Java SpringDataJPA-如何使用复合键插入子实体?
我有以下两个域模型Java SpringDataJPA-如何使用复合键插入子实体?,java,spring,hibernate,jpa,Java,Spring,Hibernate,Jpa,我有以下两个域模型 user ---- userid (PK) type like ------ likeid (PK) userid (PK) word like.userid引用User中的userid,是复合主键的一部分 并插入测试代码 @RunWith(SpringJUnit4ClassRunner.class) @SpringApplicationConfiguration(类=SpringJPatesApplication.class) @WebAppConfiguration
user
----
userid (PK)
type
like
------
likeid (PK)
userid (PK)
word
like.userid引用User中的userid,是复合主键的一部分
并插入测试代码
@RunWith(SpringJUnit4ClassRunner.class)
@SpringApplicationConfiguration(类=SpringJPatesApplication.class)
@WebAppConfiguration
@IntegrationTest(“服务器端口:0”)
公开课考试{
@自动连线
私有用户存储库用户存储库;
@org.junit.Test
public void应用程序()引发异常{
用户=新用户();
user.setType(user\u TYPE.user);
Like=新的Like();
LikeKey LikeKey=新的LikeKey();
likeKey.setliko(1L);
like.setLikeKey(likeKey);
比如。设置词(“word”);
like.setUser(user);
user.getLikes().add(like);
userRepository.save(用户);
}
}
我在运行JUnit测试时出错
Hibernate: insert into user (uno, user_type) values (null, ?)
Hibernate: insert into like (word, like_no, uno) values (?, ?, ?)
2016-07-22 13:44:36.499 WARN SqlExceptionHelper : SQL Error: 42001, SQLState: 42001
2016-07-22 13:44:36.499 ERROR SqlExceptionHelper : Syntax error in SQL statement "INSERT INTO LIKE[*] (WORD, LIKE_NO, UNO) VALUES (?, ?, ?) "; expected "identifier"; SQL statement: [42001-190]
首先,可能吗
第二,如果可能的话,我的代码中有什么错误
User.java
:
@Entity
@Table(name = "user")
@Data
public class User {
@Id
@GeneratedValue
@Column(name = "uno")
private Long userNo;
@Column(name = "user_type")
@Enumerated(EnumType.STRING)
private USER_TYPE type;
@OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
private List<Like> likes = new ArrayList<>();
public enum USER_TYPE {USER, ADMIN}
@Override
public String toString() {
return "User [userNo=" + userNo + ", type=" + type + ", likes=" + likes + "]";
}
}
@Entity
@Table(name = "like")
@Data
public class Like {
@EmbeddedId
private LikeKey likeKey;
@Column(name = "word")
private String word;
@MapsId(value = "userNo")
@ManyToOne(optional = false, targetEntity = User.class)
@JoinColumn(name = "uno")
private User user;
@Embeddable
@Data
@NoArgsConstructor
@AllArgsConstructor
public static class LikeKey implements Serializable {
private static final long serialVersionUID = 2116470670954200809L;
@Column(name = "likeNo")
private Long likeNo;
@Column(name = "uno")
private Long userNo;
}
}
我认为问题出在LIKE实体中,因为LIKE是一个Sql关键字 使用此代码:
@Entity
@Table(name = "like_table")
@Data
public class Like {
...
}
我认为问题在于LIKE实体,因为LIKE是一个Sql关键字 使用此代码:
@Entity
@Table(name = "like_table")
@Data
public class Like {
...
}