php或者mysql求一算法？-PHP 튜토리얼-php.cn

php或者mysql求一算法？

WBOY

풀어 주다： 2016-06-23 14:05:42

원래의

897명이 탐색했습니다.

这是数据库查询得到的结果。
现在需要显示结果为（price原本是直得到结果，为了方便我写成公式出来）：
price           order_id
(20/60)*700     x0045183
(40/60)*700     x0045183
  0             x0045178

计算思路为：相同的订单号，(20/60)*70，(40/60)*70 ，不同订单号直接取price值。

回复讨论(解决方案)

60 是指定的？还有70，上面结果又是 * 700 ，就看不懂了。

60 是指定的？还有70，上面结果又是 * 700 ，就看不懂了。

计算思路为：相同的订单号，(20/60)*700，(40/60)*700 ，不同订单号直接取price值。是这样的。700是相同订单price相加得到的。60为相同订单cost3相加得到的。

优选SQL完成

select cost3, (cost3/sum_cost3)*sum_price as price, a.order_id   form    tbl_name a,    (select order_id, sum(cost3) as sum_cost3, sum(price) as sum_price from tbl_name group by order_id) t  where a.order_id=t.order_id

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用 php 还麻烦些
直接查询后

while($r = mysql_fetch_assoc($rs)) {  $st[$r['order_id']['cost3'] += $r['cost3'];  $st[$r['order_id']['price'] += $r['price'];}mysql_data_seek($rs, 0);while($r = mysql_fetch_assoc($rs)) {  $res[] = ($r['cost3']/$st['cost3'])*$st['price'];}

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 select a.order_id , (select (a.cost3/sum(cost3) * sum(price))  from od where order_id=a.order_id group by order_id) as price from od a;

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上面的方法都实用，多谢哈。

select a.order_id , (select (a.cost3/sum(cost3) * sum(price)),sum(price)  from od where order_id=a.order_id group by order_id) as price from od a;

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但是为什么加了个sum(price) 提示：Operand should contain 1 column(s)

SQL code?123select a.order_id , (select (a.cost3/sum(cost3) * sum(price)),sum(price) from od where order_id=a.order_id group by order_id) as price from od a;但是为什么加了个sum(price) 提示：Operand…… 多加一个字段为什么不行？

里面的子查询有两个字段，那样写是不行的

select a.order_id , (select (a.cost3/sum(cost3) * sum(price))  from od where order_id=a.order_id group by order_id) as price,(select sum(price)  from od where order_id=a.order_id group by order_id) as sum_price from od a;

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