有人帮我看一下这个语句,要咋改才正确:
from com.kmi.expert.model.Userinfo as u left outer join (select b from com.kmi.expert.model.Userinfo a left outer join a.Usergroupinfos b where b.groupoid=1) as ug 错误信息:
org.hibernate.hql.ast.QuerySyntaxException: unexpected token: ( near line 1, column 58 [from com.kmi.expert.model.Userinfo as u left outer join (select b from com.kmi.expert.model.Userinfo a left outer join a.Usergroupinfos b where b.groupoid=1) as ug]
at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:31)
以下的略....
正确的SQL如下:
select * from userinfo a left join (
select b.* from userinfo a left outer join usergroupinfo b on a.oid=b.userinfooid where b.groupoid=1)
b on a.oid = b.userinfooid
根据SQL写一个正确对应的HQL也行
from com.kmi.expert.model.Userinfo as u left outer join (select b from com.kmi.expert.model.Userinfo a left outer join a.Usergroupinfos b where b.groupoid=1) as ug 错误信息:
org.hibernate.hql.ast.QuerySyntaxException: unexpected token: ( near line 1, column 58 [from com.kmi.expert.model.Userinfo as u left outer join (select b from com.kmi.expert.model.Userinfo a left outer join a.Usergroupinfos b where b.groupoid=1) as ug]
at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:31)
以下的略....
正确的SQL如下:
select * from userinfo a left join (
select b.* from userinfo a left outer join usergroupinfo b on a.oid=b.userinfooid where b.groupoid=1)
b on a.oid = b.userinfooid
根据SQL写一个正确对应的HQL也行
解决方案 »
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