如何创建C#类模型以获取给定格式的XML序列化请求?

时间:2011-07-13 14:42:30

标签: c# xml-serialization

我正在设计我的班级模型。类模型的序列化消息需要采用以下格式:

<?xml version="1.0" encoding="UTF-8" ?>
<Request>
  <Name>TesterScript</Name>
  <ID>CD_20110628133820576</ID>
  <Type>
    <ItemId>191_20110628T133821</ItemId>
    <ShopId>MyBCShop</ShopId>
    <MessageXml>
    <ChildMessage>
      This is my message
    </ChildMessage>
    </MessageXml>
  </Type>
  <SentTime>2011-06-30T15:27:06-07:00</SentTime>
</Request>

我如何设计课程? 另外,将建议的类模型序列化为上述XML消息的最佳方法是什么?我正在考虑使用:

// Serialize the request
XmlSerializer xs = new XmlSerializer(typeof(Request));
StringWriter sw = new StringWriter();
xs.Serialize(sw, dispatchRequest);
string xml = sw.ToString();
return new xml;

这是最合适的方式吗?

2 个答案:

答案 0 :(得分:3)

如果您已经拥有架构,我只需使用xsd.exe来生成该类。它已被标记为可序列化,除了调用基础XmlSerializer之外你不会做任何事情(正如你在第二个片段中所做的那样,或多或少。)

答案 1 :(得分:1)

我希望我的解决方案很有用......

xml内容(我认为你的帖子有问题......)

<?xml version="1.0" encoding="utf-8" ?>
<Request>
  <Name>TesterScript</Name>
  <ID>CD_20110628133820576</ID>
  <Type>
    <ItemId>191_20110628T133821</ItemId>
    <ShopId>BARCGB2L</ShopId>
    <MessageXml>
      <ChildMessage>
        This is my message
      </ChildMessage>
    </MessageXml>
  </Type>
  <SentTime>2011-06-30T15:27:06-07:00</SentTime>
</Request>

班级:

[XmlRoot("Request")]
public class SampleClass
{
    public string Name { get; set; }
    public string ID { get; set; }
    [XmlElement("Type")]
    public SubClass SC { get; set; }
    public string SentTime { get; set; }

    public class SubClass
    {
        public string ItemId { get; set; }
        public string ShopId { get; set; }
        [XmlElement("MessageXml")]
        public Sub2Class SC2 { get; set; }

        public class Sub2Class
        {
            public string ChildMessage { get; set; }
        }
    }
}

反序列化方法:

public static T DeserializeForXml<T>(string filePath)
    {
        XmlSerializer selializer = new XmlSerializer(typeof(T));
        using (Stream fs = new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.Read))
        {
            return (T)selializer.Deserialize(fs);
        }
    }

如何使用?

SampleClass sc = Utility.DeserializeForXml<SampleClass>("test.xml");