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Anonymous's avatar
Anonymous
Not applicable
7 years ago
Solved

EARLIER chalenge

Hello Power friends,   I have a table with IT tickets like that: Id_ticket, date, old_status, new_status,days_to_fix 100,01/01/2018,<blank>,"new ticket" ,0 // user report a problem opening a tick...
Need Help
  • Chihiro's avatar
    Chihiro
    7 years ago

    Greg gave you the basic construct, and link explaining the concept.

     

    You'll just need to change RETURN portion to only return result when [new_status]="fixed".

     

    So...

     

    Column =
VAR __max =
    MAXX (
        FILTER (
            ALL ( 'Table' ),
            [Id_ticket] = EARLIER ( [Id_ticket] )
                && [date] < EARLIER ( [date] )
                && [new_status] = EARLIER ( [old_status] )
        ),
        [date]
    )
RETURN
    IF ( [new_status] = "fixed", [date] - __max, 0 )

    Result

Anonymous's avatar
Anonymous
Not applicable
7 years ago

Thanks a lot Greg_Deckler and Chihiro. It worked very well. 

 

Anonymous's avatar
Anonymous
Not applicable
7 years ago

Hello Power friends,

 

I have a table with IT tickets like that:

Id_ticket, date, old_status, new_status,days_to_fix

100,01/01/2018,<blank>,"new ticket" ,0                // user report a problem opening a ticket

100,01/01/2018,"new ticket","working",0               // IT specialist start working on it

100,01/03/2018,"working","fixed", 2                      // IT specialist solved it in two days

100,01/04/2018,"fixed" , "working",0                  // user was not satisfied with solution and put the ticket on working status again

100,01/09/2018,"working","fixed", 5                    // IT specialist got new solution in five days

100,01/10/2018,"fixed","closed", 0                       // User accept this solution

 

The question is: "How can I calculate the "days_to_fix" collum in this example? Note that It has to be filled every time we have a "fixed" on new_status collum. 

 

Thanks for your time and knowledge sharing.