EclipseLink JPA：列出带有引用变量的实体

Question

我正在使用 JPA 的 eclipseLink 对我的实体执行 CRUD 操作。我面临以下问题：

我在数据库中有两个 table：

CREATE TABLE User (
   id INTEGER PRIMARY KEY AUTO_INCREMENT,
   username VARCHAR(30) NOT NULL UNIQUE,
   email VARCHAR(50) NOT NULL UNIQUE,
   password VARCHAR(255) NOT NULL,
   signUpDate timestamp NOT NULL DEFAULT NOW()
);

CREATE TABLE Friendship (
   id INTEGER PRIMARY KEY AUTO_INCREMENT,
   friendsSince timestamp NOT NULL DEFAULT NOW(),
   user1_Id INTEGER NOT NULL REFERENCES User(id),
   user2_Id INTEGER NOT NULL REFERENCES User(id)
);

对应实体

@Entity
public class User {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    private String username;
    private String email;
    private String password;
    @Temporal(value = TemporalType.DATE)
    private Date signUpDate;
    // constructors & setters & getters ...
}

@Entity
public class Friendship {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    @ManyToOne
    @JoinColumn(name="user1_Id", referencedColumnName = "id")
    private User user1;
    @ManyToOne
    @JoinColumn(name="user2_Id", referencedColumnName = "id")
    private User user2;
    @Temporal(value = TemporalType.DATE)
    private Date friendsSince;

    // constructors & setters & getters ... 
}

如果我想检索一些实体的列表，根据查询的 "WHERE" 子句，我会得到这个 "unknown state or association field [user1_Id] of class [com.filip.xxx.Friendship]" 错误。

具体来说：我尝试构建此查询：

Query query = mgr.createQuery("select f.id ,f.friendsSince, f.user1_Id, f.user2_Id from Friendship f where f.user1_Id = :user1Id and f.user2_Id = :user2Id or f.user1_Id = :user11Id and f.user2_Id = :user12Id");

并收到此异常：

java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Error compiling the query [select f.id ,f.friendsSince, f.user1_Id, f.user2_Id from Friendship f where f.user1_Id = :user1Id and f.user2_Id = :user2Id or f.user1_Id = :user11Id and f.user2_Id = :user12Id], line 1, column 31: unknown state or association field [user1_Id] of class [com.filip.xxx.Friendship].

似乎将属性映射回实体有问题，因为我对持久化这两个实体没有问题。

有趣的是，如果我运行这个查询：

Query query = mgr.createQuery("select f from Friendship f");

它 return 是我所有友谊实体的正确列表。

注意友情实体(user1, user2)中引用变量的名称与对应的table的变量名称(user1_Id,user2_Id)不同。在我在实体中使用与 table 中相同的变量名称之前，但在持久友谊实体中收到此错误：

javax.persistence.PersistenceException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.3.2.v20111125-r10461):
org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'USER1_ID' in 'field list'
Error Code: 1054
Call: INSERT INTO FRIENDSHIP (FRIENDSSINCE, USER1_ID, USER2_ID) VALUES (?, ?, ?)
bind => [3 parameters bound]

基本不明白，为什么eclipse link在创建sql的时候要重命名实体的引用变量(user1 -> USER1_ID, user2 -> USER2_ID)查询，当它有问题时将其映射回实体。

我已经尝试过这些解决方案：构建查询和 return user1_Id 列作为 user1 和 user2_Id 作为 user2

select f.id ,f.friendsSince, f.user1_Id as user1, f.user2_Id as user2 from Friendship f where f.user1_Id = :user1Id and f.user2_Id = :user2Id or f.user1_Id = :user11Id and f.user2_Id = :user12Id

但收到与上述相同的 IllegalArgumentException。

你能帮我解决这个问题吗？

谢谢

Answer 1

异常

unknown state or association field [user1_Id] of class [com.filip.xxx.Friendship]

收到是因为您使用的是 user1_Id 名称，它是一个数据库列名称。

另一方面，ElementManager.createQuery() 方法需要接受实体字段名称的 JPQL 字符串 user1。尝试将您的查询字符串替换为：

select f.id, f.friendsSince, f.user1, f.user2
from Friendship f 
where f.user1 = :user1Id and 
      f.user2 = :user2Id or 
      f.user1 = :user11Id and 
      f.user2 = :user12Id

EclipseLink JPA：列出带有引用变量的实体

EclipseLink JPA: list entities with reference variables

java

jpa

eclipselink

entitymanager