Question

因此，我为此一直苦苦挣扎了大约8个小时，这是我整个移动应用程序中的最后一部分。请帮助我将此伪代码转换为有效的HQL。我在底部的内存数据库中有一个可以在H2中运行的语句，但是我的prod数据库是oracle，它很糟糕。我正在尝试获取排行榜并添加base64图像。即使我删除了下面的唯一工作H2语句，它也仍然无法运行，并显示错误`

不一致的数据类型：预期-获得CLOB

。

声明我无法在HQL工作：

select a.*, b.image 
from (select distinct score.staffId as staffId, 
      sum(score.totalScore) as totalScore, 
      sum(score.timeTaken) as timeTaken from Score score) a 
join User b 
on a.staffId=b.staffId"

H2中的工作声明：

"select distinct score.staffId as staffId, 
        sum(score.totalScore) as totalScore, 
        sum(score.timeTaken) as timeTaken, 
        user.image as image"
+ " from Score score, User user"
+ " where score.staffId = user.staffId"
+ " group by score.staffId order by totalScore desc, timeTaken asc"

Answer 1

尽管我对HQL一无所知，但我将从另一个角度尝试回答您的问题。

将查询重写为...

select a.*, b.image
from (
        select staffId,
            sum(totalScore) as totalScore,
            sum(timeTaken) as timeTaken
        from Score
        group by staffId
    ) a
    join User b
        on a.staffId = b.staffId

...它应该可以工作。

如果仍然没有，那么您刚刚见证了称为“复杂视图合并”的Oracle查询优化器功能的奇妙之处，并且您将不得不通过no_merge提示来稍微帮助Oracle DB ... < / p>

select a.*, b.image
from (
        select --+ no_merge
            staffId,
            sum(totalScore) as totalScore,
            sum(timeTaken) as timeTaken
        from Score
        group by staffId
    ) a
    join User b
        on a.staffId = b.staffId

对于您的ORM，请勿将其用于数据库读取，而应使用普通SQL或jOOQ。 ORMs are perfect at writing stuff to DB, not at reading it.

使用休眠将Oracle db中的两个表联接在一起-一个表具有clob数据

1 个答案: