将json转换为java对象错误UnrecognizedPropertyException

jvidinwx 于 2023-03-21 发布在 Java

关注(0)|答案(2)|浏览(205)

我必须从我的json文件中创建java对象，看起来像这样：{“colors”：[“red”，“绿色”，“blue”]，“isPrimary”：true，“rgb”：{“r”：255，“g”：0，“B”：0 } }
我创建了一个带有参数的类来描述我的文件

public class Color {
    private List<String> colors;
    private boolean isPrimary;
    private HashMap<String,Integer> rgb = new HashMap<String,Integer>();

    public Color(){};
    public List<String> getColors() {
        return colors;
    }

    public void setColors(List<String> colors) {
        this.colors = colors;
    }

    public boolean isPrimary() {
        return isPrimary;
    }

    public void setPrimary(boolean primary) {
        isPrimary = primary;
    }

    public HashMap<String, Integer> getRgb() {
        return rgb;
    }

    public void setRgb(HashMap<String, Integer> rgb) {
        this.rgb = rgb;
    }

    @Override
    public String toString() {
        return "Color{" +
                "colors=" + colors +
                ", isPrimary=" + isPrimary +
                ", rgb=" + rgb +
                '}';
    }
}

主要类别：

public static void main(String[] args) throws JsonProcessingException {
        String jsonStr = "{\r\n" +
                "  \"colors\": [\"red\", \"green\", \"blue\"],\r\n" +
                "  \"isPrimary\": true,\r\n" +
                "  \"rgb\": {\r\n" +
                "    \"r\": 255,\r\n" +
                "    \"g\": 0,\r\n" +
                "    \"b\": 0\r\n" +
                "  }\r\n" +
                "}";
        ObjectMapper objectMapper = new ObjectMapper();
        Color color = objectMapper.readValue(jsonStr, Color.class);
        System.out.println(color);
    }

我是按照指示做的，但可能出了什么问题

Exception in thread "main" com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "isPrimary" (class itstep.task_5.Color), not marked as ignorable (3 known properties: "colors", "primary", "rgb"])
 at [Source: (String)"{
  "colors": ["red", "green", "blue"],
  "isPrimary": true,
  "rgb": {
    "r": 255,
    "g": 0,
    "b": 0
  }
}";

Java

来源：https://stackoverflow.com/questions/75777621/converting-json-into-java-object-error-unrecognizedpropertyexception

2条答案

按热度按时间

3xiyfsfu1#

问题是JSON字段名是isPrimary，而Java字段名是primary。
要解决此问题，您可以将JSON字段名称更改为primary，或者向Java字段添加@JsonProperty注解，以指定它应Map到isPrimary JSON字段名称：

import com.fasterxml.jackson.annotation.JsonProperty;
import java.util.HashMap;
import java.util.List;

public class Color {
    private List<String> colors;
    @JsonProperty("isPrimary")
    private boolean primary;
    private HashMap<String, Integer> rgb = new HashMap<>();

    public Color() {}

    public List<String> getColors() {
        return colors;
    }

    public void setColors(List<String> colors) {
        this.colors = colors;
    }

    public boolean isPrimary() {
        return primary;
    }

    public void setPrimary(boolean primary) {
        this.primary = primary;
    }

    public HashMap<String, Integer> getRgb() {
        return rgb;
    }

    public void setRgb(HashMap<String, Integer> rgb) {
        this.rgb = rgb;
    }

    @Override
    public String toString() {
        return "Color{" +
                "colors=" + colors +
                ", primary=" + primary +
                ", rgb=" + rgb +
                '}';
    }
}

import com.fasterxml.jackson.databind.ObjectMapper;
import java.io.IOException;

public class Main {
    public static void main(String[] args) throws IOException {
        String jsonStr = "{\r\n" +
                "  \"colors\": [\"red\", \"green\", \"blue\"],\r\n" +
                "  \"isPrimary\": true,\r\n" +
                "  \"rgb\": {\r\n" +
                "    \"r\": 255,\r\n" +
                "    \"g\": 0,\r\n" +
                "    \"b\": 0\r\n" +
                "  }\r\n" +
                "}";
        ObjectMapper objectMapper = new ObjectMapper();
        Color color = objectMapper.readValue(jsonStr, Color.class);
        System.out.println(color);
    }
}

输出：

Color{colors=[red, green, blue], primary=true, rgb={r=255, b=0, g=0}}

赞(0）回复(0）举报 2023-03-21

fnvucqvd2#

你的setter和getter有错误的名字。Jackson试图使用格式为set{fieldname}的setter方法来赋值，例如setIsPrimary。但是你的setter被称为setPrimary，所以你的类的属性“isPrimary”（属性由同名的getter和setter标识）找不到。这就是Exception试图告诉你的。
TL;DR：将setPrimary方法重命名为setIsPrimary。

赞(0）回复(0）举报 2023-03-21

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将json转换为java对象错误UnrecognizedPropertyException

2条答案

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