在 CrudRepository 中创建 JPQL 查询

Question

我正在尝试构建一个查询，该查询将使用JPQL从传递的User（想要传递登录的User）返回所有ToDoItems。

到目前为止，这是我的东西，不知道如何构建它：

public interface UserRepository extends CrudRepository<User, Long> {

   @Query(" from User u ")
   public Iterable<ToDoItem> findAllToDosForLoggedInUser(@Param("user") User user);

}

这是我的实体（省略了 setter 和 getter）： ToDoItem.java

@Entity
@Table (name = "TO_DO_ITEM")
public class ToDoItem extends BaseEntity {

   @Column(name = "TITLE", nullable = false)
   private String title;

   @Column(name = "COMPLETED")
   private boolean completed;

   @Column(name = "DUE_DATE", nullable = false)
   @Temporal(TemporalType.DATE)
   protected Calendar dueDate;

   // a ToDoItem is only associated with one user
   @ManyToOne(cascade=CascadeType.PERSIST)
   @JoinColumn(name = "USER_ID")
   private User user;


   // JPA demands empty contructor
   public ToDoItem() {}

   public ToDoItem(User user, String title, Calendar dueDate) {
      this.user = user;
      this.title = title;
      this.dueDate = dueDate;
   }
}

User.java

@Entity
@Table (name = "USERS")
public class User extends BaseEntity {

   @Column(name = "USERNAME")
   private String username;

   @Column(name = "PASSWORD")
   private String password;

   @Column(name = "EMAIL")
   private String email;

   // user can have many ToDoItems
   @OneToMany(cascade = CascadeType.ALL)
   private Set<ToDoItem> toDoItems;

   // JPA demands empty constructor
   public User() {}

   public User(String username, String password, String email) {
      this.username = username;
      this.password = password;
      this.email = email;
   }
}

有人可以检查我的实体是否正确吗？

Answer 1

您的查询应该是：

@Query(" select td from User u inner join u.toDoItems td where u = :user")

在您的用户实体中，您的映射应该具有mappedBy 属性：

@OneToMany(cascade = CascadeType.ALL, mappedBy="user")
private Set<ToDoItem> toDoItems;