Java @RequestMapping未获取参数和值。Spring framework

我正在尝试创建一个带有CRUD操作的api。为此，创建了一个用户bean和一个扩展Crudepository的用户存储库

我试图创建一个用户，但它接受空值。在数据库中创建了一个用户id，其值为null

代码：

这里出了什么问题？我是网络开发新手。请帮忙。 谢谢

编辑：

与我尝试为特定用户添加位置的方式相同。 创建了与用户实体的location bean和ManyTone关系

正在插入位置，但位置id和用户id返回null

 @Entity
public class Location {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Integer id;


    private String location;
    private double latitude;

    public Location() {}

    @ManyToOne(fetch = FetchType.LAZY)
    private User user;


       @Override
    public String toString() {
        return "Location [id=" + id + ", location=" + location + ", latitude=" + latitude + ", longitude=" + longitude
                + "]";
    }
    public double getLatitude() {
        return latitude;
    }
    public void setLatitude(double latitude) {
        this.latitude = latitude;
    }
    public double getLongitude() {
        return longitude;
    }
    public void setLongitude(double longitude) {
        this.longitude = longitude;
    }
    private double longitude;
    public String getLocation() {
        return location;
    }
    public void setLocation(String location) {
        this.location = location;
    }
    public User getUser() {
        return user;
    }
    public void setUser(User user) {
        this.user = user;
    }
    public char[] getId() {
        // TODO Auto-generated method stub
        return null;
    }
}


@Controller    // This means that this class is a Controller
@RequestMapping(path="/Locations") // This means URL's start with /demo (after Application path)

public class LocationController {

        @Autowired // This means to get the bean called userRepository
                   // Which is auto-generated by Spring, we will use it to handle the data
        private LocationRepository locationRepository;
        private UserRepository userRepository;

         @RequestMapping("/create")
         @ResponseBody
         public Location create(@RequestBody Location location) {
           String locId = "";
           Location newLocation = new Location();
           try {
               User user = userRepository(location.getUser()); //Get the parent Object

               newLocation = new Location(); //Create a new Many object
               newLocation.setLatitude(location.getLatitude());
               newLocation.setLongitude(location.getLongitude());
               newLocation.setLocation(location.getLocation());
               newLocation.setUser(user);

               locationRepository.save(newLocation);
               locId = String.valueOf(newLocation.getId());

           }
           catch (Exception ex) {
            // return "Error creating the user: " + ex.toString();
               return newLocation;
           }
           return locationRepository.save(newLocation);
         }

        private User userRepository(User user) {
            // TODO Auto-generated method stub
            return null;
        }

        @GetMapping(path="/all")
        public @ResponseBody Iterable<Location> getAllLocations() {
            // This returns a JSON or XML with the users
            return locationRepository.findAll();
        }

}



  public interface LocationRepository extends CrudRepository<Location, Long>{

}

回应

    {
    "id": null,
    "location": "miraroad",
    "latitude": 15645,
    "user": null,
    "longitude": 154645
}

这里怎么了？

---

请帮忙。谢谢。

试试这个。这应该行得通

 @RequestMapping("/create")
 @ResponseBody
 public String create(@RequestBody User user) {
   String userId = "";
   try {
        userRepository.save(user);
        userId = String.valueOf(user.getId());
    } catch (Exception ex) {
          return "Error creating the user: " + ex.toString();
     }
        return "User succesfully created with id = " + userId;
  }

---

并在用户类中创建一个无参数构造函数

试试这个。这应该行得通

 @RequestMapping("/create")
 @ResponseBody
 public String create(@RequestBody User user) {
   String userId = "";
   try {
        userRepository.save(user);
        userId = String.valueOf(user.getId());
    } catch (Exception ex) {
          return "Error creating the user: " + ex.toString();
     }
        return "User succesfully created with id = " + userId;
  }

---

并在用户类中创建一个无参数构造函数

我想将创建的用户作为json对象返回，我怎么能呢@pvpkiranInstead字符串返回类型，只返回用户对象。Spring将自动将其转换为json。只需执行

returnuserrepository.save（user）

并将方法的签名更改为returnuserThank。。如何命名返回的json对象或json数组？将

@ManyToOne（fetch=FetchType.LAZY）

更改为

@ManyToOne（fetch=FetchType.EAGER）

没有帮助。尝试更改我想将创建的用户作为json对象返回，我如何才能@pvpkiranInstead字符串返回类型，只返回用户对象。Spring将自动将其转换为json。只需执行

returnuserrepository.save（user）

并将方法的签名更改为returnuserThank。。如何命名返回的json对象或json数组？将

@ManyToOne（fetch=FetchType.LAZY）

更改为

@ManyToOne（fetch=FetchType.EAGER）

没有帮助。试图改变

    {
    "id": null,
    "location": "miraroad",
    "latitude": 15645,
    "user": null,
    "longitude": 154645
}

 @RequestMapping("/create")
 @ResponseBody
 public String create(@RequestBody User user) {
   String userId = "";
   try {
        userRepository.save(user);
        userId = String.valueOf(user.getId());
    } catch (Exception ex) {
          return "Error creating the user: " + ex.toString();
     }
        return "User succesfully created with id = " + userId;
  }