获取别名的总和

如何解决获取别名的总和

试图计算别名 studentnotreviewed 的总和，解决这个问题的最佳方法是什么，我一直在尝试不同的东西。我想获取未审核的学生别名的总数。

select x.* from
(
select s.id as School,e.id as examID,e.exam_start as examstart,e.exam_end as examend,(select count(*) from students st where e.id=st.exam_id and (st.review_flag='' or st.review_flag is null)) as studentsnotreviewed,e.exam_end + Interval 2 day as 'DueDate',(select SUM(studentsnotreviewed)) as SUM
from exams e
#left join exams on st.exam_id = e.id
left join schools s on e.school_id=s.id

where e.exam_start < now() and e.exam_start>'2021-01-01' and e.practice=0) as x
where studentsnotreviewed>0 and (studentsnotreviewed>15 or examend < now())
order by duedate asc,studentsnotreviewed desc

[样本数据]

学校	考试ID	考试开始	考试	studentsnotreviewed	截止日期	总和
343	458092	2021 年 3 月 16 日下午 3:52	2017 年 8 月 28 日上午 12:59	2	2017 年 8 月 30 日，上午 12:59	2
125	360216	2021 年 1 月 7 日上午 11:55	2018 年 12 月 13 日晚上 11:17	27	2018 年 7 月 28 日晚上 11:17	27

[想要的结果] 未审核学生的别名总和。

未审核的学生总数
29

解决方法

我相信元数据库支持 WITH，因此您可以将整个查询别名为临时数据块，然后将其作为表重用：

with x as
(
  select 
    s.id as School,e.id as examID,e.exam_start as examstart,e.exam_end as examend,(select count(*) from students st where e.id=st.exam_id and (st.review_flag='' or st.review_flag is null)) as studentsnotreviewed,e.exam_end + Interval 2 day as 'DueDate',from
    exams e
    left join exams on st.exam_id = e.id
    left join schools s on e.school_id=s.id
  where e.exam_start < now() and e.exam_start>'2021-01-01' and e.practice=0
) 

select
  x.*,(select SUM(studentsnotreviewed) from x) as SUM
from
  x
where 
  studentsnotreviewed>0 and 
  (studentsnotreviewed>15 or examend < now())
order by 
  duedate asc,studentsnotreviewed desc

您的查询似乎有一个轻微的语法错误 - 您在外部查询中提到了 # 而没有提到 students st ..我相信你可以解决这个问题

我假设您希望 29 作为在结果集中重复的值。如果你真的只想要一个 29 的值，我想它可以通过以下方式给出：

select count(*) 
from 
  exams e 
  inner join students st on e.id=st.exam_id 
where
  (st.review_flag='' or st.review_flag is null) and
  e.exam_start < now() and e.exam_start>'2021-01-01' and 
  e.practice=0