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我正在尝试使用spring CRUDRepository使用 JPA 数据库模型（ User.java，UserInfo.java ）将新对象插入数据库。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）在开头设置。

当我使用CRUDRepository ( UserRepository.java ) 创建新对象时出现错误 - 无法将新对象插入第二个模型 ( UserInfo.java )，因为id为空(第一个模型已正确添加)。我认为问题在于数据库模型中的共享/映射复合主键。我用EntityManager尝试了相同的模型，这不是错误 - 全部都添加了。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。

谁能帮我找到解决方案？如果有人想运行代码，我还会在GitHub 上添加源代码。

下面的日志：

记录CRUD 存储库

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]
...

记录实体管理器

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 (?, ?, ?, ?)

下面的代码：

主要模型：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;
    /* ... */
}

春天 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;
    }
}
4

1 回答 1

2

和User类UserPK需要稍作修改。您的字段到主键的列映射应该在UserPK类内。以下是变化，

@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;
    }
}

对存储库的相应更改，

@Repository
public interface UserRepository extends CrudRepository<User, User.UserPK> {
    User findByIdAndAndType(Long id, String type);
}
于 2017-09-17T23:17:23.580 回答