使用COUNT（）进行MySQL查询以返回字段等于COUNT值的记录

Question

我有下面的查询工作正常：

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

在表队列中，有一个int列“ attendance_pass”。 现在，我想拥有另一个类似于上面的查询，该查询仅返回COUNT(*) FROM attendance等于cohort.attendance_pass的记录。 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

如何修改第二个查询以仅获取那些记录？

Answer 1

聚合函数必须在having子句中，而不能在where子句中。

顺便说一句，您可以使用别名。

而且我认为您不能GROUP BY asc，您当然是指GROUP BY然后是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

Answer 2

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