[英]XDocument to List of object
XDocument的內容是下面的XML。
我想要一個List(),請參閱此消息的末尾。
<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; }
}
我想要一個List(),這個內容應該像這樣:
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" };
當然-所以您需要這樣的東西:
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();
請注意多個from
子句以使結果變平。
另一種選擇是只找到所有“葉子”元素並從父級獲取代碼部分:
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();
編輯:如果您的文檔使用名稱空間,那也很容易:
XNamespace ns = "http://the-uri-of-the-namespace";
var query = from data in doc.Descendants(ns + "myData")
...
這使用XName operator +(XNamespace, string)
重載運算符。
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