本文介绍了SQL Server中where子句中的累积条件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

大家好,
我写了一个查询,并得到如下所示的Output,但我希望所有记录的WorkingHours< 48.


请帮助我.我只想要大胆的结果.


查询和输出如下.

Hi All,
I wrote a query and I got an Output like below but I want all records whose WorkingHours <48.


Please help me.I want only result which is bold.


Query and Output is below.

SELECT OffEmail,EmployeeName,(Sum(DHours*60+DMinutes)-sum(MHours*60+MMinutes))/60  as WorkingHours,(Sum(DHours*60+DMinutes)-sum(MHours*60+MMinutes))%60  as WorkingMins from Attendance_Master  where LoginTime between '2011-04-01 00:00:00.000' and '2011-05-01 00:00:00.000'
group by OffEmail,EmployeeName






输出

OffEmail员工姓名WorkingHours WorkingMins

[email protected] Abhishek Kumar Gupta 204 23
[email protected] Abhishek Mishra 191 47
[email protected]管理员24 0
[email protected] Ajay Pratap Singh 219 41
[email protected]阿米特·库玛·辛格202 42
[email protected]阿米特·特瓦里0 0
[email protected] Amresh Sahu 202 24
Anjum Jeelani 173 15






Output

OffEmail EmployeeName WorkingHours WorkingMins

[email protected] Kumar Gupta20423
[email protected] Mishra 19147
[email protected] Admin 240
[email protected] Ajay Pratap Singh21941
[email protected] Amit Kumar Singh20242
[email protected] Amit Tewari 00
[email protected] Amresh Sahu 20224
[email protected] Jeelani 17315

推荐答案

SELECT OffEmail,EmployeeName,(Sum(DHours*60+DMinutes)-sum(MHours*60+MMinutes))/60  as WorkingHours,(Sum(DHours*60+DMinutes)-sum(MHours*60+MMinutes))%60  as WorkingMins from Attendance_Master  where LoginTime between '2011-04-01 00:00:00.000' and '2011-05-01 00:00:00.000'
group by OffEmail,EmployeeName
having (Sum(DHours*60+DMinutes)-sum(MHours*60+MMinutes))/60 < 48 


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08-19 16:12