Question

我有一个像这样的表结构：

USER_ID START_DATE END_DATE   MON TUE WED THU FRI SAT SUN  
======= ========== ========== === === === === === === ===
      1 2018-03-01 2018-03-15   0   1   0   1   0   0   0
      2 2018-02-20 2018-02-23   1   1   1   1   1   0   0

其中M / T / W等列为1/0，表示事件是否发生在星期一/星期二/星期三等。

所以我需要一个查询，我可以在两个日期之间为USER_ID运行，这将为该日期范围内每一天发生的事件返回一行。

因此，例如，如果用户在接下来的四周内每周一和周三都有一个事件，那么今天将有一行作为start_date，将4周时间作为end_date，其中MON和WED设置为＆＃39; 1＆＃39;

我需要的是一个查询，在这种情况下，如果我在该范围内使用三周搜索周期将返回6行，每行包含事件发生的日期。

我对如何处理此问题感到困惑，任何帮助表示赞赏！

Answer 1

在这里（假设日期范围不超过19年）：

select date_add(start_date, interval 7 * (d2.n * 100 + d1.n * 10 + d0.n) + weekday.d day) as day
  from 
    (select 1 as n union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) d2,
    (select 1 as n union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) d1,
    (select 1 as n union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) d0,
    (select 0 as d from schedule where user_id = 30 and mon = 1
     union select 1 as d from schedule where user_id = 30 and tue = 1
     union select 2 as d from schedule where user_id = 30 and wed = 1
     union select 3 as d from schedule where user_id = 30 and thu = 1
     union select 4 as d from schedule where user_id = 30 and fri = 1
     union select 5 as d from schedule where user_id = 30 and sat = 1
     union select 6 as d from schedule where user_id = 30 and sun = 1) as weekday,
    schedule s
    where user_id = 30
      and date_add(start_date, interval 7 * (d2.n * 100 + d1.n * 10 + d0.n) + weekday.d day) between s.start_date and s.end_date
  order by d2.n, d1.n, d0.n

user_id = 30 （在SQL中输入8次）：

day
==========
2018-03-02
2018-03-04
2018-03-09
2018-03-11

结果：

select date_add(start_date, interval 7 * (d2.n * 100 + d1.n * 10 + d0.n) + weekday.d day) as day
  from 
    (select 1 as n union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) d2,
    (select 1 as n union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) d1,
    (select 1 as n union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) d0,
    (select 0 as d from schedule where user_id = 31 and mon = 1
     union select 1 as d from schedule where user_id = 31 and tue = 1
     union select 2 as d from schedule where user_id = 31 and wed = 1
     union select 3 as d from schedule where user_id = 31 and thu = 1
     union select 4 as d from schedule where user_id = 31 and fri = 1
     union select 5 as d from schedule where user_id = 31 and sat = 1
     union select 6 as d from schedule where user_id = 31 and sun = 1) as weekday,
    schedule s
    where user_id = 31
      and date_add(start_date, interval 7 * (d2.n * 100 + d1.n * 10 + d0.n) + weekday.d day) between s.start_date and s.end_date
  order by d2.n, d1.n, d0.n

对于 user_id = 31 （它在SQL中输入了8次）：

day
==========
2018-02-20
2018-02-21
2018-02-22
2018-02-23

结果：

    SELECT *from posts 
    where    REPLACE(REPLACE(REPLACE(CONCAT(`primary_phone`,`workphone`,`cellphone`),'-',''),')',''),'(','')
    like '%xxxxxxxx%'

轻松自负。

MySQL查询返回两个日期范围之间的行

1 个答案: