ショッピング カートの収益は曜日によって異なり、月曜日と木曜日が少なく、水曜日と土曜日が多いようです。したがって、次のような合計を示す Web ページを表示したいと思います。
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.
次のように、特定の曜日をクエリできます。
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
これにより、過去 12 週間のすべての日曜日の 1 日の合計が得られます。したがって、必要なものを取得するために 7 つのクエリを実行できます (曜日ごとに 1 つのクエリ)。1回のクエリですべてを取得できるはずです。
クエリは何ですか?ありがとう!
編集:これは、推奨されるソリューションからの修正されたクエリです。
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