我有四个table.entities,
我的场景或关系是,在一个项目中,一个用户有很多角色。 我很困惑在上面的hibernate实体中映射关系。 需要一些帮助。
提前致谢。
答案 0 :(得分:1)
有许多方法可以实现与Hibernate的所谓三元关系,既可以利用"simple" map,也可以使用中间实体/表来实现关联。您的用例决定哪个是正确的。这取决于数据的存储方式以及您将如何(通常)读取数据。报告也是一个重要的用例。
如果您想允许许多用户 - 项目 - 角色组合,地图(例如,在用户实体上)不是真正合适的,因为您的密钥可能是项目或角色,而密钥只能出现一次。尽管如此,每个关系条目在您的系统中应该是唯一的,因此在这种情况下,我倾向于使用至少一些唯一约束的中间表。
“快速而肮脏”的方式将是一个像以下的实体:
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
import javax.persistence.UniqueConstraint;
@Entity
@Table(uniqueConstraints = @UniqueConstraint(columnNames={"user_id", "project_id", "role_id"}))
public class UserProjectRoleSimple {
@Id
@GeneratedValue
private Long id;
@ManyToOne
private User user;
@ManyToOne
private Project project;
@ManyToOne
private Role role;
// you also need constructors, getters, equals, hashcode and stuff
}
另一种(更好的)方法是将关系对象标识符用作复合键。这有点冗长,但您不需要额外的代理键,因此您的联接表更清晰。
import java.io.Serializable;
import javax.persistence.Embeddable;
import javax.persistence.EmbeddedId;
import javax.persistence.Entity;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.validation.constraints.NotNull;
import org.hibernate.annotations.Immutable;
@Entity
@Immutable
public class UserProjectRole {
protected UserProjectRole() {
}
public UserProjectRole(final User user, final Project project, final Role role) {
this.userProjectRoleId = new UserProjectRoleId(user, project, role);
this.user = user;
this.project = project;
this.role = role;
}
@EmbeddedId
protected UserProjectRoleId userProjectRoleId;
@ManyToOne
@JoinColumn(name = "userId", insertable = false, updatable = false)
private User user;
@ManyToOne
@JoinColumn(name = "projectId", insertable = false, updatable = false)
private Project project;
@ManyToOne
@JoinColumn(name = "roleId", insertable = false, updatable = false)
private Role role;
public User getUser() {
return user;
}
public Project getProject() {
return project;
}
public Role getRole() {
return role;
}
@Embeddable
static class UserProjectRoleId implements Serializable {
private static final long serialVersionUID = 7994974851694559677L;
@NotNull
private Long userId;
@NotNull
private Long projectId;
@NotNull
private Long roleId;
protected UserProjectRoleId() {
}
protected UserProjectRoleId(final User user, final Project project, final Role role) {
this.userId = user.getId();
this.projectId = project.getId();
this.roleId = role.getId();
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((projectId == null) ? 0 : projectId.hashCode());
result = prime * result + ((roleId == null) ? 0 : roleId.hashCode());
result = prime * result + ((userId == null) ? 0 : userId.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
UserProjectRoleId other = (UserProjectRoleId) obj;
if (projectId == null) {
if (other.projectId != null)
return false;
} else if (!projectId.equals(other.projectId))
return false;
if (roleId == null) {
if (other.roleId != null)
return false;
} else if (!roleId.equals(other.roleId))
return false;
if (userId == null) {
if (other.userId != null)
return false;
} else if (!userId.equals(other.userId))
return false;
return true;
}
}
}
该映射也非常简单,唯一的特殊部分是@JoinColumn(name = "...Id", insertable = false, updatable = false)
添加。有了它,你可以使用你的映射实体(用于导航),而无需存储两次。