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DateDiff function

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Posted on by Muhammad Anas Naeemi

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Hey community,

I have created a field "Resolution Time" that will be used for calculating the time between a issue created by the client and resolved by the employee using DateDiff(Issue Creation Date, Issue Close Date, TimeUnit.hours). So for e.g. if the issue is resolved in 5 days the resolution time will be 120 hours (5 days*24 hours). The thing is that employee only works for 8 hours a day. So the employee took 40 hours to resolve the issue and not 120 hours. How can I formulate that?

This formula I have created but it is not working accurately.

Like in the above case it should give 121 hours and 8 minutes in 24hours a day format and 41 hours and 8 minutes in 8hours a day format but its giving only 121 hours in 24hours a day format and 40.33 in 8hours a day format. Kindly suggest me the accurate formula for doing it. Thank you,

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Eiken on at

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RE: DateDiff function

Hi,

You can also create two new calculate field, one is calculate hours and another calculate minutes using DateDiff.

Then create a field using formula.
Naveen Ganeshe 3,393 User Group Leader on at

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RE: DateDiff function

Good to hear that.
Muhammad Anas Naeemi 386 on at

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RE: DateDiff function

Hey everyone, thank you everyone for the response. I have achieved my desired functionality.
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Naveen Ganeshe 3,393 User Group Leader on at

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RE: DateDiff function

Here is a c# code. you can convert this
TimeSpan resolutionTime = issueCloseDate - issueCreationDate; int totalDays = (int)resolutionTime.TotalDays; int workingDays = totalDays - (totalDays/7)*2; int workingHours = workingDays * 8; int minutes = resolutionTime.Minutes;
Muhammad Anas Naeemi 386 on at

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RE: DateDiff function

Hey Naveen it is not working. Can you show me code for displaying A days, B hours and C minutes
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Naveen Ganeshe 3,393 User Group Leader on at

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RE: DateDiff function

Hi Anas,

You can give it a try:

DateDiff(IssueCloseDate,IssueCreateDate,TimeUnit.Minutes) / 60 / 8

It will result in time in days
Muhammad Anas Naeemi 386 on at

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RE: DateDiff function

You can also suggest me the formula resulting as 5 days 1 hour and 8 minutes for the above given time duration.