T＆GT;名单，LT的XML序列化; - XML根 [英] XML Serialization of List<T> - XML Root

问题描述

＃2（.NET 2.0）第一个问题：

First question on Stackoverflow (.Net 2.0):

所以，我想用下面的返回列表的XML：

So I am trying to return an XML of a List with the following:

public XmlDocument GetEntityXml()
    {        
        StringWriter stringWriter = new StringWriter();
        XmlDocument xmlDoc = new XmlDocument();            

        XmlTextWriter xmlWriter = new XmlTextWriter(stringWriter);

        XmlSerializer serializer = new XmlSerializer(typeof(List<T>));

        List<T> parameters = GetAll();

        serializer.Serialize(xmlWriter, parameters);

        string xmlResult = stringWriter.ToString();

        xmlDoc.LoadXml(xmlResult);

        return xmlDoc;
    }

现在这个将用于我已经定义了多个实体。

Now this will be used for multiple Entities I have already defined.

说我想获得名单，LT的XML;猫＆GT;

该XML会是这样的：

<ArrayOfCat>
  <Cat>
    <Name>Tom</Name>
    <Age>2</Age>
  </Cat>
  <Cat>
    <Name>Bob</Name>
    <Age>3</Age>
  </Cat>
</ArrayOfCat>

有没有为我获得这些实体时，得到同样的根所有的时间？一个办法

Is there a way for me to get the same Root all the time when getting these Entities?

例如：

<Entity>
  <Cat>
    <Name>Tom</Name>
    <Age>2</Age>
  </Cat>
  <Cat>
    <Name>Bob</Name>
    <Age>3</Age>
  </Cat>
</Entity>

另外请注意，我不打算反序列化XML回列表＆LT; CAT和GT;

推荐答案

有一个更简单的方法：

public XmlDocument GetEntityXml<T>()
{
    XmlDocument xmlDoc = new XmlDocument();
    XPathNavigator nav = xmlDoc.CreateNavigator();
    using (XmlWriter writer = nav.AppendChild())
    {
        XmlSerializer ser = new XmlSerializer(typeof(List<T>), new XmlRootAttribute("TheRootElementName"));
        ser.Serialize(writer, parameters);
    }
    return xmlDoc;
}

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