Question

例如，我有2个表和一个日期范围（2015年12月1日 - 2016年1月10日）。

第一张表：USERS

id (int)   date (datetime)
 1      3-dec-2015
 2      4-dec-2015
 3      19-dec-2015
 4      20-dec-2015
 5      21-dec-2015
 6      29-dec-2015
 7      30-dec-2015

第二张桌子：BIRTHDAYS

id (int)   date (datetime)
 1      6-dec-2015
 2      8-dec-2015
 3      9-dec-2015
 4      17-dec-2015
 5      28-dec-2015

查询后的结果应如下：

[0] 1st week => 2 users, 1 birthday
[1] 2nd week => 0 users, 2 birthday
[2] 3ed week => 1 users, 1 birthday
[3] 4th week => 1 users, 0 birthday
[4] 5th week => 2 users, 1 birthday
[5] 6th week => 0 users, 0 birthday

任何想法如何实现这一结果或接近的结果？如果需要，我可以使用PHP。

Answer 1

我会从这样的事情开始：

select ((week(dateb) - week('2015-12-01')) + 1) as week_number, count(a.dateb) as userdates 
from users as a
where dateb between '2015-12-01' and '2016-01-01'
group by week(dateb)
order by week(dateb);

和

select ((week(dateb2) - week('2015-12-01')) + 1) as week_number, count(dateb2) as birthdays 
from birthdays
where dateb2 between '2015-12-01' and '2016-01-01'
group by week(dateb2)
order by week(dateb2);

演示，http://sqlfiddle.com/#!9/c83cb/21

从那里你可以摆弄PHP的输出。

另请注意，使用此方法仅返回包含填充数据的行。因此，您应该检查每行增加1的迭代。

e.g。所以对于{1}}，从第1周到第3周，你应输出users;或者你想要显示它。

mysql - 选择计数多个表，按周分组，自定义日期范围

1 个答案: