Question

我有一个xml

 <Root>
  <Parent>
    <Child1><Node1>AAA</Node1><Node2>BBB</Node2></Child1>
    <Child2><NodeX>XXX</NodeX><NodeY>YYY</NodeY></Child2>
    <Child1><Node1>EEE</Node1><Node2>FFF</Node2></Child1>
    <Child2><NodeX>GGG</NodeX><NodeY>HHH</NodeY></Child2>
    <OtherChild></OtherChild>

  </Parent>
</Root>

Child2将始终与child1一起使用。我需要知道如何使用xsl：foreach循环并创建XML输出示例。我可能有其他节点，如<OtherChild>，但我担心只是Child1和Chid2节点

    <TransformedXML>
  <Child attributefromNode1="AAA" attributefromNode2="BBB" attributefromNodeX="XXX" attributeFromNodeY="YYY"/>
  <Child attributefromnode1="EEE" attributefromNode2="FFF" attributefromNodeX="GGG" attributeFromNodeY="HHH"/>
</TransformedXML>

我的问题是我如何在XSLT中循环，考虑到Child2节点将始终跟随Child1？

Answer 1

我这样的事情可以解决问题：

  <xsl:stylesheet version="1.0"                 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns="http://www.w3.org/TR/xhtml1/strict">
  <xsl:output
   method="xml"
   indent="yes" />

  <xsl:template match="/">
<xsl:for-each select="Root/Parent/Child1">
<Child>
  <xsl:attribute name="attributefromNode1">
    <xsl:value-of select="Node1" />

  </xsl:attribute>
  <xsl:attribute name="attributefromNode2" >
    <xsl:value-of select="Node2" />

  </xsl:attribute>
  <xsl:attribute name="attributefromNodeX">
    <xsl:value-of select="following-sibling::Child2/NodeX" />

  </xsl:attribute>
  <xsl:attribute name="attributefromNodeY">
    <xsl:value-of select="following-sibling::Child2/NodeY" />

  </xsl:attribute>



</Child>   
    </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

Answer 2

你不应该循环......

XML INPUT

<Root>
  <Parent>
    <Child1><Node1>AAA</Node1><Node2>BBB</Node2></Child1>
    <Child2><NodeX>XXX</NodeX><NodeY>YYY</NodeY></Child2>
    <Child1><Node1>EEE</Node1><Node2>FFF</Node2></Child1>
    <Child2><NodeX>GGG</NodeX><NodeY>HHH</NodeY></Child2>
    <OtherChild></OtherChild>

  </Parent>
</Root>

<强> XSL

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="node()|@*">
      <xsl:apply-templates select="node()|@*"/>
  </xsl:template>

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

  <xsl:template match="Child1">
    <Child attributefromNode1="{Node1}" 
      attributefromNode2="{Node2}" 
      attributefromNodeX="{following-sibling::Child2[1]/NodeX}" 
      attributefromNodeY="{following-sibling::Child2[1]/NodeY}"/>  
  </xsl:template>

</xsl:stylesheet>

XML OUTPUT

<?xml version="1.0" encoding="UTF-8"?>
<TransformedXML>
   <Child attributefromNode1="AAA" attributefromNode2="BBB" attributefromNodeX="XXX"
          attributefromNodeY="YYY"/>
   <Child attributefromNode1="EEE" attributefromNode2="FFF" attributefromNodeX="GGG"
          attributefromNodeY="HHH"/>
</TransformedXML>

Answer 3

这是（几乎和以前一样）一个简短的解决方案：

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

 <xsl:key name="kFollowingChild1" match="*[not(self::Child1)]/*"
  use="generate-id(../preceding-sibling::Child1[1])"/>

 <xsl:template match="Parent">
  <TransformedXML>
   <xsl:apply-templates/>
  </TransformedXML>
 </xsl:template>

 <xsl:template match="Child1">
  <Child>
   <xsl:for-each select="*|key('kFollowingChild1', generate-id())">
    <xsl:attribute name="attribute{position()}">
      <xsl:value-of select="."/>
    </xsl:attribute>
   </xsl:for-each>
  </Child>
 </xsl:template>

 <xsl:template match="text()"/>
</xsl:stylesheet>

将此转换应用于提供的XML文档：

<Root>
  <Parent>
    <Child1><Node1>AAA</Node1><Node2>BBB</Node2></Child1>
    <Child2><NodeX>XXX</NodeX><NodeY>YYY</NodeY></Child2>
    <Child1><Node1>EEE</Node1><Node2>FFF</Node2></Child1>
    <Child2><NodeX>GGG</NodeX><NodeY>HHH</NodeY></Child2>
    <OtherChild></OtherChild>
  </Parent>
</Root>

产生了想要的正确结果：

<TransformedXML>
   <Child attribute1="AAA" attribute2="BBB" attribute3="XXX" attribute4="YYY"/>
   <Child attribute1="EEE" attribute2="FFF" attribute3="GGG" attribute4="HHH"/>
</TransformedXML>

Answer 4

这个XSLT 2.0样式表：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:strip-space elements="*"/>
    <xsl:template match="Parent">
        <TransformedXML>
            <xsl:apply-templates select="Child1"/>
        </TransformedXML>
    </xsl:template>
    <xsl:template match="Child1">
        <Child>
            <xsl:apply-templates select="*|following-sibling::Child2[1]/*"/>
        </Child>
    </xsl:template>
    <xsl:template match="Child1/*|Child2/*">
        <xsl:attribute name="attributefrom{name()}">
            <xsl:value-of select="."/>
        </xsl:attribute>
    </xsl:template>
</xsl:stylesheet>

输出：

<TransformedXML>
    <Child attributefromNode1="AAA" attributefromNode2="BBB" attributefromNodeX="XXX" attributefromNodeY="YYY"/>
    <Child attributefromNode1="EEE" attributefromNode2="FFF" attributefromNodeX="GGG" attributefromNodeY="HHH"/>
</TransformedXML>

xslt分组问题

4 个答案: