如何使用 IF 条件与 SQL 比较日期

Posted 2023-05-08

【中文标题】如何使用 IF 条件与 SQL 比较日期

【英文标题】：How how use IF condition for comparing date with SQL

【发布时间】：2019-05-21 06:11:58

【问题描述】：

我有一个来自考勤机的 MS ACCESS 表，它由 USERID、CHECKTIME 列组成。

通过 MS Access 中的以下 SQL 查询，我设法排序并找到 MIN（日期和时间）作为开始/打卡，MAX（日期和时间）作为每天打卡（日期/工作日期）

SELECT MIN(a.CHECKTIME) as work_start_time,  MAX(a.CHECKTIME)   as work_end_time,  b.CheckDate as work_date, WEEKDAY(b.CheckDate,2) as work_day, a.USERID  as user_id 

FROM CHECKINOUT a RIGHT JOIN (
  SELECT DISTINCT
  DateValue(CHECKTIME) as CheckDate,
  USERID FROM CHECKINOUT
  WHERE ( MONTH(CHECKTIME) = 4  )
  AND YEAR(CHECKTIME) = 2019 AND USERID =50
)      

b on a.USERID = b.USERID
and DateValue(a.CHECKTIME) = b.CheckDate GROUP BY b.checkDate, a.USERID

查询结果：

Work_start_Time        | work_end_time        | Work_date | Work_day |User_ID
1/04/2019 8:30:19 AM   | 1/04/2019 8:30:19 AM |  1/04/2019| 1        |50
2/04/2019 12:16:38 AM  | 2/04/2019 9:40:52 AM |  2/04/2019| 2        |50
3/04/2019 1:25:29 AM   | 3/04/2019 1:25:29 AM |  3/04/2019| 3        |50
4/04/2019 7:09:26 AM   | 4/04/2019 9:10:37 PM |  4/04/2019| 4        |50

我想做的是，如果第二天 (CHECKTIME +1) work_start_time 小于 5AM，那么第二天的 MIN(CHECKTIME) 就变成了 Work_end_time。

SELECT MIN(a.CHECKTIME) as work_start_time,  

IF (MIN(a.CHECKTIME+1) < 5:00:00 AM )
 MIN(a.CHECKTIME+1) as work_end_time,
ELSE
 MAX(a.CHECKTIME)   as work_end_time, 

 b.CheckDate as work_date, WEEKDAY(b.CheckDate,2) as work_day, a.USERID  as user_id 

FROM CHECKINOUT a RIGHT JOIN (
  SELECT DISTINCT
  DateValue(CHECKTIME) as CheckDate,
  USERID FROM CHECKINOUT
  WHERE ( MONTH(CHECKTIME) = 4  )
  AND YEAR(CHECKTIME) = 2019 AND USERID =50
)        

b on a.USERID = b.USERID
and DateValue(a.CHECKTIME) = b.CheckDate GROUP BY b.checkDate, a.USERID

查询结果：

Work_start_Time        | work_end_time        | Work_date | Work_day |User_ID
1/04/2019 8:30:19 AM   | 2/04/2019 12:16:38AM |  1/04/2019| 1        |50
2/04/2019 9:40:52 AM   | 3/04/2019 1:25:29 AM |  2/04/2019| 2        |50
3/04/2019 1:25:29 AM   | 3/04/2019 1:25:29 AM |  3/04/2019| 3        |50
4/04/2019 7:09:26 AM   | 4/04/2019 9:10:37 PM |  4/04/2019| 4        |50

【参考方案1】：

尝试使用IIF：

IIF(TimeValue(MIN(a.CHECKTIME)) < #05:00:00#), MIN(a.CHECKTIME+1), MAX(a.CHECKTIME)) as work_end_time, 

【讨论】：

感谢 Gustav，IIF 声明有效，但总是错误的。我假设

是的。必须排除日期部分。请查看编辑后的答案。

谢谢古斯塔夫，不知何故仍未给出我预期的逻辑。但它给了我一个开始和玩耍的点。非常感谢。真棒

您可以提供一些示例数据。但如果您觉得是，请标记为已回答。