如何获取上周和本周

Question

如何从周一时间 (00:00:01) 获取上周数据并在周日时间 (23:59:59) 结束... 与本周从周一时间 (00:00:01) 到周日时间 (23:59:59) 结束的时间相同

我试过了！

$query = "SELECT users.name,count(*) as count, 
                 campaign.campaign_name, 
                 campaign.payout_cost*count(*) as totalPrice
            FROM users 
            JOIN transactions on users.uid=transactions.uid 
            JOIN campaign on campaign.campaign_id_id=transactions.campaign_id 
           WHERE uid=$uid 
             AND `date` >= DATE_SUB(DATE(NOW()), INTERVAL DAYOFWEEK(NOW())+6 DAY)   
             AND `date` <  DATE_SUB(DATE(NOW()), INTERVAL DAYOFWEEK(NOW())-1 DAY) 
           GROUP BY campaign.campaign_name_name ";

Answer 1

有几种方法可以做到...

select data from tableName
where date between date_sub(now(),INTERVAL 1 WEEK) and now();

select data FROM tableName
wherdate >= curdate() - INTERVAL DAYOFWEEK(curdate())+6 DAY
AND date < curdate() - INTERVAL DAYOFWEEK(curdate())-1 DAY

Answer 2

您可以使用WEEK() 函数，通过添加返回给定日期的周数

AND WEEK(date-INTERVAL 1 DAY) = WEEK(NOW()) - 1 获取从周一到周日的本周数据，

和

AND WEEK(date-INTERVAL 1 DAY) = WEEK(NOW()) - 2为上周的数据

在WHERE uid=$uid之后进入WHERE条件

比如

$query = "SELECT c.campaign_name,
                 COUNT(*) as total_count,
                 SUM(c.payout_cost) as total_payout
            FROM transactions t
            JOIN campaign c
              ON c.campaign_id = t.campaign_id
           WHERE uid = $uid
             AND WEEK(date - INTERVAL 1 DAY) = WEEK(NOW()) - 1               
           GROUP BY c.campaign_name "; 

并将WEEK(NOW()) - 1替换为WEEK(NOW()) - 2，也

Demo

Answer 3

通过避免列上的week() 之类的函数，您走在了正确的轨道上——这只会弄乱优化器。另一方面，uid 参数应作为 参数 传递，而不是修改查询字符串。

您想使用weekday() 函数，因为您希望每周从星期一开始。只是 MySQL 的一些奥秘：weekday() 在星期一返回 0，而 dayofweek() 在星期一返回 2。

因此，本周的逻辑是：

date >= curdate() - interval weekday(curdate()) day and
date < curdate() + interval 7 - weekday(curdate()) day

上周，这将是：

date >= curdate() - interval 7 + weekday(curdate()) day and
date < curdate() + interval - weekday(curdate()) day

注意curdate()（或current_date）返回的当前日期不包含时间部分，因此不需要date()。