在 where 子句中使用 Select Min 和 MAx答案

【问题标题】：Use a Select Min and MAx in your where clause在 where 子句中使用 Select Min 和 MAx
【发布时间】：2018-01-15 15:06:07
【问题描述】：

我正在使用许多选择语句来推动结果，但是我有 2 个选择语句返回约会的最短时间和约会的最长时间。但是，如果最小值和最大值相同，我想从 where 子句中删除。

SELECT
  APPOINTMENTS.userid,
  users.LOCATIONID,
  MIN(APPOINTMENTTIME) AS mintime,
  MAX(APPOINTMENTTIME) AS maxtime

FROM appointments
WHERE APPOINTMENTDATE BETWEEN '2017-01-07' AND '2017-01-07'
AND NOT mintime <> maxtime
GROUP BY appointments.USERID,
         users.LOCATIONID,
         appointments.APPOINTMENTDATE

我收到错误 Mintime is not valid.

我确定我需要将其保存在新的子选择中，但不确定..

干杯

【问题讨论】：

左对齐 SQL 太难读了...

标签： sql max where-clause min

【解决方案1】：

HAVING clause 用于聚合函数，例如

HAVING NOT MIN(APPOINTMENTTIME) <> MAX(APPOINTMENTTIME)

我不知道你为什么在你的谓词中使用双重否定，根据你的描述你只需要<>

HAVING MIN(APPOINTMENTTIME) <> MAX(APPOINTMENTTIME)

【讨论】：

感谢您的快速回复，效果很好。以前没用过。将对该功能进行一些研究.. 一直在学习 :-) 感谢 Garethd

【解决方案2】：

根据 Sql Server 试试这个

        SELECT * FROM (
        select   
        APPOINTMENTS.userid, 
        users.LOCATIONID,
        min(APPOINTMENTTIME) as mintime, 
        max(APPOINTMENTTIME) as maxtime
        from appointments
        where 
        APPOINTMENTDATE between '2017-01-07' and '2017-01-08'

        group by 
        appointments.USERID,users.LOCATIONID,
        appointments.APPOINTMENTDATE
        )as t WHERE  mintime <> maxtime

【讨论】：

【解决方案3】：

请尝试

select   
APPOINTMENTS.userid, 
users.LOCATIONID,
min(APPOINTMENTTIME) as mintime, 
max(APPOINTMENTTIME) as maxtime

from appointments
where 
APPOINTMENTDATE between '2017-01-07' and '2017-01-07'

group by 
appointments.USERID,users.LOCATIONID,
appointments.APPOINTMENTDATE
having min(APPOINTMENTTIME)  <> max(APPOINTMENTTIME)

【讨论】：