首先，我同意那些评论无效JSON格式的人.您可以在此处查看示例 https://json.org/example.html

First, I'm agree with guy who commented that is not valid JSON format. You can see examples here https://json.org/example.html

第二，您需要创建一个对象JSON，其中包含所需的字段，例如:

Second, You need to create an object JSON which has fields needed for example:

  public class UserStat es implements Serializable {

         private String name;
         private long count;

         public String getName() {
             return name;
         } 

         public void setName(String name) {
             this.name = name;
         } 

         public long getCount() {
             return count;
         } 

         public void setCount(long count) {
             this.count = count;
         } 
  }

并在您的自定义查询中.根据这种方式，您的报酬看起来像这样:

And in your custom query. Based your return looks like on this way:

  @Query("SELECT u.name, count(u) FROM User u")
   public List<UserStat> findUserStat() ;