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Display Datediff parameter RRS feed

  • 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())
    Tuesday, May 28, 2019 8:40 AM

All replies

  • Hi,

    Try with this expression :

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


    Ousama EL HOR

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    Tuesday, May 28, 2019 8:57 AM
  • 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.

    • Proposed as answer by Mitarai Queen Thursday, May 30, 2019 1:14 AM
    Wednesday, May 29, 2019 1:26 AM