我需要进行以下查询,并提取总订单数和按天分组的订单总和.我正在使用时间戳存储所有内容.
SELECT COUNT(id) as order_count, SUM(price + shipping_price) as order_sum, DAY(FROM_UNIXTIME(created)) as day FROM `order` WHERE '.implode(' AND ', $where).'我需要按天分组,但是当我在过去一个周末的销售额中进行结算时,我需要将order_count设为1而不是3.如何才能按天将上述值分组?
注意:爆破仅用于定义时间段(在创建的位置> = TIMESTAMP和< = TIMESTAMP)
更新
没有GROUP BY day
Array ( [order_count] => 3 [order_sum] => 69.70 [day] => 17 )使用GROUP BY day
Array ( [order_count] => 1 [order_sum] => 24.90 [day] => 17 )我需要此查询来返回每天的销售额,多少订单以及这些销售额的总和.我在这里某处缺少一个难题....
解决方案您只是忘了在最后添加GROUP BY ...吗?
SELECT COUNT(id) as order_count, SUM(price + shipping_price) as order_sum, DAY(FROM_UNIXTIME(created)) as order_day FROM `order` WHERE '.implode(' AND ', $where).' GROUP BY order_day注意:
由于day是MySQL函数,因此不能将as day用于日列.使用类似order_day的东西.
独角兽根据@OMG Unicorn的评论,您可以使用:
DAY(FROM_UNIXTIME(created)) as `day`只要将day包装在`反引号中即可.
I need to take the following query and pull the total order counts and sum of the orders grouped by day. I'm storing everything using timestamps.
SELECT COUNT(id) as order_count, SUM(price + shipping_price) as order_sum, DAY(FROM_UNIXTIME(created)) as day FROM `order` WHERE '.implode(' AND ', $where).'I need to group by DAY but when I do for this past weekend's sales it takes my order_count and makes it 1 instead of 3. How can I pull the above values grouped by day?
NOTE: The implode is used ONLY to define the time period (WHERE created >= TIMESTAMP AND <= TIMESTAMP)
Update
Without GROUP BY day
Array ( [order_count] => 3 [order_sum] => 69.70 [day] => 17 )With GROUP BY day
Array ( [order_count] => 1 [order_sum] => 24.90 [day] => 17 )I need this query to return each day that had sales, how many orders, and the sum of those sales. I'm missing a piece of the puzzle here somewhere....
解决方案Are you just forgetting to add GROUP BY ... at the end?
SELECT COUNT(id) as order_count, SUM(price + shipping_price) as order_sum, DAY(FROM_UNIXTIME(created)) as order_day FROM `order` WHERE '.implode(' AND ', $where).' GROUP BY order_dayNOTE:
You cannot use as day for your day column because day is a MySQL function. Use something like order_day.
Of UnicornsPer @OMG Unicorn's comment, you can use:
DAY(FROM_UNIXTIME(created)) as `day`So long as wrap day in ` backticks.
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