3

这是源 XML：

<customers>
    <firstname1>Sean</firstname1>
    <lastname1>Killer</lastname1>
    <sex1>M</sex1>
    <firstname2>Frank</firstname2>
    <lastname2>Woods</lastname2>
    <sex2>M</sex2>
    <firstname3>Jennifer</firstname3>
    <lastname3>Lee</lastname3>
    <sex3>F</sex3>
</customers>

我怎样才能把它转换成这个？

<MyCustomers>
    <Customer>
        <Name> Sean Killer</Name>
        <Sex>M</Sex>
    </Customer>
    <Customer>
        <Name> Frank Woods</Name>
        <Sex>M</Sex>
    </Customer>
    <Customer>
        <Name>Jennifer Lee</Name>
        <Sex>F</Sex>
    </Customer>
</MyCustomers>
4

3 回答 3

4

As per comments:

what if elements were not in consequent orders?

In this case (assuming XSLT 1.0) you can use translate() to get the id of the elements and then search for the corresponding elements by correct name built using concat(). I would change the following-sibling:: axis to the ../ (short for parent::) to make sure to eventually catch also elements preceding the current firstname.

<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:template match="customers">
        <MyCustomers>
            <xsl:apply-templates select="*[starts-with(name(),'firstname')]"/>
        </MyCustomers>
    </xsl:template>

    <xsl:template match="*[starts-with(name(),'firstname')]">
        <xsl:variable name="id" select="translate(name(),'firstname','')"/>

        <Customer>
            <Name><xsl:value-of select="concat(.,' ',
                    ../*[name()=concat('lastname',$id)])"/></Name>
            <Sex><xsl:value-of select="../*[name()=concat('sex',$id)]"/></Sex>
        </Customer>
    </xsl:template>

</xsl:stylesheet>

Obsolete answer

Assuming the fixed input document structure as shown in the question, a fine working XSLT 1.0 transform is: 

<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:template match="customers">
        <MyCustomers>
            <xsl:apply-templates select="*[starts-with(name(),'firstname')]"/>
        </MyCustomers>
    </xsl:template>

    <xsl:template match="*[starts-with(name(),'firstname')]">
        <Customer>
            <Name><xsl:value-of select="concat(.,' ',
                    following-sibling::*[1]
                    [starts-with(name(),'lastname')])"/></Name>
            <Sex><xsl:value-of select="following-sibling::*[2]
                    [starts-with(name(),'sex')]"/></Sex>
        </Customer>
    </xsl:template>

</xsl:stylesheet>

Little explanation

You need XPath 1.0 function starts-with() because of the sad name of the tags in your XML input. You can use the following-sibling:: axis to get the required following sibling tags of any element whose name starts with firstname.

于 2011-07-11T20:34:17.937 回答
0

即使顶部元素的子元素以任意方式混洗，这种转换也会产生想要的结果：

<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:variable name="vNumCustomers"
      select="count(/*/*) div 3"/>

 <xsl:template match="/*">
     <MyCustomers>
       <xsl:for-each select=
           "*[not(position() > $vNumCustomers)]">
         <xsl:variable name="vNum" select="position()"/>

         <Customer>
          <Name>
            <xsl:value-of select=
             "concat(/*/*[name()=concat('firstname',$vNum)],
                     ' ',
                     /*/*[name()=concat('lastname',$vNum)]
                     )
             "/>
          </Name>
          <Sex>
            <xsl:value-of select=
             "/*/*[name()=concat('sex',$vNum)]
             "/>
          </Sex>
         </Customer>
       </xsl:for-each>
     </MyCustomers>
 </xsl:template>
</xsl:stylesheet>

当应用于此 XML 文档时（对提供的文档进行任意重新洗牌）：

<customers>
    <sex1>M</sex1>
    <lastname2>Woods</lastname2>
    <lastname1>Killer</lastname1>
    <sex2>M</sex2>
    <firstname3>Jennifer</firstname3>
    <firstname2>Frank</firstname2>
    <lastname3>Lee</lastname3>
    <firstname1>Sean</firstname1>
    <sex3>F</sex3>
</customers>

产生了想要的正确结果：

<MyCustomers>
   <Customer>
      <Name>Sean Killer</Name>
      <Sex>M</Sex>
   </Customer>
   <Customer>
      <Name>Frank Woods</Name>
      <Sex>M</Sex>
   </Customer>
   <Customer>
      <Name>Jennifer Lee</Name>
      <Sex>F</Sex>
   </Customer>
</MyCustomers>

说明：

  1. 我们计算提供数据的客户数量。变量$vNumCustomers保存这些数据。

  2. 对于每个客户{i} (i = 1 to $vNumCustomers)，我们创建相应的<Customer{i}>元素。为了避免使用递归，我们这里使用Piez 方法。

于 2011-07-12T13:02:15.887 回答
0

这是一个 XSLT 2.0 样式表，它将获得您正在寻找的输出，即使它们没有按顺序排列。它还按“名字”元素名称排序。

示例 XML 输入（混合以显示不同的顺序）：

<customers>
  <lastname1>Killer</lastname1>
  <sex3>F</sex3>
  <firstname2>Frank</firstname2>
  <firstname1>Sean</firstname1>
  <lastname2>Woods</lastname2>
  <sex2>M</sex2>
  <firstname3>Jennifer</firstname3>
  <sex1>M</sex1>
  <lastname3>Lee</lastname3>
</customers>

XSLT 2.0 样式表（使用 Saxon-HE 9.3 测试）：

<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:choose>
      <xsl:when test="name()[starts-with(.,'firstname')]">
        <xsl:variable name="suffix" select="substring(name(),10)"></xsl:variable>
        <xsl:message><xsl:value-of select="$suffix"/></xsl:message>
        <customer>
          <Name>
            <xsl:value-of select="concat(.,' ',/customers/*[starts-with(name(),'lastname')][ends-with(name(),$suffix)])"/>  
          </Name>
          <Sex>
            <xsl:value-of select="/customers/*[starts-with(name(),'sex')][ends-with(name(),$suffix)]"/>
          </Sex>
        </customer>
      </xsl:when>
      <xsl:when test="name()='customers'">
        <MyCustomers>
          <xsl:apply-templates>
            <xsl:sort select="name()[starts-with(.,'firstname')]"></xsl:sort>
          </xsl:apply-templates>
        </MyCustomers>
      </xsl:when>
      <xsl:otherwise>
        <xsl:apply-templates select="node()|@*"/>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>

</xsl:stylesheet>

输出：

<MyCustomers>
   <customer>
      <Name>Sean Killer</Name>
      <Sex>M</Sex>
   </customer>
   <customer>
      <Name>Frank Woods</Name>
      <Sex>M</Sex>
   </customer>
   <customer>
      <Name>Jennifer Lee</Name>
      <Sex>F</Sex>
   </customer>
</MyCustomers>
于 2011-07-11T21:09:21.977 回答