这个想法是,在给定的一天中有两种用餐选择,例如:10/02/16 meal id 4 OR meal id 12select * from meals JOIN menu ON meals.id = menu.option1 AND meals.id=menu.option2
但每次我运行这个命令时,我都会收到一条错误消息
mysql> describe holidays;| Field | Type |+----------+--------------++----------+--= holidays.date AND timesheet.location = holidays.lid, 1, 0) AS is_holi