在特殊的 xml 结构中选择具有 name 属性 name 的元素
问题描述：
下面是我的 xml 文档的结构.我只想先取每个节点的值 <attribute name="a"> .谷歌搜索没有显示任何有效的解决方案.
below is the structure of my xml document. I just want to first take the value of every node <attribute name="a"> then make a comparison of it with a given value. However I don't how to locate <attribute name="a"> of each node using xml selectnodes in c#. Google searches don't show any working solutions.
<nodes>
<node name = "node1">
<attribute name="a">This is node1 a</attribute>
<attribute name="b">This is node1 b</attribute>
</node>
<node name = "node2">
<attribute name="a">This is node2 a</attribute>
<attribute name="b">This is node2 b</attribute>
</node>
...
</nodes>
答
假设您问题中的 XML 标记代表您的整个文档，您可以这样做:
Assuming the XML markup in your question represents your whole document, you can do:
XmlNodeList attrElements
= yourDocument.SelectNodes("/nodes/node/attribute[@name='a']");