java - 如何通过 hibernate 从多表查询中获取数据？

Question

我有一个用户管理应用程序，它为每个用户分配一个团队和一个或多个对不同应用程序的访问权限。现在对于报告页面，我试图通过 Hibernate 从两个表(UserInfo 和 UserAppAccess)中获取数据，但我不能。 这是表格:

表 1(用户信息):

@Entity
@Table(name = "user_info", uniqueConstraints = { @UniqueConstraint(columnNames =   "username"), @UniqueConstraint(columnNames = "email") })
public class UserInfo implements java.io.Serializable {

    public enum UserStatus {
        active, inactive
    }

    public enum UserType {
        user, creator
    }

    private static final long serialVersionUID = 2650114334774359089L;

    @Id
    @Column(name = "id", unique = true, nullable = false, length = 100)
    private String id;

    @Column(name = "username", unique = true, nullable = false, length = 50)
    private String username;

    @Column(name = "password", nullable = false, length = 80)
    private String password;

    @Column(name = "status", nullable = false, length = 10)
    @Enumerated(EnumType.STRING)
    private UserStatus status;

    @Column(name = "type", nullable = false, length = 10)
    @Enumerated(EnumType.STRING)
    private UserType type;

    @Column(name = "phone", nullable = true, length = 30)
    private String phone;

    @Column(name = "email", nullable = true, length = 50)
    private String email;

    @Column(name = "first_name", nullable = true, length = 50)
    private String firstName;

    @Column(name = "last_name", nullable = true, length = 50)
    private String lastName;

    @Column(name = "login", nullable = true, length = 100)
    private long login;

    @Column(name = "alert", nullable= true, length=500)
    private String alert;

    @OneToOne
    @JoinColumn(name = "team_id")
    private Team team;
}

表 2(团队):

@Entity
@Table(name = "team", uniqueConstraints = { @UniqueConstraint(columnNames = "team_name"), @UniqueConstraint(columnNames = "team_code") })
public class Team implements java.io.Serializable {

    private static final long serialVersionUID = 7933770163144650730L;

    @Id
    @Column(name = "id", unique = true, nullable = false, length = 80)
    private String id;

    @Column(name = "team_name", unique = true, nullable = false, length = 100)
    private String name;

    @Column(name = "team_code", unique = true, nullable = false, length = 10)
    private String code;
}

表 3(访问):

@Entity
@Table(name = "access_def")
public class Access implements java.io.Serializable {

    private static final long serialVersionUID = 7933770163144650730L;

    @Id
    @Column(name = "id", unique = true, nullable = false, length = 80)
    private String id;

    @Column(name = "access_name", unique = true, nullable = false, length = 100)
    private String name;

    @Column(name = "access_code", unique = true, nullable = false, length = 10)
    private String code;
}

表4(应用):

@Entity
@Table(name = "application", uniqueConstraints = { @UniqueConstraint(columnNames = "name") })

public class Application implements java.io.Serializable {

    private static final long serialVersionUID = 5803631085624275364L;

    @Id
    @Column(name = "name", nullable = false, length = 100)
    private String name;
}

表 5(用户应用访问):

@Entity
@Table(name = "user_app_access")
@Embeddable
public class UserAppAccess implements java.io.Serializable {

    private static final long serialVersionUID = 7933770163144650730L;

    @Id
    @Column(name = "id", unique = true, nullable = false, length = 80)
    private String id;

    @OneToOne
    @JoinColumn(name = "user_id")
    private UserInfo userInfo;

    @Column(name = "app_name", nullable = false, length = 100)
    private String appName;

    @OneToOne
    @JoinColumn(name = "access_id")
    private Access access;
}

我有一个报告页面，允许管理员选择多个选项(例如:列出团队测试和应用程序 APP1 中的所有 Activity 用户)。

这是我获取数据的代码，但它不起作用:

public List<?> getReport(String teamId,String appName,UserStatus active,UserStatus inactive) {
    Session session = sessionFactory.getCurrentSession();
    String hql = "SELECT u.firstName,u.username,u.status,u.lastName,u.phone,u.team  From  UserInfo u,AppAccess a WHERE u.status =? OR u.status =? AND u.team.id = ? AND a.appName = :appName ";
    Query query = session.createQuery(hql);
    query.setParameter(0, active);
    query.setParameter(1, inactive);
    query.setParameter(2, teamId);
    query.setParameter("appName", appName);
    System.out.println(query.list());
    return query.list();
}

例如当我通过

• 活跃用户:活跃
• 不活跃的用户:null
• 团队:测试

• 应用程序:app1

  teamId :28f66133-26c3-442b-a071-4d19d64ec0aeappName :app1active :activeinactive:null

我正在从我的 return query.list() 中取回它；

[[Ljava.lang.Object;@2961116f, [Ljava.lang.Object;@23bfa3a2, [Ljava.lang.Object;@7a8ff303, [Ljava.lang.Object;@9b88d2, [Ljava.lang.Object;@6333934d, [Ljava.lang.Object;@4f0bd71c, [Ljava.lang.Object;@125797cf, [Ljava.lang.Object;@34afa071, [Ljava.lang.Object;@764e75bc, [Ljava.lang.Object;@1913c652, [Ljava.lang.Object;@61413e5a, [Ljava.lang.Object;@264b898, [Ljava.lang.Object;@22930462, [Ljava.lang.Object;@6204cfa9, [Ljava.lang.Object;@29dd9285, [Ljava.lang.Object;@11be6f3c, [Ljava.lang.Object;@6d78d53d, [Ljava.lang.Object;@17f7cff1, [Ljava.lang.Object;@e74e382, [Ljava.lang.Object;@1c047338, [Ljava.lang.Object;@68286fe6, [Ljava.lang.Object;@36ca9a76, [Ljava.lang.Object;@2f62d514, [Ljava.lang.Object;@1932c5a, [Ljava.lang.Object;@6544c984, [Ljava.lang.Object;@70a2d0d, [Ljava.lang.Object;@2d13b417, [Ljava.lang.Object;@6894691f, [Ljava.lang.Object;@6781a7dc, [Ljava.lang.Object;@7133919a]  

Answer 1

我建议使用 native SQL 和 JDBC 进行报告(请参阅 How should I use Hibernate Mapping while dealing with huge data table )

出于性能原因，最好从 DAO 中的结果集中创建 View 模型对象。它可能看起来像是混合了抽象级别( View 层和持久层)，但是当您需要获取大量数据并且不想进行从持久性模型到 View 模型的不必要的对象转换时，这是可以的。

如果你想坚持 hibernate ，你可以定义一个合成实体并在 View 上映射，只包含来自多个的必要列:

@Entity
@Table("V_USER_REPORT")
public class UserAppData {
   // columns from table "user"
   @Id
   @Column(name = "id", unique = true, nullable = false, length = 100)
   private String id;
   @Column(name = "username", unique = true, nullable = false, length = 50)
   private String username;

   // columns from table "user"
   @Column(name = "app_name", nullable = false, length = 100)
   private String appName;

   // columns from table "team"
   @Column(name = "team_id", unique = true, nullable = false, length = 80)
   private String team_id;
   @Column(name = "team_name", unique = true, nullable = false, length = 100)
   private String name;
   @Column(name = "team_code", unique = true, nullable = false, length = 10)
   private String code;

   // and so on...
}

然后您可以像处理普通实体一样通过参数获取此类实体。