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我需要序列化一些容器类，其HasValue属性被评估为真。

我不需要在序列化之前从容器列表中删除无效元素。序列化程序应该能够确定哪些对象需要序列化。我想自定义序列化程序可能适合我的需要，但我不知道如何解决这个问题。任何其他解决方案/最佳实践将不胜感激。

这里是我的课

public static class ContainerFactory
{
         public static Container Create()
         {
             var container = new Container();
             container.Persons.AddRange(new[]
                 {
                     new Person
                         {
                             FirstName = "Thomas"
                         },
                      new Person
                         {
                             FirstName = "Andrew",
                             LastName = "Martin",
                              Vehicles = new Vehicles
                                 {
                                    new Vehicle { Hsn = "65976GHR", Tsn = "HUZUKL"}
                                 }
                         },
                          new Person
                         {
                             FirstName = "Arnold",
                             LastName = "Beckmann",
                             Vehicles = new Vehicles
                                 {
                                    new Vehicle { Hsn = "345XXXHZ"},
                                    new Vehicle { Hsn = "659JUKI", Tsn = "787999HGF"}
                                 }
                         }

                 });
             return container;
         }
}

[Serializable]
public class Container
{
    public Container()
    {
        Persons = new Persons();
    }

    public Persons Persons { get; set; }

    public void Serialize()
    {
        var serializer = new XmlSerializer(typeof (Container));
        var streamWriter = new StreamWriter(@"C:\container.xml", false);
        serializer.Serialize(streamWriter, this);
    }
}

public class Persons: List<Person>
{
}

public class Vehicles: List<Vehicle>
{
    public Vehicles()
    {
    }

    public Vehicles(IEnumerable<Vehicle> vehicles):base(vehicles)
    {
    }
}

[Serializable]
public class Person : IHasValue
{
    public Person()
    {
        this.Vehicles = new Vehicles();
        this.Id = Guid.NewGuid().ToString();
    }

    public string Id { get; set; }
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public Vehicles Vehicles { get; set; }

    public bool HasValue
    {
        get { return !string.IsNullOrEmpty(this.FirstName) && !string.IsNullOrEmpty(this.LastName); }
    }
}

public interface IHasValue
{
    bool HasValue { get;}
}

public class Vehicle: IHasValue
{
    public string Hsn { get; set; }
    public string Tsn { get; set; }

    public bool HasValue
    {
        get { return !string.IsNullOrEmpty(Hsn) && !string.IsNullOrEmpty(Tsn); }
    }
}

//Using the .NET XMLSerializer to test my container
Container container = ContainerFactory.Create();
container.Serialize();
Console.WriteLine("Press any Key to continue...");
Console.ReadLine();

输出

<?xml version="1.0" encoding="utf-8"?>
<Container xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Persons>
    <Person>
      <Id>76cdcc18-b256-40fe-b813-cd6c60e682ca</Id>
      <FirstName>Thomas</FirstName>
      <Vehicles />
    </Person>
    <Person>
      <Id>26623bf9-d799-44d2-bc1a-7ec91292d1cd</Id>
      <FirstName>Andrew</FirstName>
      <LastName>Martin</LastName>
      <Vehicles>
        <Vehicle>
          <Hsn>65976GHR</Hsn>
          <Tsn>HUZUKL</Tsn>
        </Vehicle>
      </Vehicles>
    </Person>
    <Person>
      <Id>f645cde1-10c8-4df5-81df-9b9db7712ec3</Id>
      <FirstName>Arnold</FirstName>
      <LastName>Beckmann</LastName>
      <Vehicles>
        <Vehicle>
          <Hsn>345XXXHZ</Hsn>
        </Vehicle>
        <Vehicle>
          <Hsn>659JUKI</Hsn>
          <Tsn>787999HGF</Tsn>
        </Vehicle>
      </Vehicles>
    </Person>
  </Persons>

我怎样才能实现仅针对哪些车辆/人员进行序列化的目标HasValue == true？

4

2 回答 2

1

Persons : List < Person > 也应该是可序列化的。

public void GetObjectData( SerializationInfo info, StreamingContext context )    
{
    foreach (Person person in this)    
    {
        if(person.HasValue)
        {
            info.AddValue("Firsname", person, typeof(Person));

            info.AddValue (....);

            ..............

        }
    }
}
于 2014-01-08T15:13:34.323 回答
1

你需要使用：

  1. ISerializable 接口- 您想要序列化的每个类以及您想要在不同节点中的每个属性都需要使用适当的属性。看这里的例子
  2. 当你序列化一个类时，检查是否HasValue == true
于 2013-09-03T20:36:01.443 回答