Question

我正在使用Spring-mvc＆amp;我项目中的Spring-data-jpa。我有这两个实体

Location.java：

@Entity
@Table(name = "location")
@JsonIgnoreProperties(ignoreUnknown = true)
public class Location {
    @Id
    @GeneratedValue
    private int id;

    // other attributes...

    @ManyToOne(optional=true)
    @JoinColumn(name ="user")
    @JsonBackReference
    private User user;

    @ManyToOne(optional=true)
    @JoinColumn(name ="client")  
    @JsonBackReference
    private Client client;

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    // getters & setters

}

User.java：

@Entity
@Table(name = "users")
@JsonIgnoreProperties(ignoreUnknown = true)
public class User {

    @Id
    @GeneratedValue
    private int uid;

    // other attributes...

    @OneToMany(mappedBy="user")
    @JsonManagedReference
    private List<Location> locations;

    // getters & setters...

    @JsonIgnore
    public List<Location> getLocations() {
        return locations;
    }

    public void setLocations(List<Location> locations) {
        this.locations = locations;
    }

}

我想要做的是添加一个新位置并将其链接到现有用户。对于这两种权利，都有一个存储库和一个服务。 LocationRepository.java：

public interface LocationRepository extends CrudRepository<Location, Integer> {

    List<Location> findAll();

    //....

}

LocationService.java

@Service
public class LocationService {

    @Autowired
    private LocationRepository locationRepository;


    public List<Location> findAll() {
        return locationRepository.findAll();
    }

    public void save(Location location) {
        locationRepository.save(location);
    }

    @Secured("ROLE_ADMIN")
    public void delete(int locationId) {
        locationRepository.delete(locationId);
    }
}

ApiController.java

@Controller
@RequestMapping(value = "/rest/api")
public class ApiController {

    @Autowired
    UserService userService;
    @Autowired
    LocationService locationService;

    // Storing a new location

    @RequestMapping(value = "/locations/checkin", method = RequestMethod.POST, produces = "application/json")
    public ResponseEntity<?> checkin(@ModelAttribute("location") Location location) {
        locationService.save(location);
        List<Location> locationslist = locationService.findAll();
        return new ResponseEntity<List<Location>>(locationslist, HttpStatus.OK);
    }

}

直到现在一切似乎都很好。我将这个Restfull服务用于Android客户端，我所要做的就是创建一个新的Location对象并通过POST请求将其发送到正确的URI。

这是对象的结构：

{
    "id":1,
    ....,
    ....,
    "user":{
            "uid":1,
            "attr1":"value1",
            "attr2":"value2",
             .....
            }
}

问题：我总是遇到（org.hibernate.exception.SQLGrammarException）错误（由以下原因引起：org.postgresql.util.PSQLException：错误：语法错误在或附近＆＃34;用户＆＃34）

我做错了什么？什么是存储我的位置的最佳方法？

Answer 1

该异常表明生成了无效的SQL语句

org.postgresql.util.PSQLException: ERROR: syntax error at or near "user"

尝试重命名列＆＃39;用户＆＃39;用户＆＃39;用户＆＃39;是Postgres中的保留字，所以不是

@JoinColumn(name ="user")

使用类似

的内容

@JoinColumn(name ="user_location")

Spring MVC：使用ManyToOne / OneToMany关系存储数据

1 个答案: