急！xsl格式化XML成表格，超链接里如何取XML的属性？
---test.xml----
<?xml version="1.0" encoding="gb2312" ?>
<?xml-stylesheet type="text/xsl" href="test.xsl"?>
<report>
<operator code="844" title="陈宇宁">
<DataLevel content="Data" index="1">
<Data content="yuwen" value="90.50" index="1" />
<Data content="shuxue" value="1017.96" index="2" />
</DataLevel>
</operator>

<operator code="341" title="方晓萍">
<DataLevel content="Data" index="1">
<Data content="yuwen" value="85.00" index="1" />
<Data content="shuxue" value="98.50" index="2" />
</DataLevel>
</operator>
</report>
-----test.xsl----
<?xml version="1.0" encoding="gb2312"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:template match="/">
<table cellpadding='0' cellspacing='1' border='1' widtd='700px' align='center'>
<tr align='center'><td>姓名</td><td>语文</td><td>数学</td></tr>
<xsl:for-each select="report/operator"><tr>
<td ><a href='detail.asp?code=如何把xml里的code取出来放在这里' target='_blank'><xsl:value-of select="./@title"/></a></td>
<td><xsl:value-of select="./DataLevel/Data[@content='yuwen']/@value"/></td>
<td><xsl:value-of select="./DataLevel/Data[@content='shuxue']/@value"/></td>
</tr>
</xsl:for-each>
</table>
</xsl:template>
</xsl:stylesheet>
－－－－－问题－－－
想实现：在点击姓名时超链接参数带上code编号，比如点'陈宇宁'打开链接是：'detail.asp?code=844'
请帮帮忙，谢谢……

[解决办法]
<a href= 'detail.asp?code={@code}' target= '_blank ' >