MySQL查詢使用COUNT（）返回其中字段等於COUNT值

Question

記錄我有以下查詢工作正常：

SELECT *, COUNT(*) FROM attendance, cohort 
WHERE 
attendance.cohort_fk = cohort.cohort_pk 
AND 
attendance.cohort_fk = '$cohort' 
AND 
YEAR(attendance.attended) = '$year' 
GROUP BY attendance.person_id ASC 

在該表隊列中，有一個int的列「attendance_pass」。現在我想要有另一個類似於上面的查詢，只返回COUNT(*) FROM attendance等於cohort.attendance_pass的記錄。例如。

SELECT *, COUNT(*) FROM attendance, cohort 
WHERE 
attendance.cohort_fk = cohort.cohort_pk 
AND 
attendance.cohort_fk = '$cohort' 
AND 
YEAR(attendance.attended) = '$year' 
AND 
COUNT() = cohort.attendance_pass 
GROUP BY attendance.person_id ASC 

如何修改第二個查詢以獲取這些記錄？

Answer 1

您需要使用HAVING

SELECT *, COUNT(*) FROM attendance, cohort 
WHERE 
attendance.cohort_fk = cohort.cohort_pk 
AND 
attendance.cohort_fk = '$cohort' 
AND 
YEAR(attendance.attended) = '$year' 

GROUP BY attendance.person_id ASC 

HAVING COUNT(*) = cohort.attendance_pass 

Answer 2

聚合函數必須位於having子句中，而不是位於where子句中。

順便說一下，您可以使用別名。

而且我不認爲你可以GROUP BY ASC，你肯定意味着GROUP BY那麼ORDER BY ... ASC

select *, count(*) as cnt from attendance 
-- etc. 

where 
--etc. 
having cnt = cohort.attendance_pass 
GROUP BY attendance.person_id 
ORDER BY attendance.person_id ASC