我想使用DATE_ADD 2次的操作,但是它返回0行(雖然它應該返回行)MySQL的DATE_ADD 2次過程
下面是該過程
select av.*, ap.*,c.* from tbl_available av
left join tbl_appointment ap on av.avHours = ap.appointmenttime
and ap.calendarid = kalenderId
and ap.appointmentdate = DATE_ADD(dag, INTERVAL 6 DAY)
left join tbl_client c on ap.clientid = c.clientid
where av.avCalendarId = KalenderId
and av.avDays = DayOfweek(DATE_ADD(dag, INTERVAL 6 DAY))
order by avHours;
它的工作原理,而不date_add
在此先感謝!
//編輯
我現在擁有的一切:
select av.*, ap.*,c.*, ab.absentid from tbl_available av
left join tbl_appointment ap on av.avHours = ap.appointmenttime
and ap.calendarid = kalenderId
and ap.appointmentdate BETWEEN dag AND DATE_ADD(dag, INTERVAL 6 DAY)
and (av.avDays = DayOfweek(ap.appointmentdate) OR ap.appointmentdate IS NULL)
left join tbl_client c on ap.clientid = c.clientid
left join tbl_absent ab on av.avHours = ab.ababsent
and ab.abHoliday = dag
and ab.abCalendarID = kalenderId
where av.avCalendarId = kalenderId
order by avDays,avHours;
但ab.absentid不牽強,這是爲什麼? :(
什麼是您的樣本數據和預期輸出?乍看起來這看起來不錯。 –
同意lc,從句法上看似乎很好。這有什麼問題? –
嗨,這裏有點大,但我希望所有的tbl_available數據與(如果有的話)相應的tbl_appointment數據和相應的tbl_client數據從1天開始。這工作,但現在我想從1周,這導致0行。但如果我使用相同的日期爲1天,它會給出行,所以我確定此過程必須返回行 – Nick