请教一个xsl变换的问题,在线等,马上给分
xml文件:
<?xml version="1.0" encoding="gb2312" ?>
<menu>
<project>
<name>Project1</name>
<item>
<name>Name_1</name>
<href>Href_1</href>
</item>
<item>
<name>Name_2</name>
<href>Href_2</href>
</item>
<item>
<name>Name_3</name>
<href>Href_3</href>
</item>
</project>
</menu>
xsl文件:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/menu">
<xsl:apply-templates select="project" />
</xsl:template>
<xsl:template match="project">
<table width="100%" border="1">
<tr>
<td>
<xsl:value-of select="name" />
</td>
</tr>
<xsl:for-each select="item">
<tr>
<td>
<a href=""><xsl:value-of select="name" /></a>
</td>
</tr>
</xsl:for-each>
</table>
</xsl:template>
</xsl:stylesheet>
问题就在超链接,怎样将<href>元素的内容指定为超链的href?