加入2个subquerys

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)

它给我的结果加入2个subquerys

变化| 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)

它给我造成

完成| 210

我想有2个一个名为完成这两者的查询（上图）合并成一个查询

来源

2015-09-03 chihao

MySQL和Oracle是两种不同的产品，略有不同版本的SQL。你在使用哪一个？ – jarlh

您可以添加一些样本数据（涉及所有表格）和预期结果。 – jarlh

@jarlh我相信我正在使用oracle sql – chihao

做两个条件计数（与CASE），一个用于is null的命名改一另一个用于is not null's。

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

来源

2015-09-03 13:11:38 jarlh

谢谢，但现在如何结合您的查询（上面）和这一个： select a.phase，a.st_code ||' - '|| b.state_name， case when a.submission_received_dt is not null then'Y'else'N'end as Changes from pcspro.sdrp15_return a， pcspro.sdrp15_states_ready b 其中a.phase = b.phase和a.st_code = b.state; – chihao

??????????????? – chihao

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)

来源

2015-09-03 13:11:55

建议改进：做UNION ALL。在firts部分也选择'Complete'，在第二个部分选择'Changes'。 – jarlh

加入2个subquerys

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