Question

我正在尝试将xml文件转换为xslt

我的xml文件是这样的，

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="testxsl.xsl"?>
<doc>
<test>
    <field name="id">c8b102</field>
    <field name="created_at">2013-11-14T00:00:00Z</field>
    <field name="userid">userfirst</field>
    <field name="screenname">userfirst</field>
    <field name="status">open</field>
    <field name="contents">testing content</field>
    <field name="sentiment">positive</field>
    <field name="location">placename</field>
    <field name="source">local</field>
    <field name="url">test</field>
    <field name="keyword">test</field>
</test>
<test>
    <field name="id">c8b102</field>
    <field name="created_at">2013-11-14T00:00:00Z</field>
    <field name="userid">usersec</field>
    <field name="screenname">usersec</field>
    <field name="status">close</field>
    <field name="contents">test sec</field>
    <field name="sentiment">positive</field>
    <field name="location">place</field>
    <field name="source">local</field>
    <field name="url">test</field>
    <field name="keyword">test</field>
  </test>
  </doc>

我已经为转换编写了一个xsl代码（testxsl.xsl）

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<body>
<h2>Complaints</h2>
<table border="1">
<tr bgcolor="#9acd32">
  <th>Created Date</th>
  <th>UserID</th>
  <th>Screenname</th>
  <th>Status</th>
  <th>Contents</th>
  <th>Source</th>
  <th>URL</th>
</tr>
<xsl:for-each select="doc/test">
<tr>
  <td>
  <xsl:value-of select="date" />
  </td>
  <td>
  <xsl:value-of select="str[@name = 'userid']" />
  </td>
  <td>
  <xsl:value-of select="str[@name='screenname']" />
  </td>
  <td>
  <xsl:value-of select="str[@name='status']" />
  </td>
  <td>
  <xsl:value-of select="str[@name='contents']" />
  </td>
  <td>
  <xsl:value-of select="str[@name='source']" />
  </td>
  <td>
  <xsl:value-of select="str[@name='url']" />
  </td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>

但我只在带边框的输出中获得标题部分，如下所示，

Created Date  UserID    Screenname  Status  Contents    Source  URL

我不知道是什么问题，我的xsl文件似乎是正确的

Answer 1

输入XML中没有名为date的元素。同样，test中的元素不会被称为str，而是field。

您的XSLT代码当然必须与输入XML中的元素相对应。

使用以下样式表：

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<body>
<h2>Complaints</h2>
<table border="1">
<tr bgcolor="#9acd32">
  <th>Created Date</th>
  <th>UserID</th>
  <th>Screenname</th>
  <th>Status</th>
  <th>Contents</th>
  <th>Source</th>
  <th>URL</th>
</tr>
<xsl:for-each select="doc/test">
<tr>
  <td>
  <xsl:value-of select="field[@name='created_at']" />
  </td>
  <td>
  <xsl:value-of select="field[@name = 'userid']" />
  </td>
  <td>
  <xsl:value-of select="field[@name='screenname']" />
  </td>
  <td>
  <xsl:value-of select="field[@name='status']" />
  </td>
  <td>
  <xsl:value-of select="field[@name='contents']" />
  </td>
  <td>
  <xsl:value-of select="field[@name='source']" />
  </td>
  <td>
  <xsl:value-of select="field[@name='url']" />
  </td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>

<强>输出

<html>
   <body>
      <h2>Complaints</h2>
      <table border="1">
         <tr bgcolor="#9acd32">
            <th>Created Date</th>
            <th>UserID</th>
            <th>Screenname</th>
            <th>Status</th>
            <th>Contents</th>
            <th>Source</th>
            <th>URL</th>
         </tr>
         <tr>
           <td>2013-11-14T00:00:00Z</td>
           <td>userfirst</td>
           <td>userfirst</td>
           <td>open</td>
           <td>testing content</td>
           <td>local</td>
           <td>test</td>
        </tr>
        <!--...-->
      </table>
   </body>
</html>

呈现HTML

enter image description here

如何将XML转换为XSL

1 个答案: