java - 获取详细的 json 响应(没有 id 引用，始终是实例)

Question

我有以下对象:

public class A
{
    private String id;
    private String name;
    private List <B> bs;
}

@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
public class B
{
    private String id;
    private String name;
}

序列化后，如果我使用restcontroller执行get操作，我会得到类似这样的东西，首先是完整的实例，然后是引用:

    {
    "A": [
        {
          "id": "1",
          "bs": [{
              "id": "1",
              "name": "b1"
            },
            {
              "id": "2",
              "name": "b2"
            }]
        },
    {
          "id": "2",
          "bs": ["1","2"]
        }]
    }

是否有机会始终获得完整的实例而不是引用？例如

{
    "A": [
        {
          "id": "1",
          "bs": [{
              "id": "1",
              "name": "b1"
            },
            {
              "id": "2",
              "name": "b2"
            }]
        },
{
          "id": "2",
          "bs": [{
              "id": "1",
              "name": "b1"
            },
            {
              "id": "2",
              "name": "b2"
            }]
        }]
    }

Answer 1

由于您在评论中指出您依赖 @JsonIdentityInfo 注释进行反序列化，因此不能删除该注释。

但是，您可以做的是创建一个混合，专门覆盖此特定用例的对象标识行为:

@JsonIdentityInfo(generator = ObjectIdGenerators.None.class)
public abstract class BMixIn {}

然后将其注册到用于编写输出的 ObjectMapper 上:

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.addMixIn(B.class, BMixIn.class);