Question

我正在使用springBoot构建json数据。我想要另一种方式的数据我得到这个回应

[
    {
        "id": "Spring",
        "name": "Spring FrameWork",
        "description": "Spring Frame work descreption"
    },
    {
        "id": "java",
        "name": "java FrameWork",
        "description": "Java Frame work descreption"
    },
    {
        "id": "javascript",
        "name": "javascript FrameWork",
        "description": "java script Frame work descreption"
    }
]

我想要这样的回复

"topics":[
    {
        "id": "Spring",
        "name": "Spring FrameWork",
        "description": "Spring Frame work descreption"
    },
    {
        "id": "java",
        "name": "java FrameWork",
        "description": "Java Frame work descreption"
    },
    {
        "id": "javascript",
        "name": "javascript FrameWork",
        "description": "java script Frame work descreption"
    }
]

我的春季启动代码是：

@RequestMapping("/topics")
    public List<Topic> getAllTopics() { 
        return topicService.getAllTopics();

    }

我的主题模型：-

public class Topic {

    private String id;
    private String name;
    private String description;

    public Topic() {

    }

    public Topic(String id, String name, String description) {
        super();
        this.id = id;
        this.name = name;
        this.description = description;
    }

    public String getId() {
        return id;
    }
    public void setId(String id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getDescription() {
        return description;
    }
    public void setDescription(String description) {
        this.description = description;
    }

}

我需要在代码中添加什么，以便JSON数据应包含该列表的名称。可以使用JSON数据对象完成此操作。但是还有其他方法吗？请回答我的问题

Answer 1

创建地图并返回

@RequestMapping("/topics")
public Map<String, List<Topic>> getAllTopics() { 
   Map<String, List<Topic>> response = new HashMap<String, List<Topic>>();
   response.put("topics", topicService.getAllTopics());
   return response;
}

Answer 2

这不是有效的json。

"topics":[
    {
        "id": "Spring",
        "name": "Spring FrameWork",
        "description": "Spring Frame work descreption"
    },
    {
        "id": "java",
        "name": "java FrameWork",
        "description": "Java Frame work descreption"
    },
    {
        "id": "javascript",
        "name": "javascript FrameWork",
        "description": "java script Frame work descreption"
    }
]

根据JSON Docs，json应该以{{1}}或{开头。

正确的json是：

并在响应中实现这种格式：

创建一个新的模型类

{
  "topics": [
    {
      "id": "Spring",
      "name": "Spring FrameWork",
      "description": "Spring Frame work descreption"
    },
    {
      "id": "java",
      "name": "java FrameWork",
      "description": "Java Frame work descreption"
    },
    {
      "id": "javascript",
      "name": "javascript FrameWork",
      "description": "java script Frame work descreption"
    }
  ]
}

在public class TopicsData { List<Topic> topics; // getters and setters }对象中设置主题列表
更改TopicsData或控制器代码以返回topicService对象，而不是TopicsData。

Answer 3

您可以通过将Topics添加到另一个类（例如TopicResponse）来实现。将此作为您的控制器返回类型：

    public class TopicResponse {
         List<Topic> topics;
        //and getter / setter of topics
}

如何在jSON数组中的列表中添加名称，使用springboot发布响应

3 个答案: