如何从类型T的对象获取xml节点？

Question

我想将类型T的对象转换为xml节点，因为我想在wpf控件中显示xml节点。下面是linqpad片段，我一起工作：

[Serializable] 
public class test { 
public int int1 { get ; set ;} 
public int int2 { get ; set ;} 
public string str1 { get ; set ;} 
} 

void Main() 
{ 
test t1 = new test() ; 
t1.int1 = 12 ; 
t1.int2 = 23 ; 
t1.str1 = "hello" ; 


System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(t1.GetType()); 
StringWriter sww = new StringWriter(); 
XmlWriter writer = XmlWriter.Create(sww); 
x.Serialize(writer, t1); 
var xml = sww.XmlSerializeToXElement(); 
xml.Dump() ; 

} 

我没有得到预期的效果，相反，我得到这个：

<StringWriter xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
    <NewLine> 
</NewLine> 
</StringWriter> 

Answer 1

如果你想获得一个XElement，将xml.Root是你想要什么：

System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(t1.GetType()); 
StringWriter sww = new StringWriter(); 
XmlWriter writer = XmlWriter.Create(sww); 
x.Serialize(writer, t1); 
var xml = XDocument.Parse(sww.ToString()); 
Console.WriteLine(xml.Root); 

输出：

<test xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
    <int1>12</int1> 
    <int2>23</int2> 
    <str1>hello</str1> 
</test>