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Using xsl:variable in a xsl:foreach select statement

开发者 https://www.devze.com 2022-12-23 03:39 出处:网络
I\'m trying to iterate through an xml document using xsl:foreach but I need the select=\" \" to be dynamic so I\'m using a variable as the source. Here\'s what I\'ve tried:

I'm trying to iterate through an xml document using xsl:foreach but I need the select=" " to be dynamic so I'm using a variable as the source. Here's what I've tried:

...

<xsl:template name="SetDataPath">
  <xsl:param name="Type" />

  <xsl:variable name="Path_1">/Rating/Path1/*</xsl:variable>
  <xsl:variable name="Path_2">/Rating/Path2/*</xsl:variable>

  <xsl:if test="$Type='1'">
    <xsl:value-of select="$Path_1"/>
  </xsl:if>

  <xsl:if test="$Type='2'">
    <xsl:value-of select="$Path_2"/>
  </xsl:if>
<xsl:template>

...

    <!-- Set Data Path according to Type -->
  <xsl:variable name="DataPath">
    <xsl:call-template name="SetDataPath">
      <xsl:with-param name="Type" select="/Rating/Type" />
    </xsl:call-template> 
  </xsl:variable>

...

<xsl:for-each select="$DataPath">

...

The foreach threw an error stating: "XslTransformException - To use a result tree fragment in a path expression, first convert it to a node-set using the msxsl:node-set() function."

When I use the msxsl:node-set() function though, my results are blank.

I'm aware that I'm setting $DataPath to a string, but shouldn'开发者_如何学JAVAt the node-set() function be creating a node set from it? Am I missing something? When I don't use a variable:

<xsl:for-each select="/Rating/Path1/*">

I get the proper results.

Here's the XML data file I'm using:

<Rating>
    <Type>1</Type>
    <Path1>
       <sarah>
          <dob>1-3-86</dob>
          <user>Sarah</user>
       </sarah>
       <joe>
          <dob>11-12-85</dob>
          <user>Joe</user>
       </joe>
    </Path1>
    <Path2>
       <jeff>
          <dob>11-3-84</dob>
          <user>Jeff</user>
       </jeff>
       <shawn>
          <dob>3-5-81</dob>
          <user>Shawn</user>
       </shawn>
    </Path2>
</Rating>

My question is simple, how do you run a foreach on 2 different paths?


Try this:

   <xsl:for-each select="/Rating[Type='1']/Path1/*
                         |
                          /Rating[Type='2']/Path2/*">


Standard XSLT 1.0 does not support dynamic evaluation of xpaths. However, you can achieve your desired result by restructuring your solution to invoke a named template, passing the node set you want to process as a parameter:

<xsl:variable name="Type" select="/Rating/Type"/>
<xsl:choose>
    <xsl:when test="$Type='1'">
        <xsl:call-template name="DoStuff">
            <xsl:with-param name="Input" select="/Rating/Path1/*"/>
        </xsl:call-template>
    </xsl:when>
    <xsl:when test="$Type='2'">
        <xsl:call-template name="DoStuff">
            <xsl:with-param name="Input" select="/Rating/Path2/*"/>
        </xsl:call-template>
    </xsl:when>
</xsl:choose>

...

<xsl:template name="DoStuff">
    <xsl:param name="Input"/>
    <xsl:for-each select="$Input">
        <!-- Do stuff with input -->
    </xsl:for-each>
</xsl:template>


The node-set() function you mention can convert result tree fragments into node-sets, that's correct. But: Your XSLT does not produce a result tree fragment.

Your template SetDataPath produces a string, which is then stored into your variable $DataPath. When you do <xsl:for-each select="$DataPath">, the XSLT processor chokes on the fact that DataPath does not contain a node-set, but a string.

Your entire stylesheet seems to be revolve around the idea of dynamically selecting/evaluating XPath expressions. Drop that thought, it is neither possible nor necessary.

Show your XML input and specify the transformation your want to do and I can try to show you a way to do it.

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