如何将下面的 Json 转换为 Java Pojo 类？答案

【问题标题】：how to convert below Json To Java Pojo class?如何将下面的 Json 转换为 Java Pojo 类？
【发布时间】：2020-11-29 07:56:23
【问题描述】：

请在 Json 下方查看。我想把它转换成 POJO 类。

我已经尝试过以下课程

@Data
public class Root {
    @JsonProperty("0")      // Dont want to pass like this. it should be single list and index is the key of that array
    public List<Request> list;
    @JsonProperty("1")
    public List<Request> list1;
    @JsonProperty("2")
    public List<Request> list2;

    @Data
    private class Request {
        int id;
        String title;
        int level;
        List<Request> children;
        @JsonProperty("parent_id")
        int parentId;
    }
}

{
  "0": [
    {
      "id": 12123,
      "title": "sfsdf",
      "level": 0,
      "children": [
        
      ],
      "parent_id": null
    }
  ],
  "1": [
    {
      "id": 213,
      "title": "d",
      "level": 1,
      "children": [
        
      ],
      "parent_id": 1212312
    },
    {
      "id": 3232,
      "title": "dfsdf",
      "level": 1,
      "children": [
        
      ],
      "parent_id": 42
    },
    {
      "id": 234,
      "title": "tder",
      "level": 1,
      "children": [
        
      ],
      "parent_id": 122
    }
  ],
  "2": [
    {
      "id": 452,
      "title": "Blll",
      "level": 2,
      "children": [
        
      ],
      "parent_id": 322
    },
    {
      "id": 123,
      "title": "trrr",
      "level": 11,
      "children": [
        
      ],
      "parent_id": 1221
    },
    {
      "id": 33,
      "title": "sdfw",
      "level": 2123,
      "children": [
        
      ],
      "parent_id": 10
    }
  ]
}

【问题讨论】：

标签： java json spring spring-boot

【解决方案1】：

使用@JsonAnySetter。

例如，为了简单起见，使用 public 字段：

class Root {
    @JsonAnySetter
    public Map<String, List<Request>> lists = new LinkedHashMap<>();
}

class Request {
    public int id;
    public String title;
    public int level;
    public List<Request> children;
    @JsonProperty("parent_id")
    public Integer parentId; // 'Integer', not 'int', since it can be null
}

测试

ObjectMapper mapper = new ObjectMapper();
Root root = mapper.readValue(new File("test.json"), Root.class);
for (Entry<String, List<Request>> entry : root.lists.entrySet()) {
    System.out.println(entry.getKey() + ":");
    for (Request request : entry.getValue()) {
        System.out.println("  id=" + request.id +
                        ", title=" + request.title +
                        ", level=" + request.level +
                     ", children=" + request.children +
                     ", parentId=" + request.parentId);
    }
}

输出

0:
  id=12123, title=sfsdf, level=0, children=[], parentId=null
1:
  id=213, title=d, level=1, children=[], parentId=1212312
  id=3232, title=dfsdf, level=1, children=[], parentId=42
  id=234, title=tder, level=1, children=[], parentId=122
2:
  id=452, title=Blll, level=2, children=[], parentId=322
  id=123, title=trrr, level=11, children=[], parentId=1221
  id=33, title=sdfw, level=2123, children=[], parentId=10

【讨论】：

【解决方案2】：

只需删除第二个和第三个列表。只保留一个列表，让它保留它的名字，比如list 或requests。然后添加到该列表中。

Root root = new Root(); // whatever
root.getRequests().add(myFirstRequest); // index 0 in the array
root.getRequests().add(mySecondRequest); // index 1 in the array
root.getRequests().add(myThirdRequest); // index 2 in the array

String json = // however you were making json out of Root before

【讨论】：

我想将这个 Root 对象传递给请求正文中的其余控制器。

【解决方案3】：

通过在控制器请求正文中传递Map<String, List<Request>> lists。我的问题已经解决了。我得到了预期的 JSON。

【讨论】：