Aar*_*ron 1 java port hibernate hql
我正在将一个KodoJDO应用程序移植到Hibernate.我有一个查询,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中具有权利的用户.
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'
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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']
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db的结构如下:
User
id
UserEntitlement
userid
entitlementid
Entitlement
id
systemid
System
id
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java代码结构如下:
class User
{
String id;
Set<Entitlement> entitlements;
}
class Entitlement
{
String id;
System system;
}
class System
{
String id;
}
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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";
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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.
具有隐式内部联接的另一种变体,用于授予系统权利
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";
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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
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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.
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