我正在将KodoJDO的应用程序移植到Hibernate。 我有一个查询,它遍历db中的4个表,以及java代码中的3个对象。
在英语中,查询是查找在系统X中拥有权利的用户。
我在用户对象上调用的JDOQL where子句是 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']
db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
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,并且会将该代码发布到另一天。
另一种带有隐式内部联接的变体,用于系统
的权利 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";
答案 0 :(得分:6)
你应该使用连接:
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,您必须考虑对象之间的对象和关系,而不是在表,外键和连接表方面。