XDocument的内容是下面的XML。
我想获得一个List(),请参阅此消息的末尾。
<myXml>
<myDatas code="01">
<myVar name="myvar" value="A" />
<myData name="A" value="A1" />
<myData name="B" value="B1" />
</myDatas>
<myDatas code="02">
<myVar name="myvar" value="B" />
<myData name="A" value="A2" />
<myData name="D" value="D2" />
</myDatas>
</myXml>
public class MyData
{
public string MainCode { get; set; }
public string Code { get; set; }
public string Value { get; set; }
}
我想要一个List(),这个内容应该是这样的:
new MyData { MainCode = "01"; Code = "A"; Value = "A1" };
new MyData { MainCode = "01"; Code = "B"; Value = "B1" };
new MyData { MainCode = "02"; Code = "A"; Value = "A2" };
new MyData { MainCode = "02"; Code = "D"; Value = "D2" };
答案 0 :(得分:6)
当然 - 所以你需要这样的东西:
var query = from datas in doc.Root.Elements("myDatas")
let code = (string) datas.Attribute("code")
from data in datas.Elements("myData")
select new MyData {
MainCode = code,
Code = (string) data.Attribute("name"),
Value = (string) data.Attribute("value"),
};
var list = query.ToList();
请注意多个from
子句以展平结果。
另一种选择是找到所有“leaf”元素并从父级中获取代码部分:
var query = from data in doc.Descendants("myData")
select new MyData {
MainCode = (string) data.Parent.Attribute("code"),
Code = (string) data.Attribute("name"),
Value = (string) data.Attribute("value"),
};
var list = query.ToList();
编辑:如果您的文档使用名称空间,那也很容易:
XNamespace ns = "http://the-uri-of-the-namespace";
var query = from data in doc.Descendants(ns + "myData")
...
这使用XName operator +(XNamespace, string)
重载运算符。