HQL没有使用多个值“in”工作

时间:2014-10-28 07:05:29

标签: java hibernate

我有hibernate,就像这样。

String queryString = "from MFunction as mFunction where mFunction.functionKey in " 
            + "((select mRole.enterprise  from MRole as mRole where mRole.roleKey=:role), " 
            + "(select mRole2.project from MRole as mRole2 where mRole2.roleKey=:role ), " 
            + "(select mRole3.technology  from MRole  as mRole3 where mRole3.roleKey=:role ))";

但是hibernate查询只接受第一个值。

Hibernate: select mfunction0_.FunctionKey as Function1_73_, mfunction0_.`Add` as Add2_73_, mfunction0_.`Audit` as Audit3_73_, mfunction0_.ClientKey as ClientKey73_, mfunction0_.CreatedBy as CreatedBy73_, mfunction0_.CreatedTs as CreatedTs73_, mfunction0_.`Delete` as Delete7_73_, mfunction0_.`Edit` as Edit8_73_, mfunction0_.`Financial` as Financial9_73_, mfunction0_.FunctionName as Functio10_73_, mfunction0_.`General` as General11_73_, mfunction0_.Level as Level73_, mfunction0_.LevelKey as LevelKey73_, mfunction0_.LogicalDeleteTms as Logical14_73_, mfunction0_.UpdatedBy as UpdatedBy73_, mfunction0_.UpdatedTs as UpdatedTs73_, mfunction0_.`View` as View17_73_ from appanalytixdb.M_Function mfunction0_ where mfunction0_.FunctionKey in (select mrole1_.Enterprise from appanalytixdb.M_Role mrole1_ where mrole1_.RoleKey=?)

Hibernate版本:4.1.8.Final

1 个答案:

答案 0 :(得分:1)

您的问题是您创建了一个列表列表,您应该创建一个功能键列表。

我只看到两个选项:

  1. 更改一个' in' 3个与或相关的陈述。
  2. String queryString ="从MFunction作为mFunction,其中mFunction.functionKey(从MRole选择mRole.enterprise作为mRole,其中mRole.roleKey =:role)或mFunction.functionKey in(从MRole选择mRole2.project作为mRole2)其中mRole2.roleKey =:role)或mFunction.functionKey in(从MRole选择mRole3.technology为mRole3,其中mRole3.roleKey =:role)&#34 ;;

    1. 最佳解决方案是预先选择所有项目,技术,企业并将它们结合在一起(以避免重复)。但根据这篇文章Hibernate Union alternatives,hibernate还不支持union。 如果你的表中没有那么多条目,我会创建一个视图,并在其上使用我的查询。在这种情况下,视图可以非常便宜且易于使用。