Display Datediff parameter

Question

I need to show to show a parameter using datediff function, where user should be able to select the number from that.
this is my calculation 'DateDiff("d",Fields!NOV_JOB_CREATION_DATE.Value,Today())'
i was able to add this field to view, but i want this t custom parameter here user should be able to select number.
i tried multiple ways but no luck.
any help would be appreciated.
DateDiff("d",Fields!NOV_JOB_CREATION_DATE.Value,Today())

Answer 1

Hi,

Try with this expression :

DateDiff(DateInterval.Day, Fields!NOV_JOB_CREATION_DATE.Value, Today)

---

Ousama EL HOR

[If a post helps to resolve your issue, please click the "Mark as Answer"  of that post or click "Vote as helpful"  button of that post. By marking a post as Answered or Helpful, you help others find the answer faster. ]

 [User Page]     [MSDN Page]     [Blog]     [Linkedin]

Answer 2

Hi  PavanKumar664

According to your description , you could try to add one more calculated field for your dataset ,and then add the calculated field  for the parameter available value .

See:

Hope it can help you.

Best Regards,

Eric Liu

---

Best Regards, Eric Liu MSDN Community Support Please remember to click Mark as Answer if the responses that resolved your issue, and to click Unmark as Answer if not. This can be beneficial to other community members reading this thread.