从XML的不同部分选择XSL中的数据

I am new to XSLT.

我是XSLT的新手。

I have an XML-

我有一个XML-

<root>
  <fiberList>
              <fiber>
                     <A>abcd</A>
                     <B>bcde</B>
                     <C>cdef</C>

              </fiber>
    </fiberList>
           <generatorList>
              <generator>
                     <D>defg</D>
                     <E>efgh</E>
              </generator>
 </root>

I want to transform this XML into another XML through XSL. The resulting XML should be something like this-

我想通过XSL将这个XML转换为另一个XML。生成的XML应该是这样的 -

<parent>
          <child>
                  <M>abcd</M>
                  <N>bcde</N>
                  <O>efgh</O>
          </child>
</parent>

But I am getting the following XML after transformation-

但我在转换后得到以下XML -

<parent>
          <child>
                  <M>abcd</M>
                  <N>bcde</N>
                  <O/>
          </child>
</parent>

My XSL is the following-

我的XSL如下 -

<xsl:template match="/">
<xsl:element name="parent">
      <xsl:apply-templates select="root/fiberList/fiber"/>
</xsl:element>
</xsl:template>
<xsl:template match="root/fiberList/fiber">
    <xsl:element name="child">
        <xsl:element name="M">
            <xsl:value-of select="A"/>
        </element>
        <xsl:element name="N">
            <xsl:value-of select="B"/>
        </element>
       <xsl:element name="O">
           <xsl:value-of select="root/generatorList/generator/E"/>
       </element>
   </element>
</template>
</stylesheet>

Please help.

2 个解决方案

#1

i haven't tried it .. but i think a leading '/' is missing here :

我还没试过..但我认为这里缺少一个领先的'/':

 <xsl:value-of select="root/generatorList/generator/E"/>

try this please :

请试试这个:

 <xsl:value-of select="/root/generatorList/generator/E"/>

otherwise the xslt engine is trying to find a relative path "root/generatorList/generator/E" starting at the matched element (/root/fiberList/fiber) of the template but there is no /root/fiberList/fiber/root/generatorList/generator/E.

否则xslt引擎试图从模板的匹配元素(/ root / fiberList / fiber)开始寻找相对路径“root / generatorList / generator / E”,但没有/ root / fiberList / fiber / root / generatorList /发电机/ E。

#2

As an alternate, you could use this stylesheet too:

作为替代方案,您也可以使用此样式表:

XSLT-1.0:

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

    <xsl:template match="/root">
        <parent>
            <child>
                <xsl:apply-templates select="fiberList/fiber/A | fiberList/fiber/B | generatorList/generator/E"/>
            </child>
        </parent>
    </xsl:template>

    <xsl:template match="A">
        <M>
            <xsl:value-of select="."/>
        </M>
    </xsl:template>
    <xsl:template match="B">
        <N>
            <xsl:value-of select="."/>
        </N>
    </xsl:template>
    <xsl:template match="E">
        <O>
            <xsl:value-of select="."/>
        </O>
    </xsl:template>

</xsl:stylesheet>

EDIT:

In XSLT-2.0 you can simplify the xsl:apply-templates instruction to the following:

在XSLT-2.0中,您可以将xsl:apply-templates指令简化为以下内容:

<xsl:apply-templates select="fiberList/fiber/(A | B) | generatorList/generator/E"/>

#1