本文介绍了如何在两个日期字段之间获得超过3分钟的时间的处理方法,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





如何在两个日期字段之间获得超过3分钟



submit_date - dtm_AlarmStartTime是两个字段的日期





选择Incident_Number,dateadd(s,submit_date +'19800','19700101')为'submit_date',dateadd( S,dtm_AlarmStartTime + '19800', '19700101'),为 'dtm_AlarmStartTime',

(submit_date - dtm_AlarmStartTime),如从HPD_Help_Desk差分

其中DATEADD(S,submit_date + '19800' ,'19700101')> CONVERT(date,sysdatetime())

Hi,

How to get morethan 3minutes between two date fields

submit_date - dtm_AlarmStartTime are two fields date


select Incident_Number, dateadd(s,submit_date+'19800','19700101') as 'submit_date',dateadd(s,dtm_AlarmStartTime+'19800','19700101') as 'dtm_AlarmStartTime' ,
(submit_date - dtm_AlarmStartTime) as Difference
from HPD_Help_Desk where dateadd(s,submit_date+'19800','19700101') >CONVERT(date,sysdatetime())

推荐答案

SELECT * FROM MyTable WHERE DATEDIFF(mi, startDate, endDate) > 3


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10-19 22:07