select Pending = PD.Qty - (select sum(GRND.Qty)
from tbl_GRN GRN inner join tbl_GRND GRND on
GRN.grnId = GRND.grnId where GRN.pivID=PD.pivID and GRND.prodId=PD.prodId)
from tbl_PD PD where PD.pivId
假设我有一个这样的表： | id | date | name | value |
|----|------------|------|-------|
| 0 | 2017-01-14 | foo | one |
| 1 | 2017-01-17 | bar | two |
| 2 | 2017-01-18 | john | five |
| 3 | 2017-01-19 | doe
有人可以帮助我理解相关查询。以下是我试图理解的查询，但无法从中获得任何信息。请帮助。谢谢。 --Step 1
select e1.Name as 'Employee', e1.Salary
from Employee e1
where 3 >
(
select count(distinct e2.Salary)
from Employe
我想在我的数据库上执行这个查询，这是查询基于第三个表的结果的两个表。 SELECT *
FROM ads_user AS u
INNER JOIN ads_medium AS m
ON u.id = m.owner_id
WHERE m.id IN (SELECT medium_id,
Count(*) AS count
SELECT *,
IF(users_posts.uid IN (SELECT puid FROM post_ups WHERE post_ups.uid = UID LIMIT 400) AND users_posts.uid <> UID ,10,0)
FROM users_posts
即时得到这个错误 该版本的MySQL还不支持“LIMIT & IN/ALL/ANY/SOME
我有2个表PORT和发货。我必须选择对应于最大（计数）的名称。它会为别名“A”引发错误。这段代码或其他可用的替代品是否有错误？ select name from
(
select name, count(name) as countval from
(
select p.name from port p
inner join shipment s on