Question

我正在尝试使用 spring CRUDRepository 使用JPA数据库模型将新对象插入数据库（ User.java ， UserInfo.java ）。数据库模型与复合主键（ UserPK.java ）相关，其中一个是自动生成的（字段名为 id ），第二个（字段名为 type < / strong>）在开头设置。

使用 CRUDRepository （ UserRepository.java ）创建新对象时出错 - 无法将新对象插入第二个模型（ UserInfo.java < / strong>），因为 id 为null（正确添加了第一个模型）。我认为问题在于在数据库模型中共享/映射复合主键。我用 EntityManager 尝试了相同的模型并且它不是错误 - 所有都被添加了。接下来我使用了 @PrimaryKeyJoinColumns 注释，但结果与上面相同（但我不确定我是否正确使用它） - CRUDRepository 失败并且 EntityManager 成功。

任何人都可以帮我找到解决方案吗？如果有人想要运行代码，我还会在GitHub上添加源代码。

以下日志：

记录 CRUDRepository

Hibernate: select user0_.id as id1_0_1_, user0_.type as type2_0_1_, user0_.email as email3_0_1_, user0_.login as login4_0_1_, userinfo1_.id as id3_1_0_, userinfo1_.type as type4_1_0_, userinfo1_.name as name1_1_0_, userinfo1_.surname as surname2_1_0_ from user user0_ left outer join user_info userinfo1_ on user0_.id=userinfo1_.id and user0_.type=userinfo1_.type where user0_.id=? and user0_.type=? Hibernate: call next value for seq_id Hibernate: select userinfo0_.id as id3_1_0_, userinfo0_.type as type4_1_0_, userinfo0_.name as name1_1_0_, userinfo0_.surname as surname2_1_0_ from user_info userinfo0_ where userinfo0_.id=? and userinfo0_.type=? Hibernate: insert into user (email, login, id, type) values (?, ?, ?, ?) Hibernate: insert into user_info (name, surname, id, type) values (?, ?, ?, ?) WARN 17653 --- [nio-8080-exec-3] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 23502, SQLState: 23502 ERROR 17653 --- [nio-8080-exec-3] o.h.engine.jdbc.spi.SqlExceptionHelper : NULL not allowed for column "ID"; SQL statement: insert into user_info (name, surname, id, type) values (?, ?, ?, ?) [23502-196] ...

记录 EntityManager

Hibernate: call next value for seq_id Hibernate: insert into user (email, login, id, type) values (?, ?, ?, ?) Hibernate: insert into user_info (name, surname, id, type) values (?, ?, ?, ?)

以下代码：

主要型号： User.java

import com.fasterxml.jackson.annotation.JsonManagedReference; import javax.persistence.*; import javax.validation.constraints.NotNull; import java.io.Serializable; @Entity @Table(name = "USER") @IdClass(UserPK.class) public class User implements Serializable { @OneToOne(cascade = CascadeType.ALL, mappedBy = "user") @JsonManagedReference private UserInfo info; @Id @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "SEQ_ID") @SequenceGenerator(name = "SEQ_ID", sequenceName = "SEQ_ID", allocationSize = 1) @NotNull @Column(name = "ID") private Long id; @Id @NotNull @Column(name = "TYPE") private String type; @NotNull @Column(name = "LOGIN") private String login; @NotNull @Column(name = "EMAIL") private String email; /* ... */ }

第二个模型： UserInfo.java

import com.fasterxml.jackson.annotation.JsonBackReference; import javax.persistence.*; import java.io.Serializable; @Entity @Table(name = "USER_INFO") public class UserInfo implements Serializable { @Id @OneToOne(cascade = CascadeType.ALL) @JoinColumns({ @JoinColumn(name = "id", referencedColumnName = "id"), @JoinColumn(name = "type", referencedColumnName = "type") }) @JsonBackReference @MapsId private User user; @Column(name = "NAME") private String name; @Column(name = "SURNAME") private String surname; /* ... */ }

组合主键： UserPK.java

import java.io.Serializable; public class UserPK implements Serializable { private Long id; private String type; /* ... */ }

spring CRUDRepository ： UserRepository.java

import org.springframework.data.repository.CrudRepository; import org.springframework.stereotype.Repository; @Repository public interface UserRepository extends CrudRepository<User, UserPK> { User findByIdAndAndType(Long id, String type); }

使用 EntityManager 的存储库： UserRepositoryEM.java

import org.springframework.stereotype.Repository; import javax.persistence.EntityManager; import javax.persistence.PersistenceContext; import javax.transaction.Transactional; @Repository @Transactional public class UserRepositoryEM { @PersistenceContext private EntityManager entityManager; public User findByKey(UserPK key) { return entityManager.find(User.class, key); } public User save(User user) { entityManager.persist(user); entityManager.flush(); return user; } }

Answer 1

name = input(...) if name in student: del student[name] else: print("Invalid name")和User类需要稍加修改。主键的字段到列映射应位于UserPK类中。以下是更改，

UserPK

对存储库的相应更改

@Entity
@Table(name = "USER")
public class User implements Serializable {

    @OneToOne(cascade = CascadeType.ALL, mappedBy = "user")
    @JsonManagedReference
    private UserInfo info;

    @EmbeddedId
    private UserPK id;

    @Column(name = "LOGIN", nullable = false)
    private String login;

    @Column(name = "EMAIL", nullable = false)
    private String email;
    /* ... */

    @Embeddable
    @SequenceGenerator(name = "SEQ_ID",  initialValue=1, allocationSize=100)
    public static class UserPK implements Serializable {

        @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "SEQ_ID")
        @Column(name = "ID", nullable = false)
        private Long id;

        @Column(name = "TYPE", nullable = false)
        private String type;
    }
}

Spring CrudRepository无法在复合主键中共享生成的id

1 个答案: