Question

我是xslt的新手，我很难弄清楚如何转换下面的xml。我知道我必须遍历行，但我不知道如何将列名转换为元素。任何帮助将不胜感激。

这是从webservice收到的xml

<?xml version="1.0" encoding="utf-8" ?> 
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <soap:Body>
   <xmlns="http://www.someservice.com/webservices/">
    <GetResultsResult>
    <Columns>
      <WSColumn>
         <Name>triptype</Name> 
      </WSColumn>
      <WSColumn>
         <Name>description</Name> 
      </WSColumn>
      <WSColumn>
         <Name>id</Name> 
      </WSColumn>
    </Columns>
    <Rows>
      <WSRow>
        <Cell>
          <anyType xsi:type="xsd:string">vacation</anyType> 
          <anyType xsi:type="xsd:string">Trip to Bahamas</anyType> 
          <anyType xsi:type="xsd:int">89</anyType> 
        </Cell>
       </WSRow>
       <WSRow>
        <Cell>
          <anyType xsi:type="xsd:string">vacation</anyType> 
          <anyType xsi:type="xsd:string">Trip to California</anyType> 
          <anyType xsi:type="xsd:int">75</anyType> 
         </Cell>
        </WSRow>
        <WSRow>
         <Cell>
           <anyType xsi:type="xsd:string">business</anyType> 
           <anyType xsi:type="xsd:string">Trip to Chicago</anyType> 
           <anyType xsi:type="xsd:int">82</anyType> 
          </Cell>
         </WSRow>
       </Rows>
       <HasErrors>false</HasErrors> 
       <ErrorMessage /> 
        </GetResultsResult>
      </GetResultsResponse>
    </soap:Body>
</soap:Envelope>

这是转型后的理想结果

<Trips>
   <Trip>
     <triptype>vacation</triptype>
     <description>Trip to Bahamas</description>
     <id>89</id> 
    </Trip>
    <Trip>
      <triptype>vacation</triptype>
      <description>Trip to California</description>
      <id>75</id>
     </Trip>
     <Trip>
       <triptype>business</triptype>
       <description>Trip to Bahamas</description>
       <id>82</id>
     </Trip>
</Trips>

提前谢谢!!! 马塞洛

Answer 1

这应该这样做：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:gr="http://www.someservice.com/webservices/"
                exclude-result-prefixes="gr">

  <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
  <xsl:key name="kColumn" match="gr:WSColumn/gr:Name"
           use="count(../preceding-sibling::gr:WSColumn) + 1" />

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

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

  <xsl:template match="gr:WSRow/gr:Cell">
    <Trip>
      <xsl:apply-templates select="*"/>
    </Trip>
  </xsl:template>

  <xsl:template match="gr:Cell/*">
    <xsl:element name="{key('kColumn', position())}">
      <xsl:value-of select="."/>
    </xsl:element>
  </xsl:template>
</xsl:stylesheet>

在您的示例输入上运行时（在修复了损坏的GetResultsResponse标记之后），这会产生：

<Trips>
  <Trip>
    <triptype>vacation</triptype>
    <description>Trip to Bahamas</description>
    <id>89</id>
  </Trip>
  <Trip>
    <triptype>vacation</triptype>
    <description>Trip to California</description>
    <id>75</id>
  </Trip>
  <Trip>
    <triptype>business</triptype>
    <description>Trip to Chicago</description>
    <id>82</id>
  </Trip>
</Trips>

如何使用xslt在xml中转换数据集

1 个答案: