当嵌套 JSON 中出现相同的字段名称时，将 JSON 解析为 java 对象

Question

以下是我的JSON：{  "time":{    "date":{      "year":2017,      "month":3,      "day":12     },    "time":{      "hour":10,      "minute":42,      "second":42,      "nano":810000000     }   },"name":"Jon","message":{"product":"orange""price":2000}}“时间”字段有一个嵌套的“时间”字段。我如何使用杰克逊解析它到java对象。任何人都可以告诉我正确的方法吗？

Answer 1

您可以创建如下类：

class JavaObject {

    private TimeObject time;

    private String name;

    //other fields

    //getters and setters

}

class TimeObject {

    private Date date;

    private Time time;

    //getters and setters

}

class Date {

    private int year;

    private int month;

    private int day;

    //getters and setters

}

class Time {

    private int hour;

    private int minute;

    private int second;

    private long nano;

    //getters and setters

}

完成后，您可以使用 将 String 反序列化为对象，例如：JacksonjsonJavaObject

ObjectMapper objectMapper = new ObjectMapper();

JavaObject javaObject = objectMapper.readValue("{\n" + 

        "  \"time\":{\n" + 

        "    \"date\":{\n" + 

        "      \"year\":2017,\n" + 

        "      \"month\":3,\n" + 

        "      \"day\":12\n" + 

        "     },\n" + 

        "    \"time\":{\n" + 

        "      \"hour\":10,\n" + 

        "      \"minute\":42,\n" + 

        "      \"second\":42,\n" + 

        "      \"nano\":810000000\n" + 

        "     }\n" + 

        "   },\n" + 

        "\"name\":\"Jon\"}", JavaObject.class);

System.out.println(javaObject);

Answer 2

如果你只需要内部对象，你可以用一种快速的方式做到这一点：time

// your POJO class

// (public fields sould be private with getter & setter, of course)

public class Pojo {

    public int hour;

    public int minute;

    public int second;

    public long nano;

    @Override

    public String toString() {

        return hour + ":" + minute + ":" + second + ":" + nano;

    }

}

然后：

//your json string

String jsonString = "{\"time\":{\"date\":{\"year\":2017,\"month\":3,\"day\":12},"

                + "\"time\":{\"hour\":10,\"minute\":42,\"second\":42,\"nano\":810000000}},"

                + "\"name\":\"Jon\",\"message\":{\"product\":\"orange\",\"price\":2000}}";

ObjectMapper mapper = new ObjectMapper();

JsonNode jsonRoot = mapper.readTree(jsonString); //parse string to JsonNode

Pojo pojo = mapper.treeToValue(jsonRoot.at("/time/time"), Pojo.class); //create Pojo instance from inner time object

System.out.println(pojo); //see if it worked

这打印：

10:42:42:810000000

Answer 3

使用杰克逊库

ObjectMapper jsonMapper= new ObjectMapper();

YourCorrespondingObject object = jsonMapper.readValue("your json as string...", YourCorrespondingObject.class);

首先构建对象，填充它并转换为字符串更容易，以确保它等于现有字符串，例如：

String jsonInString = mapper.writeValueAsString(object);