已知如下xml 文件，如何循环列表出来各级属性为name的值

[阿土伯土]

已知如下xml 文件，如何循环列表出来各级属性为name的值


<?xml version="1.0" encoding="gb2312"?>
<data>

<item name="a" value="1" />

<item name="b" value="2">
<sub name="xx" url="u1" />
</item>

<item name="c" value="3">
<sub name="cxx1" url="u1" />
<sub name="cxx2" url="u1" />
</item>

<item name="d" value="4" />

</data>

[patchclass]

<xsl:for-each select="//item">
<xsl:value-of select="@name"/>
</xsl:for-each>

[阿土伯土]

patchclass的更好，asp中已经解决

y.asp

<%
set objXML = server.createobject("microsoft.xmldom")
objxml.load(server.mappath("y.xml"))
if objxml.readyState = 4 then

Set Root = objxml.getElementsByTagName("item")
'Response.Write Root.length'菜单个数

For i=0 To Root.length-1'读取一级子菜单
Response.Write Root(i).getAttribute("name")'属性name

If Root(i).childNodes.length>0 Then
'response.write "拥有子节点" 'Root(II).getAttribute("sub")
ReadChildItem(Root(i))
end if

Response.Write "<br>"
Next

else
response.write "加载XML文件失败"
end if

sub ReadChildItem(ParentNode)'读取子节点函数

For j=0 To ParentNode.childNodes.length-1

'response.write "["&ParentNode.childNodes.length-1&"]"
response.write "["& ParentNode.childNodes(j).getAttribute("name") &"]"
'If ParentNode.childNodes(j).getAttribute("name") <> "-" Then
'If ParentNode.childNodes(j).childNodes.length=0 Then
'ParentNode.childNodes(j).getAttribute("name")
'If Not IsNull(ParentNode.childNodes(j).getAttribute("name")) then Response.Write ParentNode.childNodes(j).getAttribute("name")

next

end sub
%>

y.xml
<?xml version="1.0" encoding="gb2312"?>
<data>
<item name="a" value="1" />
<item name="b" value="2">
<sub name="xx" url="u1" />
</item>
<item name="c" value="3">
<sub name="cxx1" url="u1">sub1</sub>
<sub name="cxx1" url="u1">sub2</sub>
<sub name="cxx2" url="u1" />
</item>
<item name="d" value="4" />
</data>