到目前为止,我一直在使用forloops来构建包含记录的表格,但是我无法完成表格,因为我无法找到将剩余数据插入第三列的方法在另一个节点内。
我得到了什么
我想要什么
以下代码
XML
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="grades.xsl"?>
<school>
<subject Id="3311">
<className>English</className>
<studentList>
<student id="1001">Lisa Simpson</student>
<student id="1002">Barney Rubble</student>
<student id="1003">Donald Duck</student>
</studentList>
<classwork>
<assignment name="Final Exam">
<mark studId="1001">38</mark>
<mark studId="1002">21</mark>
<mark studId="1003">20.5</mark>
</assignment>
</classwork>
</subject>
</school>
XSL(迄今为止)
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<body>
<h2>My CD Collection</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th style="text-align:left">Student</th>
<th style="text-align:left">ID</th>
<th style="text-align:left">Mark</th>
</tr>
<xsl:for-each select="//student">
<tr>
<td><xsl:value-of select="."/></td>
<td><xsl:value-of select="./@id"/></td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
答案 0 :(得分:0)
您可以尝试以下样式表:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">
<xsl:strip-space elements="*"/>
<xsl:output indent="yes"/>
<!-- key for the target node for the mark column -->
<xsl:key name="kID" match="mark" use="@studId"/>
<xsl:template match="/">
<table>
<!-- set table headers -->
<tr>
<th>Student</th>
<th>ID</th>
<th>Mark</th>
</tr>
<!-- Loop through each target node -->
<xsl:for-each select="school/subject/studentList/student">
<tr>
<td><xsl:value-of select="."/></td>
<td><xsl:value-of select="@id"/></td>
<!-- get the value using keys -->
<td><xsl:value-of select="key('kID', @id)"/></td>
</tr>
</xsl:for-each>
</table>
</xsl:template>
</xsl:stylesheet>