我尝试:(查看代码),但它不起作用。
List<User> users = query.getResultList();
if (users.isEmpty()) {
System.out.println("tworzę konto admina");
User admin = new User();
admin.set(...)
Role role = new Role();
role.setName("Develop");
role.set...
// the first method dont work
admin.getRoles().add(role); /// but admin.getRoles() IS Null!!! (I get nullPointerExeption)
// the second method dont work too
// List<Role> roles = new ArrayList<Role>();
//roles.add(role);
//admin.setRoles(roles);
entityManager.persist(role);
entityManager.persist(admin);
}
用户类别:
public class User {
@ManyToMany
@JoinTable(
name = "role_has_user",
joinColumns = {
@JoinColumn(name = "user_id")
},
inverseJoinColumns = {
@JoinColumn(name = "role_id")
}
)
}
在第二种方法中我得到:
<小时/>Local Exception Stack:
Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.5.0.v20130507-3faac2b): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'rw.SEQUENCE' doesn't exist
Error Code: 1146
Call: UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT + ? WHERE SEQ_NAME = ?
bind => [2 parameters bound]
Query: DataModifyQuery(name="SEQUENCE" sql="UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT + ? WHERE SEQ_NAME = ?")
at org.eclipse.persistence.exceptions.DatabaseException.sqlException(DatabaseException.java:331)...
最佳答案
您的实体无法持久保存,因为您的 JPA 提供程序无法为它们生成 Id。您可能使用了 @GenerateValue 注释,但没有在底层数据库上创建表 SEQUENCE 。您使用模式生成吗?如果不是,则创建包含 SEQ_COUNT
SEQ_NAME
列的适当表。
关于java - JPA,如何插入具有ManyToMany关系的新实体?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24804586/