我如何移植从JDOQL到HQL的多个表中的联接进行端口查询

Question

我正在将KodoJDO的应用程序移植到Hibernate。我有一个查询，该查询跨越数据库中的4个表和Java代码中的3个对象。

用英语查询是查找在系统X中具有权利的用户。

我在用户对象上调用的JDOQL where子句是其中entitlements.contains（ent）&&（upper（ent.system.id）='EVPN'）

执行查询的一些SQL是：

 select unique(u.id)
 from USER u, USERENTITLEMENT ue, ENTITLEMENT e, SYSTEM s
 where u.id = ue.userid
 and ue.entitlementid = e.id
 and e.systemid = s.id
 and s.id = 'evpn'

我对HQL的最佳猜测给了我一个例外

org.hibernate.hql.ast.QuerySyntaxException: unexpected AST node: ( [select user from com.ebig.entity.User as user, com.ebig.entity.Entitlement as ent, com.ebig.entity.System as sys where  entitlements.contains(ent) and ent.system = sys and sys.id  = 'evpn']

数据库的结构如下：

 User
   id

 UserEntitlement
   userid
   entitlementid

 Entitlement
   id
   systemid

 System
   id

Java代码的结构如下：

 class User 
 {
    String id;
    Set<Entitlement> entitlements;
 }

 class Entitlement
 {
    String id;
    System system;
 }

 class System
 {
    String id;
 }

更新我的最终查询有效

            hqlQuery = "select distinct user from User as user "+
        "inner join user.entitlements as entitlement inner join entitlement.system as system "+
        "where  system.id  = 'evpn'  AND mod(user.flags, 2) = 0  AND source = 1";

是的，我知道我应该使用参数，但是我有很多问题要解决，并且将代码推迟一天。

具有隐式内部联接以获取系统权利的另一种形式

        hqlQuery = "select distinct user from User as user "+
        "inner join user.entitlements as entitlement "+
        "where  entitlement.system.id  = 'evpn'  AND mod(user.flags, 2) = 0  AND source = 1";

Answer 1

select distinct u.id from User u
inner join u.entitlements as entitlement
inner join entitlement.system as system
where system.id = :evpn

其中：evpn是您必须绑定的命名参数。

[在执行HQL时必须考虑对象和对象之间的关系，而不要考虑表，外键和联接表。