Question

select count(cou_code) as Changes
from sdrp15_cosd    
where sd_code in
(select sd_code from sdrp15_submission_log
where QA_date is  null)

它给了我结果

更改| 629

select count(cou_code) as Complete
from sdrp15_cosd    
where sd_code in
(select sd_code from sdrp15_submission_log
where QA_date is not null)

它给了我结果

完成| 210

我希望有2列一个名为的更改，一个名为complete，其中两个查询（上图）合并为一个查询

Answer 1

进行两次条件计数（使用CASE），一次用于is null，一次用于is not null。

select count(case when sd_code in (select sd_code from sdrp15_submission_log
                                   where QA_date is null) then 1 end) as Changes,
       count(case when sd_code in (select sd_code from sdrp15_submission_log
                                   where QA_date is not null) then 1 end) as Complete
from sdrp15_cosd

Answer 2

尝试UNION

select count(cou_code) as Changes from sdrp15_cosd where sd_code in
(select sd_code from sdrp15_submission_log where QA_date is  null) 
UNION
select count(cou_code) as Complete from sdrp15_cosd where sd_code in
(select sd_code from sdrp15_submission_log where QA_date is not null)

加入2个subquerys

2 个答案: