C# 是否可以将 XML 反序列化为 List<T>？

Is it possible to deserialize XML into List<T>?

提问by Daniel Schaffer

Given the following XML:

给定以下 XML：

<?xml version="1.0"?>
<user_list>
   <user>
      <id>1</id>
      <name>Joe</name>
   </user>
   <user>
      <id>2</id>
      <name>John</name>
   </user>
</user_list>

And the following class:

以及以下课程：

public class User {
   [XmlElement("id")]
   public Int32 Id { get; set; }

   [XmlElement("name")]
   public String Name { get; set; }
}

Is it possible to use XmlSerializerto deserialize the xml into a List<User>? If so, what type of additional attributes will I need to use, or what additional parameters do I need to use to construct the XmlSerializerinstance?

是否可以用于XmlSerializer将 xml 反序列化为List<User>? 如果是这样，我需要使用什么类型的附加属性，或者我需要使用哪些附加参数来构造XmlSerializer实例？

An array ( User[]) would be acceptable, if a bit less preferable.

数组 ( User[]) 是可以接受的，如果不太可取的话。

采纳答案by Marc Gravell

You can encapsulatethe list trivially:

您可以简单地封装列表：

using System;
using System.Collections.Generic;
using System.Xml.Serialization;

[XmlRoot("user_list")]
public class UserList
{
    public UserList() {Items = new List<User>();}
    [XmlElement("user")]
    public List<User> Items {get;set;}
}
public class User
{
    [XmlElement("id")]
    public Int32 Id { get; set; }

    [XmlElement("name")]
    public String Name { get; set; }
}

static class Program
{
    static void Main()
    {
        XmlSerializer ser= new XmlSerializer(typeof(UserList));
        UserList list = new UserList();
        list.Items.Add(new User { Id = 1, Name = "abc"});
        list.Items.Add(new User { Id = 2, Name = "def"});
        list.Items.Add(new User { Id = 3, Name = "ghi"});
        ser.Serialize(Console.Out, list);
    }
}

回答by JaredPar

Not sure about List<T> but Arrays are certainly do-able. And a little bit of magic makes it really easy to get to a List again.

不确定 List<T> 但数组肯定是可行的。一点点魔法使再次访问列表变得非常容易。

public class UserHolder {
   [XmlElement("list")]
   public User[] Users { get; set; }

   [XmlIgnore]
   public List<User> UserList { get { return new List<User>(Users); } }
}

回答by Coincoin

Yes, it will serialize and deserialize a List<>. Just make sure you use the [XmlArray] attribute if in doubt.

是的，它将序列化和反序列化 List<>。如果有疑问，请确保使用 [XmlArray] 属性。

[Serializable]
public class A
{
    [XmlArray]
    public List<string> strings;
}

This works with both Serialize() and Deserialize().

这适用于 Serialize() 和 Deserialize()。

回答by Nemo

Yes, it does deserialize to List<>. No need to keep it in an array and wrap/encapsulate it in a list.

是的，它确实反序列化为 List<>。无需将其保存在数组中并将其包装/封装在列表中。

public class UserHolder
{
    private List<User> users = null;

    public UserHolder()
    {
    }

    [XmlElement("user")]
    public List<User> Users
    {
        get { return users; }
        set { users = value; }
    }
}

Deserializing code,

反序列化代码，

XmlSerializer xs = new XmlSerializer(typeof(UserHolder));
UserHolder uh = (UserHolder)xs.Deserialize(new StringReader(str));

回答by tudor.iliescu

I think I have found a better way. You don't have to put attributes into your classes. I've made two methods for serialization and deserialization which take generic list as parameter.

我想我找到了更好的方法。您不必将属性放入类中。我已经制作了两种以通用列表作为参数的序列化和反序列化方法。

Take a look (it works for me):

看看（它对我有用）：

private void SerializeParams<T>(XDocument doc, List<T> paramList)
    {
        System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(paramList.GetType());

        System.Xml.XmlWriter writer = doc.CreateWriter();

        serializer.Serialize(writer, paramList);

        writer.Close();           
    }

private List<T> DeserializeParams<T>(XDocument doc)
    {
        System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(List<T>));

        System.Xml.XmlReader reader = doc.CreateReader();

        List<T> result = (List<T>)serializer.Deserialize(reader);
        reader.Close();

        return result;
    }

So you can serialize whatever list you want! You don't need to specify the list type every time.

所以你可以序列化你想要的任何列表！您不需要每次都指定列表类型。

        List<AssemblyBO> list = new List<AssemblyBO>();
        list.Add(new AssemblyBO());
        list.Add(new AssemblyBO() { DisplayName = "Try", Identifier = "243242" });
        XDocument doc = new XDocument();
        SerializeParams<T>(doc, list);
        List<AssemblyBO> newList = DeserializeParams<AssemblyBO>(doc);

回答by PRJ

How about

怎么样

XmlSerializer xs = new XmlSerializer(typeof(user[]));
using (Stream ins = File.Open(@"c:\some.xml", FileMode.Open))
foreach (user o in (user[])xs.Deserialize(ins))
   userList.Add(o);    

Not particularly fancy but it should work.

不是特别花哨，但它应该工作。

回答by richaux

If you decorate the Userclass with the XmlTypeto match the required capitalization:

如果您使用 装饰User类XmlType以匹配所需的大写：

[XmlType("user")]
public class User
{
   ...
}

Then the XmlRootAttributeon the XmlSerializerctor can provide the desired root and allow direct reading into List<>:

然后ctorXmlRootAttribute上的XmlSerializer可以提供所需的根并允许直接读入 List<>：

    // e.g. my test to create a file
    using (var writer = new FileStream("users.xml", FileMode.Create))
    {
        XmlSerializer ser = new XmlSerializer(typeof(List<User>),  
            new XmlRootAttribute("user_list"));
        List<User> list = new List<User>();
        list.Add(new User { Id = 1, Name = "Joe" });
        list.Add(new User { Id = 2, Name = "John" });
        list.Add(new User { Id = 3, Name = "June" });
        ser.Serialize(writer, list);
    }

...

...

    // read file
    List<User> users;
    using (var reader = new StreamReader("users.xml"))
    {
        XmlSerializer deserializer = new XmlSerializer(typeof(List<User>),  
            new XmlRootAttribute("user_list"));
        users = (List<User>)deserializer.Deserialize(reader);
    }

Credit: based on answerfrom YK1.

信用：基于答案来自YK1。