xslt排序输出xml [英] xslt sort output xml

问题描述

我正在尝试找到以下问题的解决方案.

I'm trying to find a solution to the following problem.

我正在开发XSLT转换(现在大约40KB)，该转换将相当复杂的XML转换为一个非常简单的结构，如下所示:

I'm developing XSLT transformation (which is now about 40KB big) that is transforming quite complex XMLs into a quite simple structure which would like this:

<Records>
<Record key="XX">
</Record> 
<Record key="XX1">
</Record>
<Record key="XX2">
</Record>
<Record key="XX3">
</Record>
</Records>

我想让此输出XML根据Records/Record/@key值排序. 问题是我的XSLT产生的输出未排序，由于其复杂性，我无法在那里进行排序. 是否可以在输出XML上应用xsl:sort?我知道我可以准备另一个XSLT转换，但就我而言，这不是解决方案，因为我仅限于一个XSLT..请帮助！...

I would like to have this output XML sorted according to Records/Record/@key values. The problem is that my XSLT produces this output unsorted and due to its complexity I am unable to sort it there. Is it possible to apply xsl:sort on the output XML? I know that I can prepare another XSLT transform, but in my case that's not the solution, as I'm limited to only one XSLT.. Please, help!...

推荐答案

是否可以在输出XML上应用xsl:sort?

Is it possible to apply xsl:sort on the output XML?

是的，可以进行多遍处理 ，尤其是在XSLT 2.0中，您甚至不需要对结果应用xxx:node-set()扩展名，因为臭名昭著的RTF类型不再存在:

Yes, multipass processing is possible, and especially in XSLT 2.0 you don't even need to apply an xxx:node-set() extension on the result, because the infamous RTF type does no longer exist:

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

 <xsl:template match="/">
  <xsl:variable name="vPass1">
   <!--
        Put/Invoke your cirrent code here   
        to generate the following           
-->
    <Records>
      <Record key="XX3">
      </Record>
      <Record key="XX2">
      </Record>
      <Record key="XX4">
      </Record>
      <Record key="XX1">
      </Record>
    </Records>
  </xsl:variable>

  <xsl:apply-templates select="$vPass1/*"/>
 </xsl:template>

 <xsl:template match="Records">
  <Records>
   <xsl:perform-sort select="*">
    <xsl:sort select="@key"/>
   </xsl:perform-sort>
  </Records>
 </xsl:template>
</xsl:stylesheet>

在任何XML文档(未使用/忽略)上执行此转换时，都会生成想要的，正确的，已排序的结果:

<Records>
   <Record key="XX1"/>
   <Record key="XX2"/>
   <Record key="XX3"/>
   <Record key="XX4"/>
</Records>

在XSLT 1.0中，它与将结果从RTF类型额外转换为普通树几乎相同:

<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">
   <!--
        Put/Invoke your cirrent code here   
        to generate the following           
-->
    <Records>
      <Record key="XX3">
      </Record>
      <Record key="XX2">
      </Record>
      <Record key="XX4">
      </Record>
      <Record key="XX1">
      </Record>
    </Records>
  </xsl:variable>

  <xsl:variable name="vPass1"
                select="ext:node-set($vrtfPass1)"/>

  <xsl:apply-templates select="$vPass1/*"/>
 </xsl:template>

 <xsl:template match="Records">
  <Records>
   <xsl:for-each select="*">
    <xsl:sort select="@key"/>

    <xsl:copy-of select="."/>
   </xsl:for-each>
  </Records>
 </xsl:template>
</xsl:stylesheet>

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