我英语说得不太好,所以我用了translate.google)。
我有mysql数据库:
以及我的实体:
角色.java
@Entity
public class Role {
private Integer id;
private String roleName;
@Id
@Column(name = "id", nullable = false)
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
@Basic
@Column(name = "role_name", nullable = false, length = 255)
@GeneratedValue(strategy = GenerationType.IDENTITY)
public String getRoleName() {
return roleName;
}
public void setRoleName(String roleName) {
this.roleName = roleName;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Role role = (Role) o;
return Objects.equals(id, role.id) &&
Objects.equals(roleName, role.roleName);
}
@Override
public int hashCode() {
return Objects.hash(id, roleName);
}
}
用户.java
@Entity
public class User {
private Integer id;
private String login;
private String email;
private String password;
@Id
@Column(name = "id", nullable = false)
@GeneratedValue(strategy = GenerationType.IDENTITY)
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
@Basic
@Column(name = "login", nullable = false, length = 255)
public String getLogin() {
return login;
}
public void setLogin(String login) {
this.login = login;
}
@Basic
@Column(name = "email", nullable = false, length = 50)
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
@Basic
@Column(name = "password", nullable = true, length = 255)
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
User user = (User) o;
return Objects.equals(id, user.id) &&
Objects.equals(login, user.login) &&
Objects.equals(email, user.email) &&
Objects.equals(password, user.password);
}
@Override
public int hashCode() {
return Objects.hash(id, login, email, password);
}
public User() {
}
public User(String login, String email, String password) {
this.login = login;
this.email = email;
this.password = password;
}
}
用户角色.java
@Entity
@Table(name = "user_role", schema = "liverpool_site", catalog = "")
public class UserRole {
private Integer id;
private Integer userId;
private Integer roleId;
@Id
@Column(name = "id", nullable = false)
@GeneratedValue(strategy = GenerationType.IDENTITY)
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
@Basic
@Column(name = "user_id", nullable = false)
public Integer getUserId() {
return userId;
}
public void setUserId(Integer userId) {
this.userId = userId;
}
@Basic
@Column(name = "role_id", nullable = false)
public Integer getRoleId() {
return roleId;
}
public void setRoleId(Integer roleId) {
this.roleId = roleId;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
UserRole userRole = (UserRole) o;
return Objects.equals(id, userRole.id) &&
Objects.equals(userId, userRole.userId) &&
Objects.equals(roleId, userRole.roleId);
}
@Override
public int hashCode() {
return Objects.hash(id, userId, roleId);
}
public UserRole() {
}
public UserRole(Integer userId, Integer roleId) {
this.userId = userId;
this.roleId = roleId;
}
}
好吧,我的问题是:
现在看看这些表是如何连接的,它可以用作jpa存储库,如下所示:
public class UserRoleJpa{
private Integer idUser;
private String login;
private String email;
private String roleName;
//getter and setter methods
}
public interface UserRoleJpa extends JpaRepository<UserRoleJpa, String> {
//for example
List<UserRoleJpa> findAll();
}
我将非常高兴的任何建议和方法如何实施这一点。
如果可能,使用spring数据(jpa)
1条答案
按热度按时间2nbm6dog1#
我认为最好定义两个条目user和role,然后用
ManyToMany
将创建包含用户id和角色的表user\u role的注解您只需将其添加到您的用户实体
而且不需要userrole实体