Open always most recent file
Posted: Tue Aug 23, 2016 3:55 pm
Hello Oleg,
Short: Is there any procedure / action which always opens the most recent file in a certain folder?
Long: Every day we need to open a file like SDP {Day}-{MonthNo}-{year}. When it opens on a day, it should open the most recent one. In most of the case this is made the day before, so C:\Robotask\SDP\SDP {IncDays(-1)}-{MonthNo}-{Year} will do.
On a monday, he should open the file from friday, C:\Robotask\SDP\SDP {IncDays(-3)}-{MonthNo}-{Year} will do here as well. Works perfectly.
But, it could be that the person (who makes these reports now by hand) is on holiday. Then a -1, -3 or any other value can be wrong. In that case, he should open the most recent version, but...how can I do that in Robotask?
Short: Is there any procedure / action which always opens the most recent file in a certain folder?
Long: Every day we need to open a file like SDP {Day}-{MonthNo}-{year}. When it opens on a day, it should open the most recent one. In most of the case this is made the day before, so C:\Robotask\SDP\SDP {IncDays(-1)}-{MonthNo}-{Year} will do.
On a monday, he should open the file from friday, C:\Robotask\SDP\SDP {IncDays(-3)}-{MonthNo}-{Year} will do here as well. Works perfectly.
But, it could be that the person (who makes these reports now by hand) is on holiday. Then a -1, -3 or any other value can be wrong. In that case, he should open the most recent version, but...how can I do that in Robotask?