Php ？使用表中的计数和另一个表中的总和以及相关数据

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

如果您看到，我的报价中有3个用户1报价中有2个用户2报价中有1个用户3

我如何计算“2014-07-14”中有多少用户将向我支付多少

我需要像这样的2014-07-14的结果

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

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，这并不是说他们会付钱。只有数据表明他们会

您可以删除where子句以获取所有日期的结果。

这如何

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;

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

当然，这并不是说他们会付钱。只有数据表明他们会

您可以删除where子句以获取所有日期的结果。

这如何

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;

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

当然，这并不是说他们会付钱。只有数据表明他们会

您可以删除where子句以获取所有日期的结果。

这如何

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;

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

当然，这并不是说他们会付钱。只有数据表明他们会

---

您可以删除where子句以获取所有日期的结果。

如果效果良好，请告诉我：从offer o中选择SUMo.price、COUNTu.id、u.paymentDate，用户u where u.offerID=offer.id和u.paymentDate='2014-07-14'按u.paymentDate分组；让我知道，这是否有效：从offer o中选择SUMo.price、COUNTu.id、u.paymentDate、用户u，其中u.offerID=offer.id和u.paymentDate='2014-07-14'按u.paymentDate分组；让我知道，这是否有效：从offer o中选择SUMo.price、COUNTu.id、u.paymentDate、用户u，其中u.offerID=offer.id和u.paymentDate='2014-07-14'按u.paymentDate分组；让我知道，这是否有效：从offer o中选择SUMo.price、COUNTu.id、u.paymentDate、用户u，其中u.offerID=offer.id和u.paymentDate='2014-07-14'按u.paymentDate分组；

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