解析xml文件,C# 获取所有节点的属性值
<?xml version="1.0" encoding="utf-8"?>
<Location>
<CountryRegion Code="1" Name="中国">
<State Code="11" Name="北京">
<City Code="1" Name="东城"/>
<City Code="2" Name="西城"/>
</State>
<State Code="12" Name="天津">
<City Code="1" Name="和平"/>
</State><
/CountryRegion><
/Location>收起
<CountryRegion Code="1" Name="中国">
<State Code="11" Name="北京">
<City Code="1" Name="东城"/>
<City Code="2" Name="西城"/>
</State>
<State Code="12" Name="天津">
<City Code="1" Name="和平"/>
</State><
/CountryRegion><
/Location>收起
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using System.Xml.Linq; XElement element = XElement.Load( @"e:\txt.xml" ); foreach ( var item in element.DescendantsAndSelf()) { if (item.Attributes().Count() > 0) { foreach ( var attr in item.Attributes()) { Console.WriteLine( "名称:{0};值:{1}" , attr.Name, attr.Value); } } } |