C# 如何将xml反序列化为对象
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/10518372/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to deserialize xml to object
提问by user829174
<StepList>
<Step>
<Name>Name1</Name>
<Desc>Desc1</Desc>
</Step>
<Step>
<Name>Name2</Name>
<Desc>Desc2</Desc>
</Step>
</StepList>
I have this XML,
How should i model the Class so i will be able to deserialize it using XmlSerializerobject?
我有这个 XML，我应该如何对类建模，以便能够使用XmlSerializer对象反序列化它？
采纳答案by dknaack
Your classes should look like this
你的课程应该是这样的
[XmlRoot("StepList")]
public class StepList
{
[XmlElement("Step")]
public List<Step> Steps { get; set; }
}
public class Step
{
[XmlElement("Name")]
public string Name { get; set; }
[XmlElement("Desc")]
public string Desc { get; set; }
}
Here is my testcode.
这是我的测试代码。
string testData = @"<StepList>
<Step>
<Name>Name1</Name>
<Desc>Desc1</Desc>
</Step>
<Step>
<Name>Name2</Name>
<Desc>Desc2</Desc>
</Step>
</StepList>";
XmlSerializer serializer = new XmlSerializer(typeof(StepList));
using (TextReader reader = new StringReader(testData))
{
StepList result = (StepList) serializer.Deserialize(reader);
}
If you want to read a text file you should load the file into a FileStream and deserialize this.
如果要读取文本文件，则应将文件加载到 FileStream 中并对其进行反序列化。
using (FileStream fileStream = new FileStream("<PathToYourFile>", FileMode.Open))
{
StepList result = (StepList) serializer.Deserialize(fileStream);
}
回答by Jay Jacobs
The comments above are correct. You're missing the decorators. If you want a generic deserializer you can use this.
上面的评论是正确的。你缺少装饰器。如果你想要一个通用的解串器，你可以使用它。
public static T DeserializeXMLFileToObject<T>(string XmlFilename)
{
T returnObject = default(T);
if (string.IsNullOrEmpty(XmlFilename)) return default(T);
try
{
StreamReader xmlStream = new StreamReader(XmlFilename);
XmlSerializer serializer = new XmlSerializer(typeof(T));
returnObject = (T)serializer.Deserialize(xmlStream);
}
catch (Exception ex)
{
ExceptionLogger.WriteExceptionToConsole(ex, DateTime.Now);
}
return returnObject;
}
Then you'd call it like this:
然后你会这样称呼它：
MyObjType MyObj = DeserializeXMLFileToObject<MyObjType>(FilePath);
