如何使用xslt合并这两个xml文件

时间:2012-09-14 11:33:50

标签: xslt xslt-1.0

请问我想知道如何使用xslt将两个xml文件合并到一个xml文件中。 我有这两个xml文件,我想合并到xml文件中,如预期的输出所示。我想将第二个文件中的每个节点 Hn 包含在相应的块中。档案1。

文件1:

<Test>
  <R1>
    <Th1>
        here are some instruction.
    </Th1>
  </R1>

  <R2>
    <Th2>
        here are some instruction.
    </Th2>
  </R2>

  <R3>
    <Th3>
        here are some instruction.
    </Th3>
  </R3>
</Test>

文件2:

<test1>
  <H1>
    here are some instruction.
  </H1>

  <H2>
    here are some instruction.
  </H2>

  <H3>
    here are some instruction.
  </H3>
</test1>    

这是预期的输出:

<test2>
  <R1>
    <H1>
        here are some instruction.
    </H1>
    <Th1>
        here are some instruction.
    </Th1>
  </R1>

  <R2>
    <H2>
        here are some instruction.
    </H2>
    <Th2>
        here are some instruction.
    </Th2>
  </R2>

  <R3>
    <H3>
        here are some instruction.
    </H3>
    <Th3>
        here are some instruction.
    </Th3>
  </R3>

</test2>

感谢您的帮助。

1 个答案:

答案 0 :(得分:2)

<强>予。这个XSLT 1.0转换:

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

 <xsl:variable name="vDigits" select="'0123456789'"/>
 <xsl:variable name="vDoc2" select="document('file:///c:/temp/delete/file2.xml')"/> 

 <xsl:template match="/*">
     <test2><xsl:apply-templates/></test2>
 </xsl:template>

 <xsl:template match="/*/*">
  <xsl:copy>
   <xsl:text>&#xA;</xsl:text>
   <xsl:copy-of select=
    "$vDoc2/*/*
      [translate(name(),translate(name(),$vDigits,''),'')
      =
       translate(name(current()),translate(name(current()),$vDigits,''),'')
      ]"/>
    <xsl:copy-of select="node()"/>
   </xsl:copy>
 </xsl:template>
</xsl:stylesheet>

应用于第一个XML文档(file1.xml)时:

<Test>
  <R1>
    <Th1>
        here are some instruction.
    </Th1>
  </R1>

  <R2>
    <Th2>
        here are some instruction.
    </Th2>
  </R2>

  <R3>
    <Th3>
        here are some instruction.
    </Th3>
  </R3>
</Test>

并将第二个XML文档驻留在c:\temp\delete\file2.xml

<test1>
  <H1>
    here are some instruction.
  </H1>

  <H2>
    here are some instruction.
  </H2>

  <H3>
    here are some instruction.
  </H3>
</test1>

会产生想要的正确结果:

<test2>
  <R1>
<H1>
    here are some instruction.
  </H1>
    <Th1>
        here are some instruction.
    </Th1>
  </R1>

  <R2>
<H2>
    here are some instruction.
  </H2>
    <Th2>
        here are some instruction.
    </Th2>
  </R2>

  <R3>
<H3>
    here are some instruction.
  </H3>
    <Th3>
        here are some instruction.
    </Th3>
  </R3>
</test2>

<强>解释

正确使用Michael Kay首次演示的“双翻译”方法。


<强> II。 XSLT 2.0解决方案:

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

 <xsl:variable name="vDigits" select="'0123456789'"/>
 <xsl:variable name="vDoc2" select="document('file:///c:/temp/delete/file2.xml')"/> 

 <xsl:template match="/*">
     <test2><xsl:apply-templates/></test2>
 </xsl:template>

 <xsl:template match="/*/*">
  <xsl:if test="not(matches(name(), '^.+\d+$'))">
    <xsl:message terminate="yes">
     Element Name doesn't end with number: <xsl:sequence select="name()"/>
    </xsl:message>
  </xsl:if>
  <xsl:copy>
   <xsl:text>&#xA;</xsl:text>
   <xsl:copy-of select=
    "$vDoc2/*/*
      [replace(name(),'^.+(\d+)$', '$1')
      =
       replace(name(current()),'^.+(\d+)$', '$1')
      ]"/>
    <xsl:copy-of select="node()"/>
   </xsl:copy>
 </xsl:template>
</xsl:stylesheet>

<强>解释

正确使用标准XPath 2.0函数 matches() replace()