Question

我的购物车收入似乎每周都有所不同，星期一和星期四是低天，星期三和星期六是高天。因此，我想显示一个显示如下总数的网页：

Week #    Sun    Mon    Tue    Wed    Thu    Fri    Sat
Week 1   $5.00  $1.00  $3.00  $9.00  $1.00  $3.00  $9.00
Week 2   $5.23  $1.07  $2.98  $8.75  $0.02  $3.14  $7.51
Week 3   etc.

我可以像这样查询一周中的某一天：

SELECT count( id ) AS orders, 
       order_date, 
       date_format( order_date, '%a' ) AS weekday, 
       sum( total) AS revenue
FROM `ss_orders`
WHERE dayofweek( order_date ) = 1
      AND order_date >= date_add( now( ) , INTERVAL -83 DAY )
GROUP BY order_date
ORDER BY order_date DESC

它给出了过去12周内所有星期日的每日总数。因此，我可以进行7次查询以获得我需要的信息（每周一天的1个查询）。看起来我应该能够在一个查询中得到整个东西。

查询应该是什么？谢谢！

编辑：以下是首选解决方案中的更正查询。

SELECT 
  week( o.order_date ) as WkNumber,
  sum( if( weekday( o.order_date ) = 6, 1, 0 ) * o.total ) as SalesSun,
  sum( if( weekday( o.order_date ) = 6, 1, 0 )) as OrdersSun,
  sum( if( weekday( o.order_date ) = 0, 1, 0 ) * o.total ) as SalesMon,
  sum( if( weekday( o.order_date ) = 0, 1, 0 )) as OrdersMon,
  sum( if( weekday( o.order_date ) = 1, 1, 0 ) * o.total ) as SalesTue,
  sum( if( weekday( o.order_date ) = 1, 1, 0 )) as OrdersTue,
  sum( if( weekday( o.order_date ) = 2, 1, 0 ) * o.total ) as SalesWed,
  sum( if( weekday( o.order_date ) = 2, 1, 0 )) as OrdersWed,
  sum( if( weekday( o.order_date ) = 3, 1, 0 ) * o.total ) as SalesThu,
  sum( if( weekday( o.order_date ) = 3, 1, 0 )) as OrdersThu,
  sum( if( weekday( o.order_date ) = 4, 1, 0 ) * o.total ) as SalesFri,
  sum( if( weekday( o.order_date ) = 4, 1, 0 )) as OrdersFri,
  sum( if( weekday( o.order_date ) = 5, 1, 0 ) * o.total ) as SalesSat,
  sum( if( weekday( o.order_date ) = 5, 1, 0 )) as OrdersSat,
  sum( o.total ) as SalesWeek,
  sum( 1 ) as OrdersWeek
from
  ss_orders o
where
  o.order_date > date_add( now(), INTERVAL -13 WEEK )
group by
  week( o.order_date )
order by o.order_date desc

Answer 1

按周分组？

SELECT count( id ) AS orders, 
       date_format( order_date, '%a' ) AS weekday, 
       WEEK(order_date) AS week_number,
       sum( total) AS revenue
FROM 
       `ss_orders`
WHERE 
       order_date >= date_add( now( ) , INTERVAL -83 DAY )
GROUP BY 
       WEEK(orderdate), DAY(orderdata)
ORDER BY 
       order_date DESC

应该给你这样的结果：

orders | weekday | week_number | revenue
5        Sun       1             20
6        Mon       1             15
...

Answer 2

您实际上正在寻找一种数据透视表查询。

SELECT 
      week( o.order_date ) as WkNumber,
      sum( if( weekday( o.order_date ) = 6, 1, 0 ) * o.total ) as SalesSun,
      sum( if( weekday( o.order_date ) = 6, 1, 0 ) as OrdersSun,
      sum( if( weekday( o.order_date ) = 0, 1, 0 ) * o.total ) as SalesMon,
      sum( if( weekday( o.order_date ) = 0, 1, 0 ) as OrdersMon,
      sum( if( weekday( o.order_date ) = 1, 1, 0 ) * o.total ) as SalesTue,
      sum( if( weekday( o.order_date ) = 1, 1, 0 ) as OrdersTue,
      sum( if( weekday( o.order_date ) = 2, 1, 0 ) * o.total ) as SalesWed,
      sum( if( weekday( o.order_date ) = 2, 1, 0 ) as OrdersWed,
      sum( if( weekday( o.order_date ) = 3, 1, 0 ) * o.total ) as SalesThu,
      sum( if( weekday( o.order_date ) = 3, 1, 0 ) as OrdersThu,
      sum( if( weekday( o.order_date ) = 4, 1, 0 ) * o.total ) as SalesFri,
      sum( if( weekday( o.order_date ) = 4, 1, 0 ) as OrdersFri,
      sum( if( weekday( o.order_date ) = 5, 1, 0 ) * o.total ) as SalesSat,
      sum( if( weekday( o.order_date ) = 5, 1, 0 ) as OrdersSat,
      sum( o.total ) as SalesWeek,
      sum( 1 ) as OrdersWeek
   from
      ss_orders o
   where
      o.order_date > date_add( now(), INTERVAL -12 WEEK )
   group by
      week( o.order_date )

如果你在一周中，你可能需要调整WHERE子句以真正获得12个星期...例如将NOW（）改为类似 date_add（now（），interval -weekday（now（））），但是我必须在那个部分考虑更多。

MySQL每日总计按星期几计算

2 个答案: