Serialize/Deserialize 嵌套 POCO 属性 而没有嵌套在 json.net 中
Serialize/Deserialize nested POCO property without nesting in json.net
考虑:
public class Foo
{
public string FooName { get; set; } = "FooName";
public Bar Bar { get; set; } = new Bar();
}
public class Bar
{
public string BarName { get; set; } = "BarName";
}
如果我们序列化 Foo() 输出是:
{"FooName":"FooName","Bar":{"BarName":"BarName"}}
我要:
{"FooName":"FooName", "BarName":"BarName" }
实现此目的最简洁的方法是什么?
您可以使用自定义转换器来完成此操作,例如,这是一个快速而肮脏的示例:
public class BarConverter : JsonConverter
{
public override bool CanConvert(Type objectType) => typeof(Bar) == objectType;
public override object ReadJson(JsonReader reader, Type objectType, object existingValue,
JsonSerializer serializer)
{
throw new NotImplementedException();
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
writer.WriteValue(((Bar)value).BarName);
value.Dump();
}
}
然后像这样装饰你的模型:
public class Foo
{
public string FooName { get; set; } = "FooName";
[JsonConverter(typeof(BarConverter))]
public Bar Bar { get; set; } = new Bar();
}
考虑:
public class Foo
{
public string FooName { get; set; } = "FooName";
public Bar Bar { get; set; } = new Bar();
}
public class Bar
{
public string BarName { get; set; } = "BarName";
}
如果我们序列化 Foo() 输出是:
{"FooName":"FooName","Bar":{"BarName":"BarName"}}
我要:
{"FooName":"FooName", "BarName":"BarName" }
实现此目的最简洁的方法是什么?
您可以使用自定义转换器来完成此操作,例如,这是一个快速而肮脏的示例:
public class BarConverter : JsonConverter
{
public override bool CanConvert(Type objectType) => typeof(Bar) == objectType;
public override object ReadJson(JsonReader reader, Type objectType, object existingValue,
JsonSerializer serializer)
{
throw new NotImplementedException();
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
writer.WriteValue(((Bar)value).BarName);
value.Dump();
}
}
然后像这样装饰你的模型:
public class Foo
{
public string FooName { get; set; } = "FooName";
[JsonConverter(typeof(BarConverter))]
public Bar Bar { get; set; } = new Bar();
}