我有点卡住了,无法理解这个MySQL。以下是我想查询的简略表:
print(printID,eventID,printTime)
销售(saleID,eventID,saleTime)
我希望获得每个活动的最后打印时间,然后选择具有该eventID且销售时间大于上次打印时间的销售ID。
我尝试了许多方法,但我无法理解。请帮忙!
答案 0 :(得分:4)
select s.saleID, s.eventID, s.saleTime, lp.LastPrintTime
from (
select eventID, max(printTime) as LastPrintTime
from print
group by eventID
) lp
inner join sales s on lp.eventID = s.eventID
and saleTime > lp.LastPrintTime
答案 1 :(得分:0)
要获得上次打印时间:
SELECT MAX(p.printTime), p.eventID FROM print p
您可以在子查询中使用它来获得所需内容:
SELECT s.saleID, s.eventID
FROM sales s
INNER JOIN (SELECT MAX(p.printTime) AS lastPrintTime, p.eventID
FROM print p
GROUP BY p.eventID) print_time
WHERE print_time.eventID = s.eventID
AND s.saleTime > print_time.lastPrintTime
这有帮助吗?
答案 2 :(得分:0)
内部查询:
SELECT * FROM sales AS s INNER JOIN
(
SELECT eventID, max(printTime) as maxPrintTime FROM [print] GROUP BY eID
) as p ON p.eventID = s.eventIDAND s.saleTime > maxPrintTime