尝试连接两个表时总和不正确

Question

使用连接时，我很难从qty列中获得正确的总和。当我尝试从paymentType表得到timestamp和orders和paymentType表对rowid加入这些，然后分组对天timestamp出现该问题(day(from_unixtime(paymentType.timestamp)))

我想要的qty总和通过使用timestamp小时paymentType表，唯一的链接是rowid（这是codeigniter的购物车模块的rowid）。逻辑问题（至少对我来说）是它在orders表中存在更多的行（因为这是每个产品）而不是paymentType表（这只是为了跟踪是否使用借记卡或现金）。当我将这些表加在一起时，每个小时的总和乘以orders.rowid <--> paymentType.rowid中的每个命中。

如果解释不好，我很抱歉，但我希望能够得到这个问题的帮助是可以理解的。

我至少有10次试过，但没有一次看起来像我想要的那样工作。

以下是我orders表

+---------+----+-------+-----+----------+------------------+----------------------------------+ 
| orderID | id | price | qty | subtotal |  name  |    rowid    | 
+---------+----+-------+-----+----------+------------------+----------------------------------+ 
|  3 | 49 | 35 | 1 |  35 | Red Bull Stor | f457c545a9ded88f18ecee47145a72c0 | 
|  4 | 24 | 35 | 1 |  35 | Monster Energy | 1ff1de774005f8da13f42943881c655f | 
|  5 | 49 | 35 | 1 |  35 | Red Bull Stor | f457c545a9ded88f18ecee47145a72c0 | 
|  6 | 19 | 20 | 1 |  20 | Sprite 0.5L  | 1f0e3dad99908345f7439f8ffabdffc4 | 
|  7 | 1 | 25 | 1 |  25 | Pringles   | c4ca4238a0b923820dcc509a6f75849b | 
|  8 | 43 | 20 | 1 |  20 | Lån av stekovn | 17e62166fc8586dfa4d1bc0e1742c08b | 
|  9 | 46 | 35 | 1 |  35 | Burn    | d9d4f495e875a2e075a1a4a6e1b9770f | 
|  10 | 49 | 35 | 3 |  105 | Red Bull Stor | f457c545a9ded88f18ecee47145a72c0 | 
|  11 | 49 | 35 | 1 |  35 | Red Bull Stor | f457c545a9ded88f18ecee47145a72c0 | 
|  12 | 29 | 25 | 1 |  25 | Potetskruer  | 6ea9ab1baa0efb9e19094440c317e21b | 
|  13 | 16 | 20 | 1 |  20 | Coca-Cola 0.5L | c74d97b01eae257e44aa9d5bade97baf | 
|  14 | 46 | 35 | 1 |  35 | Burn    | d9d4f495e875a2e075a1a4a6e1b9770f | 
|  15 | 1 | 25 | 1 |  25 | Pringles   | c4ca4238a0b923820dcc509a6f75849b | 
|  16 | 18 | 20 | 1 |  20 | Eventyrbrus 0.5L | 6f4922f45568161a8cdf4ad2299f6d23 | 
|  17 | 16 | 20 | 1 |  20 | Coca-Cola 0.5L | c74d97b01eae257e44aa9d5bade97baf | 
|  18 | 15 | 30 | 1 |  30 | Coca-Cola 1.5L | 9bf31c7ff062936a96d3c8bd1f8f2ff3 | 
|  19 | 19 | 20 | 1 |  20 | Sprite 0.5L  | 1f0e3dad99908345f7439f8ffabdffc4 | 
|  20 | 50 | 20 | 1 |  20 | Stratos bar  | c0c7c76d30bd3dcaefc96f40275bdc0a | 
+---------+----+-------+-----+----------+------------------+----------------------------------+ 

这是paymentType表

+-----------+-------------+------------+----------------------------------+ 
| paymentID | paymentType | timestamp |    rowid    | 
+-----------+-------------+------------+----------------------------------+ 
|   3 | Kort  | 1424447799 | f457c545a9ded88f18ecee47145a72c0 | 
|   4 | Kort  | 1424448791 | 1ff1de774005f8da13f42943881c655f | 
|   5 | Kort  | 1424452822 | f457c545a9ded88f18ecee47145a72c0 | 
|   6 | Kort  | 1424454483 | c4ca4238a0b923820dcc509a6f75849b | 
|   7 | Kort  | 1424454665 | d9d4f495e875a2e075a1a4a6e1b9770f | 
|   8 | Kontant  | 1424454799 | f457c545a9ded88f18ecee47145a72c0 | 
|   9 | Kontant  | 1424454825 | f457c545a9ded88f18ecee47145a72c0 | 
|  10 | Kort  | 1424454870 | 6ea9ab1baa0efb9e19094440c317e21b | 
|  11 | Kontant  | 1424455510 | d9d4f495e875a2e075a1a4a6e1b9770f | 
|  12 | Kort  | 1424455847 | c4ca4238a0b923820dcc509a6f75849b | 
|  13 | Kontant  | 1424456025 | 6f4922f45568161a8cdf4ad2299f6d23 | 
|  14 | Kontant  | 1424456099 | c74d97b01eae257e44aa9d5bade97baf | 
|  15 | Kontant  | 1424456148 | 9bf31c7ff062936a96d3c8bd1f8f2ff3 | 
|  16 | Kontant  | 1424456242 | c0c7c76d30bd3dcaefc96f40275bdc0a | 
|  17 | Kort  | 1424456266 | c74d97b01eae257e44aa9d5bade97baf | 
|  18 | Kort  | 1424456445 | c0c7c76d30bd3dcaefc96f40275bdc0a | 
|  19 | Kort  | 1424456964 | 70efdf2ec9b086079795c442636b55fb | 
|  20 | Kort  | 1424457701 | 1ff1de774005f8da13f42943881c655f | 
+-----------+-------------+------------+----------------------------------+ 

编辑： SQL查询我试过到目前为止，存在更多，但这些都是最新的。我认为这些是最“正确的”。

select orders.rowid, concat(convert(paymentType.timestamp,CHAR(11))) timestamp, orders.qty, orders.name 
from orders 
join paymentType 
on orders.rowid = paymentType.rowid 
order by paymentType.timestamp; 

select orders.rowid, hour(from_unixtime(concat(convert(paymentType.timestamp,CHAR(11))))), orders.qty, orders.name 
from orders 
join paymentType 
on orders.rowid = paymentType.rowid 
#where orders.name = '".stripslashes($name)."' 
order by paymentType.timestamp 
; 
select orders.qty, orders.name, orders.rowid, paymentType.rowid, paymentType.timestamp 
from orders, paymentType 
where orders.rowid = paymentType.rowid; 

select qty, name, hour(from_unixtime(timestamp)) hour, day(from_unixtime(timestamp)) day 
from orders_w_time 
where name = 'Red Bull Stor' 
; 

select sum(qty) from orders 
inner join (select distinct rowid from paymentType) pt 
on orders.rowid = pt.rowid 
where orders.name = 'Pølse'; 

select sum(orders.qty) totalqty, orders.name, pt.timestamp timestamp from orders ord 
inner join (select timestamp from paymentType where paymentType.rowid = ord.rowid) pt 
on orders.rowid = pt.rowid 
where orders.name = 'Red Bull Stor'; 
select * from 
(
select rowid, timestamp from paymentType 
group by hour(from_unixtime(timestamp)) 
) pt 
left join 
(
select sum(qty), name, rowid from orders 
) ord 
on ord.rowid = pt.rowid 
; 
Select 
    paymentType.rowid, 
    orders.name, 
    orders.qty, 
    paymentType.timestamp 
From 
    orders, 
    paymentType 
Group By 
    day(from_unixtime(paymentType.timestamp)); 

select sum(orders.qty) Total 
from orders 
left join 
(
select rowid,timestamp 
from paymentType 
) as paymet on orders.rowid = paymet.rowid 
group by day(from_unixtime(paymet.timestamp)) 
; 

select paymentType.rowid, ord.qty, timestamp 
from paymentType 
left join 
(
select orders.rowid, qty 
from orders 
) as ord on ord.rowid = paymentType.rowid 
; 

预期的结果是对按小时分组的每个日期的'Red Bull Stor'（即。）的数量列求和。

Answer 1

你可以尝试这样的事情：

select o.name 
    , pt.rowid 
    , sum(o.qty) 
    , hour(from_unixtime(pt.timestamp)) 
    , day(from_unixtime(pt.timestamp)) 
from orders o 
join paymentType pt using(rowid) 
where o.name = 'Red Bull Stor' 
group by o.name 
     , o.rowid 
     , hour(from_unixtime(pt.timestamp)) 
     , day(from_unixtime(pt.timestamp)); 

SQLFiddle