如何使用Jackson将Color java类解析为JSON? [英] How I parse Color java class to JSON with Jackson?

问题描述

我试图用Jackson从JSON反序列化Color类，但是它抛出异常:

I am trying to deserialise the Color class from JSON with Jackson but it throws exception:

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: 无法识别的字段"colorSpace"(类java.awt.Color)，未标记为 可忽略的.

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "colorSpace" (class java.awt.Color), not marked as ignorable.

我做错了什么? 这是我的代码:

What i'm doing wrong? This is my code:

File act = new File(new File().getAbsolutePath());

ObjectMapper om = new ObjectMapper();
File f = new File(act, "123.JSON");

om.writeValue(f, new person());
person per = om.readValue(f, person.class);
System.out.println(per);

这是我的人物课程:

public class person implements Serializable {
    //it include getters, setters and builder

   String nombe = "Pepe";
   String CI = "12345678978";
   Color c = Color.red;
}

推荐答案

java.awt.Color类不是常规的POJO或Enum.如果要以JSON格式存储，则需要实现自定义序列化程序和反序列化程序. Color类可以用其RGB表示形式表示，并且可以将其存储为数字:

java.awt.Color class is not a regular POJO or Enum. You need to implement custom serialiser and deserialiser if you want to store it in JSON format. Color class can be represented by its RGB representation and you can store it as a number:

class ColorJsonSerializer extends JsonSerializer<Color> {

    @Override
    public void serialize(Color value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
        if (value == null) {
            gen.writeNull();
            return;
        }
        gen.writeNumber(value.getRGB());
    }
}

class ColorJsonDeserializer extends JsonDeserializer<Color> {

    @Override
    public Color deserialize(JsonParser p, DeserializationContext ctxt) throws IOException {
        return new Color(p.getValueAsInt());
    }
}

简单用法:

import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.module.SimpleModule;

import java.awt.*;
import java.io.IOException;

public class JsonPathApp {

    public static void main(String[] args) throws Exception {
        SimpleModule awtModule = new SimpleModule("AWT Module");
        awtModule.addSerializer(Color.class, new ColorJsonSerializer());
        awtModule.addDeserializer(Color.class, new ColorJsonDeserializer());

        ObjectMapper mapper = new ObjectMapper();
        mapper.registerModule(awtModule);

        String json = mapper.writeValueAsString(new Person());
        System.out.println(json);

        System.out.println(mapper.readValue(json, Person.class));
    }
}

上面的代码显示:

{"nombe":"Pepe","c":-65536,"ci":"12345678978"}
Person{nombe='Pepe', CI='12345678978', c=java.awt.Color[r=255,g=0,b=0]}

看看类似的问题，其中Color被存储为JSON对象:

Take a look on similar question where Color is stored as JSON object:

• 无法使用jackson反序列化器反序列化java.awt.color

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