技術標籤：sqlmysql

1. 先統計當前日期一週七天的日期

   	select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) + 0 DAY) as weekday
   	UNION
   	select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 1 DAY)
   	UNION
   	select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 2 DAY)
   	UNION
   	select date_sub(curdate(),INTERVAL WEEKDAY(
    
   curdate()) - 3 DAY)
   	UNION
   	select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 4 DAY)
   	UNION
   	select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 5 DAY)
   	UNION
   	select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 6 DAY)
   

2. 然後統計每天的數量

   SELECT
   	a.weekday,
   	COUNT(nm.id) count
   FROM
   	(
   		select
    
    date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 0 DAY) as weekday
   		UNION
   		select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 1 DAY)
   		UNION
   		select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 2 DAY)
   		UNION
   		select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 3 DAY)
   		UNION
   		select
    
    date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 4 DAY)
   		UNION
   		select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 5 DAY)
   		UNION
   		select date_sub(curdate(),INTERVAL WEEKDAY(curdate()) - 6 DAY)
   	) as a
   	LEFT JOIN need_medical as nm on nm.is_delete = 0 AND date_format(nm.appointment_date, '%Y-%m-%d')=a.weekday
   GROUP BY
   	a.weekday