如何将以下JSON字符串转换为POJO

Question

我想将以下JSON字符串转换为Java对象：

{
  "user": {
    "0": {
      "firstName": "Monica",
      "lastName": "Belluci"
    },
    "1": {
      "firstName": "John",
      "lastName": "Smith"
    },
    "2": {
      "firstName": "Owen",
      "lastName": "Hargreaves"
    }
  }
}

要将其转换为Java对象，我创建了以下类：

class User {
    private Map<String, MyObject> user = new HashMap<>();
    //Getter and Setter is here
}

class MyObject {
    private String firstName;
    private String lastName;
    //Getters and Setters are here
}

我使用Jackson库转换JSON到Java。以下是我如何使用杰克逊转换：

ObjectMapper mapper = new ObjectMapper();
User user = mapper.readValue(jsonString, User.class);

问题是，在转换上面的Map中User对象始终是空的。我做错了什么？

提前谢谢。

Answer 1

你可以尝试下面的代码，它工作正常。.

public class User {

    private Map<String, Map<String, String>> user;

    public Map<String, Map<String, String>> getUser() {
        return user;
    }

    public void setUser(Map<String, Map<String, String>> user) {
        this.user = user;
    }

}


public class JsonCast {

    public static void main(String args[]) {

        String response = "{\"user\" : {\"0\": {\"firstName\": \"Monica\",\"lastName\": \"Belluci\"},\"1\": { \"firstName\": \"John\",\"lastName\": \"Smith\"}}}";

        ObjectMapper mapper = new ObjectMapper();
        try {

            User user = mapper.readValue(response, User.class);

            System.out.println(user.getUser().get("0"));

        } catch (IOException e) {

            e.printStackTrace();
        }
    }

}

Answer 2

使用可以在gson库的帮助下完成。

import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import com.google.gson.Gson;
import com.google.gson.GsonBuilder;

public class JsonToJava {

    public static void main(String[] args) throws IOException {
        try(Reader reader = new InputStreamReader(JsonToJava.class.getResourceAsStream("/Server2.json"), "UTF-8")){
            Gson gson = new GsonBuilder().create();
            Person p = gson.fromJson(reader, YourPOJOClass.class);
            System.out.println(p);
        }
    }
}

访问此链接希望这有助于：）

Answer 3

我觉得应该可以。我已经执行了这段代码，它工作正常。这是我的例子。

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

public class TestJackson {

public static void main(String[] args) throws IOException {
        ObjectMapper mapper = new ObjectMapper();
        String testJson = "{\n" + "  \"user\": {\n" + "    \"0\": {\n" + "      \"firstName\": \"Monica\",\n" + "      \"lastName\": \"Belluci\"\n" + "    },\n" + "    \"1\": {\n" + "      \"firstName\": \"John\",\n" + "      \"lastName\": \"Smith\"\n" + "    },\n" + "    \"2\": {\n" + "      \"firstName\": \"Owen\",\n" + "      \"lastName\": \"Hargreaves\"\n" + "    }\n" + "  }\n" + "}";
        User readValue = mapper.readValue(testJson, User.class);
        System.out.println("readValue = " + readValue);
    }
}

以及用户。类别：

import java.util.HashMap;
import java.util.Map;

class User {
    private Map<String, MyObject> user = new HashMap<String, MyObject>();

    public Map<String, MyObject> getUser() {
        return user;
    }

    public void setUser(Map<String, MyObject> user) {
        this.user = user;
    }

    @Override
    public String toString() {
        return "User{" +
                "user=" + user +
                '}';
    }
}

class MyObject {
    private String firstName;
    private String lastName;

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    @Override
    public String toString() {
        return "MyObject{" +
                "firstName='" + firstName + '\'' +
                ", lastName='" + lastName + '\'' +
                '}';
    }
}