Looking for a little help. Need to take date data from (YEAR) field in (YYYY-MM-DDTHH:MM:SS) format and move it to (RELEASETIME) field in (DD-MM-YYYY) format. Something like "2017-04-16T07:00:00Z" to "16-04-2017" Any advice? [:huh:]

[image] owlcatz: 3/2/1998 There are several ways to get the leading zeroes to your date. Assuming, your date 3/2/1998 comes from the tag DATE: $regexp(%DATE%,'\b(\d)(?=\/)',0$1)

The YEAR field looks a lot like the proper format for field RELEASETIME. I would not change it at all. RELEASETIME has both: the day and the hour. And it needs it in the sequence of year-month-day. So, if you change the order then it might be that you get suprising results esp. when sorting as you…

[image] Solance: Looking for a little help. ... Do have a look into ... e. g. ... /t/16556/1 ... and follow the ISO 8601 date time standard format,... especially for the tagfield RELEASETIME. DD.20170421.1336.CEST

ok so this (DD-MM-YYYY) format is "broken" for (RELEASETIME) field. but it always could be added in (COMMENT) or somewhere else, so is there a way to convert date somehow?

Create an action of the type "Format value" for COMMENT or somewhere else. Format string: $regexp('2017-04-16T07:00:00Z',(\d+)-(\d+)-(\d+)T.*,$3-$2-$1) Instead of the '2017-04-16T07:00:00Z' part of the expression set the source field name, e.g. %year%.

[image] ohrenkino: Create an action of the type "Format value" for COMMENT or somewhere else. Format string: $regexp('2017-04-16T07:00:00Z',(\d+)-(\d+)-(\d+)T.*,$3-$2-$1) Instead of the '2017-04-16T07:00:00Z' part of the expression set the source field name, e.g. %year%. perfect, exactly wh…

Here's a different variation I'm trying to solve. How do I change format of type: 1/13/2001 -> 2001-01-13 3/2/1998 -> 1998-03-02 I feel like I'm close, but no cigar yet... Thanks!

Could you show us what you have tried so far - it is much easier to build on that.

Well To be honest, no so far really - Just testing diff variations of the above but without dashes in the input field - something like : $regexp(%date%',(\d+)\/(\d+)\/(\d+),$3-$2-$1) , I'm also using an online regex builder but in all honesty, trying to figure this out is very painful to my brain…

I think in [image] owlcatz: $3-$2-$1) the order is the wrong one - it should be $3-$1-$2 to turn 1/13/2001 into 2001-01-13 In a second step you could add the zero like: $regexp($regexp('1/13/2001',(\d+)/(\d+)/(\d+),$3-$1-$2),(\d\d\d\d)-(\d)-(\d\d),$1-0$2-$3)

Thanks man, that's much closer. I think I can get there from here, but if not I'll post back later. Love this community - where else can you post on a 5 year old thread and get an answer in minutes? You guys rock, thanks @ohrenkino & @LyricsLover

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