首页 \ 问答 \ 如何告诉DataContractJsonSerializer不包含“__type”属性(How do I tell DataContractJsonSerializer to not include the “__type” property)

如何告诉DataContractJsonSerializer不包含“__type”属性(How do I tell DataContractJsonSerializer to not include the “__type” property)

我需要将KnownType添加到下面的代码中,以便序列化成功。 当我这样做时,生成的JSON如下所示:

JSON form of Adult with 1 child: {"age":42,"name":"John","children":[{"__type":"
Child:#TestJson","age":4,"name":"Jane","fingers":10}]}

我怎么能不包含“__type”:“Child:#TestJson”? 我们在一些查询中返回了数百个这些元素,并且额外的文本会加起来。

完整代码:

using System;
using System.Collections.Generic;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Json;

namespace TestJson
{
    class Program
    {
        static void Main(string[] args)
        {
            Adult parent = new Adult {name = "John", age = 42};

            MemoryStream stream1 = new MemoryStream();
            DataContractJsonSerializer ser = new DataContractJsonSerializer(typeof(Adult));
            ser.WriteObject(stream1, parent);

            stream1.Position = 0;
            StreamReader sr = new StreamReader(stream1);
            Console.Write("JSON form of Adult with no children: ");
            Console.WriteLine(sr.ReadToEnd());


            Child child = new Child { name = "Jane", age = 4, fingers=10 };

            stream1 = new MemoryStream();
            ser = new DataContractJsonSerializer(typeof(Child));
            ser.WriteObject(stream1, child);

            stream1.Position = 0;
            sr = new StreamReader(stream1);
            Console.Write("JSON form of Child with no parent: ");
            Console.WriteLine(sr.ReadToEnd());


            // now connect the two
            parent.children.Add(child);

            stream1 = new MemoryStream();
            ser = new DataContractJsonSerializer(typeof(Adult));
            ser.WriteObject(stream1, parent);

            stream1.Position = 0;
            sr = new StreamReader(stream1);
            Console.Write("JSON form of Adult with 1 child: ");
            Console.WriteLine(sr.ReadToEnd());
        }
    }

    [DataContract]
    [KnownType(typeof(Adult))]
    [KnownType(typeof(Child))]
    class Person
    {
        [DataMember]
        internal string name;

        [DataMember]
        internal int age;
    }

    [DataContract]
    class Adult : Person
    {
        [DataMember] 
        internal List<Person> children = new List<Person>();
    }

    [DataContract]
    class Child : Person
    {
        [DataMember]
        internal int fingers;
    }
}

I need to add KnownType to the below code for it to serialize successfully. When I do, the generated JSON is as follows:

JSON form of Adult with 1 child: {"age":42,"name":"John","children":[{"__type":"
Child:#TestJson","age":4,"name":"Jane","fingers":10}]}

How do I have it not include the "__type":"Child:#TestJson"? We return hundreds of these elements on some queries and that extra text will add up.

Full code:

using System;
using System.Collections.Generic;
using System.IO;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Json;

namespace TestJson
{
    class Program
    {
        static void Main(string[] args)
        {
            Adult parent = new Adult {name = "John", age = 42};

            MemoryStream stream1 = new MemoryStream();
            DataContractJsonSerializer ser = new DataContractJsonSerializer(typeof(Adult));
            ser.WriteObject(stream1, parent);

            stream1.Position = 0;
            StreamReader sr = new StreamReader(stream1);
            Console.Write("JSON form of Adult with no children: ");
            Console.WriteLine(sr.ReadToEnd());


            Child child = new Child { name = "Jane", age = 4, fingers=10 };

            stream1 = new MemoryStream();
            ser = new DataContractJsonSerializer(typeof(Child));
            ser.WriteObject(stream1, child);

            stream1.Position = 0;
            sr = new StreamReader(stream1);
            Console.Write("JSON form of Child with no parent: ");
            Console.WriteLine(sr.ReadToEnd());


            // now connect the two
            parent.children.Add(child);

            stream1 = new MemoryStream();
            ser = new DataContractJsonSerializer(typeof(Adult));
            ser.WriteObject(stream1, parent);

            stream1.Position = 0;
            sr = new StreamReader(stream1);
            Console.Write("JSON form of Adult with 1 child: ");
            Console.WriteLine(sr.ReadToEnd());
        }
    }

    [DataContract]
    [KnownType(typeof(Adult))]
    [KnownType(typeof(Child))]
    class Person
    {
        [DataMember]
        internal string name;

        [DataMember]
        internal int age;
    }

    [DataContract]
    class Adult : Person
    {
        [DataMember] 
        internal List<Person> children = new List<Person>();
    }

    [DataContract]
    class Child : Person
    {
        [DataMember]
        internal int fingers;
    }
}

原文:https://stackoverflow.com/questions/17815772
更新时间:2024-03-09 06:03

最满意答案

您可以尝试的一个解决方案是:

{
  xtype: 'textfield',
  fieldLabel: 'DNR TYPE',
  name: 'DNR_TYPE,
  listeners: {
    beforerender: function() {
       this.setValue(Ext.getCmp('dnr-type').getRawValue());
    }
  }
}

如果Ext.getCmp('dnr-type').getRawValue()返回值,则将其分配给textfield,否则如果它是空字符串,则textfield将包含空值,并且您将看不到textfield中的值。

编辑:

当用户从组合框中选择一条记录时,我应该将该值分配给相关的文本字段。

---->你可以在combo select事件上更改textfield的值,如:

{
    xtype: 'combo'
    displayField: 'foo',
    valueField: 'bar',
    lsiteners: {
       select: function( combo, records) {
          var rec = records[0]; // records will contain selected records(1 or more)
          var textField = // get textfield using Ext.getCmp()
          textField.setValue(rec.get('modelName'));
       }
    }
}

现在,有什么方法可以在用户从表单中选择记录时再次刷新/重新加载表单吗?

---->我不清楚这个。 如果你告诉我这个场景,我可以帮你解决这个问题


One solution that you can try is :

{
  xtype: 'textfield',
  fieldLabel: 'DNR TYPE',
  name: 'DNR_TYPE,
  listeners: {
    beforerender: function() {
       this.setValue(Ext.getCmp('dnr-type').getRawValue());
    }
  }
}

If Ext.getCmp('dnr-type').getRawValue() returns value, it will be assigned to textfield, otherwise if it is blank string then textfield will contain blank value and you wouldn't see value in textfield.

EDIT:

When user select a record from the combobox, I should assign that value to related textfield.

----> you can change value of textfield on select event of combo like :

{
    xtype: 'combo'
    displayField: 'foo',
    valueField: 'bar',
    lsiteners: {
       select: function( combo, records) {
          var rec = records[0]; // records will contain selected records(1 or more)
          var textField = // get textfield using Ext.getCmp()
          textField.setValue(rec.get('modelName'));
       }
    }
}

Now, is there any method to refresh/reload form again when user select a record from the form?

----> I am not clear with this. May be I can help you with this if you tell me the scenario

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