本文介绍了mysql 从选择中选择的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我有这个问题:
SELECT DATE( a.created_at ) AS order_date, count( * ) as cnt_order
FROM `sales_order_item` AS a
WHERE MONTH( a.created_at ) = MONTH( now())-1
GROUP BY order_date
这将返回类似这样的结果(仅快照，否则将每 31 天返回一次):
which will return result something like this (snapshot only otherwise will return per 31 days):
order_date cnt_order
2012-08-29 580
2012-08-30 839
2012-08-31 1075
我的完整查询是根据上述选择进行选择的:
and my full query is selecting based on above selection:
SELECT order_date
, MAX(cnt_order) AS highest_order
FROM (
SELECT DATE (a.created_at) AS order_date
, count(*) AS cnt_order
FROM `sales_order_item` AS a
WHERE MONTH(a.created_at) = MONTH(now()) - 1
GROUP BY order_date
) AS tmax
但结果是:
order_date highest_order
2012-08-01 1075
哪个日期有误，总是选择它假设为 2012 年 8 月 31 日的第一行日期.也许这是一个我不知道的简单错误.那么如何让日期正确指向 2012-08-31?任何帮助都会很棒.
Which has the date wrong and always pick the first row of date where it suppose 2012-08-31. Maybe this is a simple error that I dont know. So how to get the date right point to 2012-08-31? Any help would be great.
推荐答案
可以尝试排序子查询结果集；类似:
You could try ordering the sub query result set; something like:
SELECT
DATE (a.created_at) AS order_date,
COUNT(*) AS cnt_order
FROM
`sales_order_item` AS a
WHERE
MONTH(a.created_at) = MONTH(now()) - 1
GROUP BY
order_date
ORDER BY
cnt_order DESC
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