num" mode="pass2">
  <num><xsl:value-of select=". * ."/></num>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the following XML document:

<nums>
  <num>01</num>
  <num>02</num>
  <num>03</num>
  <num>04</num>
  <num>05</num>
  <num>06</num>
  <num>07</num>
  <num>08</num>
  <num>09</num>
  <num>10</num>
</nums>

the wanted result ( the content of any num elements is doubled in the first pass, then squared in the second pass) is produced:

<num>4</num>
<num>16</num>
<num>36</num>
<num>64</num>
<num>100</num>
<num>144</num>
<num>196</num>
<num>256</num>
<num>324</num>
<num>400</num>

II. In XSLT 2.0

The RTF "type" was abolished, so it is much more easier to specify multi-pass processing, which becomes almost indistinguishable from functional composition, as the following equivalent transformation shows:

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:my="my:my" exclude-result-prefixes="my">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/*">
  <xsl:sequence select="my:squareAll(my:doubleAll(*))"/>
 </xsl:template>

 <xsl:function name="my:doubleAll" as="element()*">
  <xsl:param name="pElems" as="element()*"/>

  <xsl:for-each select="$pElems">
   <xsl:copy>
    <xsl:sequence select=". + ."/>
   </xsl:copy>
  </xsl:for-each>
 </xsl:function>

 <xsl:function name="my:squareAll" as="element()*">
  <xsl:param name="pElems" as="element()*"/>

  <xsl:for-each select="$pElems">
   <xsl:copy>
    <xsl:sequence select=". * ."/>
   </xsl:copy>
  </xsl:for-each>
 </xsl:function>
</xsl:stylesheet>

when this transformation is applied on the same XML document (above), the same correct result is produced:

<num>4</num>
<num>16</num>
<num>36</num>
<num>64</num>
<num>100</num>
<num>144</num>
<num>196</num>
<num>256</num>
<num>324</num>
<num>400</num>
like image 129
Dimitre Novatchev Avatar answered Nov 07 '22 16:11

Dimitre Novatchev