Question

我在过去10天内使用以下内容来计算新用户：

SELECT days.day, count(u.user_id)
FROM
(select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day) days
left join users u
on days.day = DATE(u.dateadded)
group by
days.day

哪个有效，但是应该给出date_format，2014-03-18 - 理想情况下，我希望日期格式为例如'Tues 18th Mar'

这就是我正在尝试但仅返回2行，并且在天列下显示null和BLOB

SELECT days.day, count(u.user_id)
FROM
(select DATE_FORMAT(curdate(), '%a %D %b') as day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 1 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 2 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 3 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 4 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 5 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 6 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 7 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 8 day
union select DATE_FORMAT(curdate(), '%a %D %b') - interval 9 day) days
left join users u
on days.day = DATE_FORMAT(u.dateadded, '%a %D %b')
group by
days.day

有更好的方法吗？

我正在使用MySQL和PHP

Answer 1

JOIN当然必须保持不变。无需在那里使用DATE_FORMAT()。

SELECT DATE_FORMAT(days.day, '%a %D %b') AS "day", count(u.user_id)
FROM
(select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day) days
left join users u
on days.day = DATE(u.dateadded)
group by
days.day

Answer 2

尝试此选项：

SELECT DATE_FORMAT(dateadded, '%a %D %b'), count(1)
  FROM users 
 WHERE dateadded > put_here_your_limit
 GROUP BY 1
 ORDER BY 1 DESC;

我希望它适合你。

Answer 3

你的问题是在date_format的结果上使用- interval而不是它的参数（导致为null，因为在格式化之后，它看起来不像要减去的有效日期 - 只有一个null所有union select的结果，因为您没有说union all）。这可行：

select DATE_FORMAT(curdate(), '%a %D %b') as day
union select DATE_FORMAT(curdate() - interval 1 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 2 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 3 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 4 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 5 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 6 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 7 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 8 day , '%a %D %b')
union select DATE_FORMAT(curdate() - interval 9 day , '%a %D %b')

但是因为您对用户的加入只是在例如“Tue 18th Mar”，你今天会增加用户，但也会在其他年份增加用户，例如2008年3月18日星期二。我猜这不是你的意图。

在SELECT中格式化curdate（）

3 个答案: