Jackson将JSON解析为Map< String，TypeA> [英] Jackson parse JSON to Map<String, TypeA>

问题描述

我有以下JSON:

{  
    "data": {  
        "1": {  
            "id":"1",  
            "name":"test1"  
         },  
        "2": {   
            "id":"2",  
            "name":"test2"  
         }  
    }  
}  

我想用Jackson将数据"解析为一个对象.如果我将其解析为Map<String, Object>，则效果很好，而将"1"，"2"(...)用作Key，并将相应数据作为值，再次由Map表示.

I want to parse the "data" into an Object with Jackson. If I parse it as Map<String, Object> it works well, whereas "1", "2" (...) are used as Key with the respective data as value, represented by a Map again.

现在我想将此JSON解析为Map<String, TypeA>，而类TypeA将具有两个字段，即id和name.

Now I want to parse this JSON to Map<String, TypeA> whereas class TypeA would have two fields, id and name.

有人可以给我提示如何做到这一点吗? 我总是收到以下错误:

Can someone give me a hint how to to that? I always get the following error:

无法读取JSON:找不到类型[简单的合适的构造函数 类型，类TypeA]:无法从JSON对象实例化(需要 添加/启用类型信息?)

Could not read JSON: No suitable constructor found for type [simple type, class TypeA]: can not instantiate from JSON object (need to add/enable type information?)

非常感谢，
鞋

Thanks a lot in advance,
tohoe

推荐答案

以下应为您解决.

public class MyDataObject {
    private final Map<String, TypeA> data;

    @JsonCreator
    public MyDataObject(@JsonProperty("data") final Map<String, TypeA> data) {
        this.data = data;
    }

    public Map<String, TypeA> getData() {
        return data;
    }
}

public class TypeA {
    private final String id;
    private final String name;

    @JsonCreator
    public TypeA(@JsonProperty("id") final String id, 
                 @JsonProperty("name") String name) {

        this.id = id;
        this.name = name;
    }

    public String getId() {
        return id;
    }

    public String getName() {
        return name;
    }
}

@JsonCreator用于描述如何创建对象以及属性@JsonProperty的名称.即使它们是嵌套的.

The @JsonCreator is used for describing how to create your objects together with the name of the properties @JsonProperty. Even if they are nested.

要反序列化整个事情:

ObjectMapper mapper = new ObjectMapper();
final MyDataObject myDataObject = mapper.readValue(json, MyDataObject.class);

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