Question

表＃1名称：用户

> ID  |  NAME  |  offerID  |  paymentDate
  1   |  user1 |  1        |  2014-07-14
  2   |  user2 |  2        |  2014-07-14
  3   |  user3 |  2        |  2014-07-30
  4   |  user4 |  1        |  2014-07-14
  5   |  user5 |  3        |  2014-07-14
  6   |  user6 |  1        |  2014-07-30

表＃2名称：报价

> ID | NAME  |  PRICE
  1  | offer1|  25
  2  | offer2|  45
  3  | offer3|  75

如果你看到我有（3）用户在offer1（2）用户在offer2（1）用户在offer3

我如何计算有多少用户会付钱给我＃2014; 2014-07-14＆＃39;多少钱？，

我需要的结果将是＆＃39; 2014-07-14＆＃39;喜欢这个

> paymentDate  |  usersCount  |  Totalprice
  2014-07-14   |  4           |  175

Answer 1

SELECT sum(offer.price) as Totalprice, count(users.id), paymentDate 
FROM users JOIN offers ON users.offerID = offers.ID  
WHERE paymentDate = '2014-07-14'
GROUP BY paymentDate;

Answer 2

这个怎么样？

select u.paymentDate, count(*), sum(o.price)
from users u join
     offers o
     on u.offerId = o.id
where u.date = '2014-07-14'
group by u.paymentDate;

当然，这并不是说他们将支付。只有数据表明他们会这样做。

您可以删除where子句以获取所有日期的结果。

Answer 3

SELECT U.PAYMENTDATE, COUNT(*) AS UserCount, SUM( O.PRICE) 
FROM USERS AS U
JOIN OFFERS AS O
  ON O.ID = U.OfferId
WHERE U.PAYMENTDATE = '2014-07-14'
GROUP BY U.PAYMENTDATE

Answer 4

Select  PaymentDate, Count(Distinct U.Id) usersCount, Sum(O.Price) Totalprice
From    Users U
Join    Offer O On O.Id = U.OfferId
Where   U.PaymentDate = '2014-07-14'
Group By U.PaymentDate

？使用表中的COUNT和来自另一个表的SUM以及相关数据

4 个答案: