Hibernate Composite key Criteria Join

Question

我正在嘗試在復合鍵上執行多個連接。 我正在使用別名強制創建連接，但似乎連接不是由Hibernate生成的。 我不知道為什么會這樣。 我可以使用本機SQL查詢，但不能使用條件。

我懷疑它可能與復合鍵定義的映射方式有關（參見BusinessServiceUser上的associationOverrides）

以下是我的域模型類和查詢信息。 歡迎任何想法:)

商業服務

@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;
  }
...
}

用戶

@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標准查詢

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生成的查詢

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

錯誤

java.sql.SQLException: ORA-00904: "ROLEALIAS3_"."NAME": invalid identifier

使用本機SQL查詢

 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();

眼鏡

• Hibernate 3.6.10.Final
• JPA 2.0
• Spring 4.0.0
• Oracle JDBC驅動程序版本10.2.0.3.0

Answer 1

首先，你為什么不試着減少簡約的例子呢？ 您的樣本涉及許多實體和關系，為什么不減少它，即使只是為了您自己的故障排除時間？

其次，您的代碼不完整，它錯過了User和其他實體的id。 為了回答目的，我假設id已定義在某處。

我將提供沒有業務服務和角色的答案，我想類似的解決方案將適用。

我們如何解決這個問題？

首先，減少到最簡單的標准和實體集。 例如，對BusinessServiceUser.User.emailAddress的限制：

Criteria criteria = session.createCriteria(
            BusinessServiceUser.class, "bu");
criteria.setFetchMode("bu.pk.user", FetchMode.JOIN);
criteria.createAlias("bu.pk.user", "userAlias", Criteria.LEFT_JOIN);
criteria.add(Restrictions.eq("userAlias.emailAddress", "test@test.com"));

生成的SQL查詢：

select
    this_.BUSINESS_SERVICE_ID as BUSINESS3_33_0_,
    this_.USER_ID as USER2_33_0_,
    this_.master as master33_0_ 
from
    REL_BUSINESS_SERVICE_USER this_ 
where
    useralias1_.email_address=?

顯然，缺少預期的連接（因此您不需要復雜的示例來重現問題）。

查看BusinessServiceUserId，它使用@Embedded和@ManyToOne。 注意這是Hibernate特定的擴展，通常你不應該在@Embedded中使用@ManyToOne。 讓我們嘗試簡單查詢而不是標准：

    Query q = session.createQuery("from BusinessServiceUser as u left outer join u.pk.user where u.pk.user.emailAddress='test@test'");
    q.list();

生成的SQL：

select
    businessse0_.BUSINESS_SERVICE_ID as BUSINESS2_33_0_,
    businessse0_.USER_ID as USER3_33_0_,
    user1_.id as id54_1_,
    businessse0_.master as master33_0_,
    user1_.email_address as email2_54_1_,
    user1_.first_name as first3_54_1_,
    user1_.last_name as last4_54_1_ 
from
    REL_BUSINESS_SERVICE_USER businessse0_ 
left outer join
    USERS user1_ 
        on businessse0_.USER_ID=user1_.id 
where
    user1_.email_address='test@test'

哇，加入就在那里。 所以你在這里至少有一個解決方案 - 使用查詢而不是標准。 可以使用獲取連接等制作更復雜的查詢。

現在到了標准。 首先，讓我們檢查傳統的標准映射。 使用標准映射，您無法引用@Embedded中定義的@ManyToOne。 讓我們將映射添加到BusinessServiceUser類本身而不是@Transient

 @ManyToOne(fetch=FetchType.LAZY)
 public User getUser() {
   return getPk().getUser();
 }

請注意，此額外映射不會花費您。

Criteria criteria = session.createCriteria(
            BusinessServiceUser.class, "bu");
criteria.setFetchMode("bu.user", FetchMode.JOIN);
criteria.createAlias("bu.user", "userAlias", Criteria.LEFT_JOIN);
criteria.add(Restrictions.eq("userAlias.emailAddress", "test@test.com"));

生成的SQL：

select
    this_.BUSINESS_SERVICE_ID as BUSINESS3_33_1_,
    this_.USER_ID as USER2_33_1_,
    this_.master as master33_1_,
    this_.user_id as user2_33_1_,
    useralias1_.id as id54_0_,
    useralias1_.email_address as email2_54_0_,
    useralias1_.first_name as first3_54_0_,
    useralias1_.last_name as last4_54_0_ 
from
    REL_BUSINESS_SERVICE_USER this_ 
left outer join
    USERS useralias1_ 
        on this_.user_id=useralias1_.id 
where
    useralias1_.email_address=? 

所以在這里你得到了標准的解決方案2。 在實體中添加映射並在標准中使用它們而不是復雜的pk。

雖然我不知道將@EmbeddedId pk與@AssotiationOverride一起使用的設置，條件和連接提取的方式與您嘗試的完全相同，但可能它不是最好的方法。