XML을 목록으로 직렬화 해제 할 수 있습니까?

XML을 목록으로 직렬화 해제 할 수 있습니까??

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다음과 같은 XML이 제공됩니다.

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

그리고 다음 수업 :

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

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

XmlSerializerXML을 deserialize하는 데 사용할 수 List<User>있습니까? 그렇다면 어떤 유형의 추가 속성을 사용해야합니까, 또는 XmlSerializer인스턴스 를 구성하는 데 어떤 추가 매개 변수를 사용해야 합니까?

User[]조금 덜 바람직한 경우 배열 ( )을 사용할 수 있습니다.

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목록을 간단하게 캡슐화 할 수 있습니다 .

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);
    }
}

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필요한 대문자와 일치하도록 User클래스 를 장식하면 XmlType:

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

그런 다음 XmlRootAttribute온 XmlSerializerctor에 원하는 루트를 제공하고 목록 <>에 직접 읽기를 허용 할 수 있습니다 :

    // 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);
    }

크레딧 : YK1의 답변 을 바탕으로 합니다.

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

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

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

---

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.

Take a look (it works for me):

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.

        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);

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

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

    public UserHolder()
    {
    }

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

Deserializing code,

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

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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.

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

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

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How about

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.

참고URL : https://stackoverflow.com/questions/608110/is-it-possible-to-deserialize-xml-into-listt