我正在尝试通过复合键执行多个连接。 我使用别名来强制创建连接，但似乎Hibernate没有生成连接。我不知道为什么会这样。我可以通过本地SQL查询使其工作，但在使用条件时无法使其工作。
我怀疑这可能与映射复合键定义的方式有关（请参阅
以下是我的域模型类和查询信息。 欢迎任何想法 :) BusinessService
我怀疑这可能与映射复合键定义的方式有关（请参阅
BusinessServiceUser上的associationOverrides）。以下是我的域模型类和查询信息。 欢迎任何想法 :) BusinessService
@Entity
@Table(name = "business_services")
public class BusinessService extends AbstractEntity implements Serializable {
@Column(name = "name", unique = true, nullable = false, length = 255)
private String name;
@OneToMany(fetch = FetchType.LAZY, mappedBy = "pk.businessService", cascade = CascadeType.ALL)
@ForeignKey(name = "FK_BUSINESS_SERVICE_USERS")
private Set<BusinessServiceUser> businessServiceUsers = new HashSet<BusinessServiceUser>();
...
}
BusinessServiceUser
@Entity
@Table(name = "REL_BUSINESS_SERVICE_USER")
@AssociationOverrides({
@AssociationOverride(name = "pk.businessService", joinColumns = @JoinColumn(name = "BUSINESS_SERVICE_ID")),
@AssociationOverride(name = "pk.user", joinColumns = @JoinColumn(name = "USER_ID")) })
public class BusinessServiceUser implements Serializable {
private BusinessServiceUserId pk = new BusinessServiceUserId();
private Boolean master;
public BusinessServiceUser() {
}
@EmbeddedId
public BusinessServiceUserId getPk() {
return pk;
}
public void setPk(BusinessServiceUserId pk) {
this.pk = pk;
}
@Transient
public User getUser() {
return getPk().getUser();
}
public void setUser(User user) {
getPk().setUser(user);
}
@Transient
public BusinessService getBusinessService() {
return getPk().getBusinessService();
}
public void setBusinessService(BusinessService businessService) {
getPk().setBusinessService(businessService);
}
public boolean isMaster() {
return master;
}
public void setMaster(boolean master) {
this.master = master;
}
...
}
BusinessServiceUserId
@Embeddable
public class BusinessServiceUserId implements Serializable {
private User user;
private BusinessService businessService;
@ManyToOne
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
@ManyToOne
public BusinessService getBusinessService() {
return businessService;
}
public void setBusinessService(BusinessService businessService) {
this.businessService = businessService;
}
...
}
用户
@Entity
@Table(name = "USERS")
public class User extends AbstractEntity implements Serializable {
@Column(name = "first_name", nullable = false, length = 50)
private String firstName;
@Column(name = "last_name", nullable = false, length = 100)
private String lastName;
@Column(name = "email_address", unique = true, nullable = false, length = 150)
private String emailAddress;
@ManyToMany(fetch = FetchType.LAZY, cascade = CascadeType.DETACH, targetEntity = Role.class)
@JoinTable(name = "REL_USER_ROLE", joinColumns = @JoinColumn(name = "USER_ID", nullable = false) , inverseJoinColumns = @JoinColumn(name = "ROLE_ID", nullable = false) )
@ForeignKey(name = "FK_USER_ROLE")
private Set<Role> roles = new HashSet<Role>(0);
@OneToMany(fetch = FetchType.LAZY, mappedBy = "pk.user")
@ForeignKey(name = "FK_USER_BUSINESS_SERVICE")
private Set<BusinessServiceUser> businessServiceUsers = new HashSet<BusinessServiceUser>(0);
...
}
角色
@Entity
@Table(name = "role")
public class Role extends AbstractEntity implements Serializable {
@Enumerated(EnumType.STRING)
@Column(name = "name", unique = true, nullable = false)
private RoleType name;
@Column(name = "code", unique = true, nullable = false)
private String code;
@ManyToMany(fetch = FetchType.LAZY, mappedBy = "roles")
@ForeignKey(name = "FK_ROLE_USERS")
private List<User> users = new ArrayList<User>(0);
...
}
DAO条件查询
Criteria criteria = getSession().createCriteria(
BusinessServiceUser.class);
criteria.setFetchMode("pk.user", FetchMode.JOIN);
criteria.createAlias("pk.user", "userAlias", Criteria.LEFT_JOIN);
criteria.setFetchMode("pk.businessService", FetchMode.JOIN);
criteria.createAlias("pk.businessService", "bsAlias", Criteria.LEFT_JOIN);
criteria.setFetchMode("userAlias.roles", FetchMode.JOIN);
criteria.createAlias("userAlias.roles", "roleAlias");
criteria.add(Restrictions.eq("bsAlias.name", businessService.getName()));
criteria.add(Restrictions.eq("roleAlias.name", RoleType.ROLE1));
criteria.addOrder(Order.asc("master"));
return criteria.list();
SQL生成的查询
DEBUG org.hibernate.SQL -
select
this_.BUSINESS_SERVICE_ID as BUSINESS2_3_0_,
this_.USER_ID as USER3_3_0_,
this_.master as master3_0_
from
REL_BUSINESS_SERVICE_USER this_
where
bsalias2_.name=?
and rolealias3_.name=?
order by
this_.master asc
Hibernate:
select
this_.BUSINESS_SERVICE_ID as BUSINESS2_3_0_,
this_.USER_ID as USER3_3_0_,
this_.master as master3_0_
from
REL_BUSINESS_SERVICE_USER this_
where
bsalias2_.name=?
and rolealias3_.name=?
order by
this_.master asc
错误
java.sql.SQLException: ORA-00904: "ROLEALIAS3_"."NAME": invalid identifier
工作中的本地SQL查询
List<Object[]> result = getSession()
.createSQLQuery(
"select "
+ " bsu.BUSINESS_SERVICE_ID as bsId, "
+ " bsu.USER_ID as userId, "
+ " bsu.master as master, "
+ " bs.name as business_service, "
+ " u.first_name as first_name, "
+ " u.last_name as last_name, "
+ " u.email_address as email, "
+ " r.name as role "
+ "from "
+ " REL_BUSINESS_SERVICE_USER bsu "
+ " left outer join users u ON bsu.user_id = u.id "
+ " left outer join business_services bs ON bsu.business_service_id = bs.id "
+ " left outer join rel_user_role rur ON u.id = rur.user_id "
+ " left outer join role r ON rur.role_id = r.id "
+ "where "
+ " bs.name = '" + businessService.getName() + "' "
+ " and r.name like '" + RoleType.ROLE1 + "' "
+ "order by master asc")
.list();
规格
- Hibernate 3.6.10.Final
- JPA 2.0
- Spring 4.0.0
- Oracle JDBC驱动程序版本10.2.0.3.0