JSON Jackson将多个密钥反序列化到同一字段中 [英] JSON Jackson deserialization multiple keys into same field

问题描述

我正在尝试将JSON转换为POJO. 我曾与Jackson一起转换标准JSON文件. 在这种特殊情况下，我想将键值覆盖为默认"类/变量.在这种情况下，有多个要替换的键值(即数百个，并且要替换的键值未知).

I am trying to convert JSON into POJO. I have worked with Jackson to convert standard JSON file. In this particular case, I would like to overwrite the key value to "default" class/variable. In this case, there are multiple key value to be replaced (ie. hundreds, and the key values to be replaced are unknown).

这可能吗?我本来打算将其存储到Map中，然后将每个对象迭代并存储到POJO中，但是想知道是否存在其他选择，因为我不熟悉将JSON存储到Map中.

Is this possible? I thought of storing it into Map, then iterate and store each into POJO, but wondering if there is different option, since I am not that familiar with storing JSON to Map.

要处理的JSON示例:

Example of the JSON to be processed:

"People" : {
    "person1" : { 
      "name" : "john doe",
      "address" : "123 main st",
      "email" : "[e-mail]"
    },
    "person2" : { 
      "name" : "bob cat",
      "address" : "234 dog st",
      "email" : "[e-mail]"
    },
    "person3" : { 
      "name" : "foo bar",
      "address" : "111  1st ave",
      "email" : "[e-mail]"
    },
    "person8" : { 
      "name" : "james bono",
      "address" : "999 alaska st",
      "email" : "[e-mail]"
    }
}

是否可以按以下结构生成类?主要问题是要替换成百上千的价值，并且假设它们是未知的，我不能使用这种方法.

Is it possible to generate the class in the following structure? The main issue is there are hundreds of value to be replaced and assuming they are unknown, I can't use this approach.

@JsonIgnoreProperties(ignoreUnknown = true)
public class People { 

  @JsonAlias({"person1", "person2"})
  private List<Details> person; // --> this should be the default replacing person1, person2, and so on 

  private class Details { 
    String name;
    String address;
    String email;
  }
}

推荐答案

您可以使用

You can use JsonAnySetter annotation for all properties personXYZ. See below example:

import com.fasterxml.jackson.annotation.JsonAnySetter;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.ObjectMapper;

import java.io.File;
import java.util.ArrayList;
import java.util.List;

public class JsonApp {

    public static void main(String[] args) throws Exception {
        File jsonFile = new File("./resource/test.json").getAbsoluteFile();

        ObjectMapper mapper = new ObjectMapper();
        mapper.enable(DeserializationFeature.UNWRAP_ROOT_VALUE);

        System.out.println(mapper.readValue(jsonFile, People.class).getPersons());
    }
}

class People {

    private List<Details> persons = new ArrayList<>();

    @JsonAnySetter
    public void setPerson(String name, Details person) {
        this.persons.add(person);
    }

    public List<Details> getPersons() {
        return persons;
    }

    public static class Details {
        String name;
        String address;
        String email;

        // getters, setters, toString
    }
}

对于上面的JSON代码，打印:

For your JSON above code prints:

[Details{name='john doe', address='123 main st', email='[e-mail]'}, Details{name='bob cat', address='234 dog st', email='[e-mail]'}, Details{name='foo bar', address='111  1st ave', email='[e-mail]'}, Details{name='james bono', address='999 alaska st', email='[e-mail]'}]

如果使用内部类，请记住将其设置为public static，以使其对Jackson实例化过程可见.

In case you use inner class remember to make it public static to make it visible to Jackson instantiation process.

另请参阅:

• 如何为Json使用动态属性名称对象
• Jackson注释示例

• How to use dynamic property names for a Json object
• Jackson Annotation Examples

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