SELECT COUNT(*) AS 'NumOfConsultations'
SELECT consultationDate DISTINCT(*) AS 'NumOfDays'
FROM consultation
SUM('NumOfDays' / 'NumOfCosultations' = 'AveragePatientsPerDay');我正在试着找出平均每天的会诊次数。有什么建议吗?
发布于 2014-10-16 11:35:14
假设您使用的是SQL Server,则可以执行以下操作:
SELECT
AVG(consultationId) AS 'Average Consultations Per Day',
COUNT(consultationId) AS 'Total Consultations Per Day'
FROM consultation
GROUP BY CAST(consultationDate AS DATE)发布于 2014-10-16 13:25:24
据我所知,平均每天的诊症次数如下:
select count(*) / count(distinct consultationDate)
from consultation;如果consultationDate实际上有时间组件(除了日期之外),则需要将其转换为日期。这取决于数据库,但类似于:
select count(*) / count(distinct cast(consultationDate as date))
from consultation;或者:
select count(*) / count(distinct trunc(consultationDate))
from consultation;或者:
select count(*) / count(distinct date(consultationDate))
from consultation;或许能行得通。
发布于 2019-08-27 03:24:07
选择count(*) / count(distinct serv_date)作为生产中的平均值;
https://stackoverflow.com/questions/26395732
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