text.xsl文件内容为:
<?xml version="1.0" encoding="gb2312"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
 <xsl:template match="/">
  <title>培训中心成绩单</title>
  <h2 align="center">学生成绩单 </h2>
  <table border="1" cellpadding="0" align="center">
   <tr align="center" bgcolor="#dadada">
    <td>编号</td>
    <td>姓名</td>
    <td>性别</td>
    <td>生日</td>
    <td>成绩</td>
    <td>评价</td>
   </tr>
   
 <xsl:for-each select="/roster/student"> 
   <tr>
    <td>
    <xsl:value-of select="@ID"/>
    </td>
    <td>                     <xsl:value-of select="name"/>
    </td>
    <td>
    <xsl:value-of select="sex"/>
    </td>
    <td>
    <xsl:value-of select="birthday"/>
    </td>
    <td>
    <xsl:value-of select="score"/>
    </td>
    <td>
    <xsl:choose>
       <xsl:when test="score &lt; 60 ">不及格</xsl:when>
       <xsl:when test="score &lt; 80 ">一般</xsl:when>
       <xsl:when test="score &gt; 90 ">优秀</xsl:when>
       <xsl:otherwise>良好</xsl:otherwise>
    </xsl:choose>
    </td>
   </tr>
                                   </xsl:for-each>
  </table>
 </xsl:template>
</xsl:stylesheet>

结果只显示了ID为101,名字叫李华的第1个student,错误应该是第2处红色地方,如果改成]<xsl:for-each select="/roster/*">才会显示所有student,但roster下如果有除了student以外的节点也会混淆,另外改成//student,/roster//student都只会显示第1个student