[英]Use specific property name with Json.NET when serializing
在C#中使用Json.NET,在序列化类时遇到麻烦,在该类中我需要自定义属性名称。
这就是我现在所拥有的:
{
"baseName": {
"subName1": {
"name": null,
"longName": null,
"asd1": null,
"asd2": null
},
"subName2": [
{
"id": "ID_NUMBER",
"info": {
"someInfo1": "asd",
"someInfo2": "asd2"
}
}
]
}
}
这就是我想要的 :
{
"baseName": {
"subName1": {
"name": null,
"longName": null,
"asd1": null,
"asd2": null
},
"subName2": [
{
"ID_NUMBER": {
"someInfo1": "asd",
"someInfo2": "asd2"
}
}
]
}
}
即,我希望ID_NUMBER
成为包含键someInfo1
和someInfo2
的对象的键,而不是具有id
和值ID_NUMBER
的键。
顶级JSON代码使用以下类生成(对那些不好的名字表示抱歉):
class JSONTestClass
{
public JSONBaseTestClass baseName;
}
class JSONBaseTestClass
{
public JSONSubTestClass1 subName1;
public List<JSONSubTestClass2> subName2;
}
class JSONSubTestClass1
{
public string name;
public string longName;
public string asd1;
public string asd2;
}
class JSONSubTestClass2
{
public string id;
public JSONInfoTestClass info;
}
class JSONInfoTestClass
{
public string someInfo1;
public string someInfo2;
}
和这个:
private void MakeJSON()
{
// This value can be changed at runtime
string specificId = "ID_NUMBER";
JSONInfoTestClass jitc = new JSONInfoTestClass();
jitc.someInfo1 = "asd";
jitc.someInfo2 = "asd2";
JSONTestClass jtc = new JSONTestClass();
JSONBaseTestClass jbtc = new JSONBaseTestClass();
JSONSubTestClass1 jstc1 = new JSONSubTestClass1();
JSONSubTestClass2 jstc2 = new JSONSubTestClass2();
jstc2.id = specificId;
jstc2.info = jitc;
List<JSONSubTestClass2> list = new List<JSONSubTestClass2>();
list.Add(jstc2);
jbtc.subName1 = jstc1;
jbtc.subName2 = list;
jtc.baseName = jbtc;
// Convert to JSON
string json = JsonConvert.SerializeObject(jtc, Formatting.Indented);
tbxJSONOutput.Text = json;
}
需要进行哪些更改,以便我可以获得与上述第二个JSON响应相对应的JSON输出?
您可以通过为JSONSubTestClass2
类创建自定义JsonConverter
来获得所需的输出,如下所示:
class JSONSubTestClass2Converter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return (objectType == typeof(JSONSubTestClass2));
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
JSONSubTestClass2 jstc2 = (JSONSubTestClass2)value;
JObject jo = new JObject();
jo.Add(jstc2.id, JObject.FromObject(jstc2.info));
jo.WriteTo(writer);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
然后,像这样序列化您的类:
JsonSerializerSettings settings = new JsonSerializerSettings();
settings.Converters.Add(new JSONSubTestClass2Converter());
settings.Formatting = Formatting.Indented;
// Convert to JSON
string json = JsonConvert.SerializeObject(jtc, settings);
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