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Convert seconds to minutes


recursive search for a pattern, then for each match print out the specific SEQUENCE: line number, file name, and no file contentsExtract Values from simple html file via grep/awkStrategy to extract movies's name from this ununiform dataset?Print a line in stdout that matches an expression if the output contains another expressionUsing Awk to Search Bro Log for IPs/Keywords Found in Online Text FileHow to limit printed output based on number of occurences (AWK)stream editing tools: output what's scrapedText processing - Building a slurm topology.conf file from ibnetdiscover outputBash script command to search and highlight text in terminal bufferSumming minutes and seconds with awk






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2















I currently have a function that prints the position and duration from cmus and formats it like "1/500". The issue I'm having is that I would like the position and duration data to be presented in minutes as opposed to seconds (0:01/8:20 instead of 1/500) but I'm out of ideas on how to achieve this.



Currently the relevant part of the function looks like this:



print_music()
 grep --text '^duration'





bash awk







share|improve this question
























  • Is this stackoverflow.com/q/13422743/4023950 what you need?

    – αғsнιη
    Apr 2 at 20:02











  • Using date -d@$variable +%-M%S does the conversion but unlike @glenn jackman solution, date will wrap to 0:00 at 3,600 secs.

    – bu5hman
    Apr 2 at 20:18


















2















I currently have a function that prints the position and duration from cmus and formats it like "1/500". The issue I'm having is that I would like the position and duration data to be presented in minutes as opposed to seconds (0:01/8:20 instead of 1/500) but I'm out of ideas on how to achieve this.



Currently the relevant part of the function looks like this:



print_music()
 grep --text '^duration'





bash awk







share|improve this question
























  • Is this stackoverflow.com/q/13422743/4023950 what you need?

    – αғsнιη
    Apr 2 at 20:02











  • Using date -d@$variable +%-M%S does the conversion but unlike @glenn jackman solution, date will wrap to 0:00 at 3,600 secs.

    – bu5hman
    Apr 2 at 20:18














2












2








2


0






I currently have a function that prints the position and duration from cmus and formats it like "1/500". The issue I'm having is that I would like the position and duration data to be presented in minutes as opposed to seconds (0:01/8:20 instead of 1/500) but I'm out of ideas on how to achieve this.



Currently the relevant part of the function looks like this:



print_music()
 grep --text '^duration'





bash awk







share|improve this question
















I currently have a function that prints the position and duration from cmus and formats it like "1/500". The issue I'm having is that I would like the position and duration data to be presented in minutes as opposed to seconds (0:01/8:20 instead of 1/500) but I'm out of ideas on how to achieve this.



Currently the relevant part of the function looks like this:



print_music()
 grep --text '^duration'





bash awk




bash awk






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 2 at 20:05









αғsнιη

17.1k103069




17.1k103069










asked Apr 2 at 19:09









JoeJoe

223




223












  • Is this stackoverflow.com/q/13422743/4023950 what you need?

    – αғsнιη
    Apr 2 at 20:02











  • Using date -d@$variable +%-M%S does the conversion but unlike @glenn jackman solution, date will wrap to 0:00 at 3,600 secs.

    – bu5hman
    Apr 2 at 20:18


















  • Is this stackoverflow.com/q/13422743/4023950 what you need?

    – αғsнιη
    Apr 2 at 20:02











  • Using date -d@$variable +%-M%S does the conversion but unlike @glenn jackman solution, date will wrap to 0:00 at 3,600 secs.

    – bu5hman
    Apr 2 at 20:18

















Is this stackoverflow.com/q/13422743/4023950 what you need?

– αғsнιη
Apr 2 at 20:02





Is this stackoverflow.com/q/13422743/4023950 what you need?

– αғsнιη
Apr 2 at 20:02













Using date -d@$variable +%-M%S does the conversion but unlike @glenn jackman solution, date will wrap to 0:00 at 3,600 secs.

– bu5hman
Apr 2 at 20:18






Using date -d@$variable +%-M%S does the conversion but unlike @glenn jackman solution, date will wrap to 0:00 at 3,600 secs.

– bu5hman
Apr 2 at 20:18











1 Answer
1






active

oldest

votes


















6














This will help you:



sec2min() printf "%d:%02d" "$((10#$1 / 60))" "$((10#$1 % 60))"; 




$ sec2min 500
8:20
$ sec2min 1
0:01





share|improve this answer























  • Thanks! That's working perfectly :)

    – Joe
    Apr 2 at 22:08











  • Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

    – RubberStamp
    Apr 3 at 0:41











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














This will help you:



sec2min() printf "%d:%02d" "$((10#$1 / 60))" "$((10#$1 % 60))"; 




$ sec2min 500
8:20
$ sec2min 1
0:01





share|improve this answer























  • Thanks! That's working perfectly :)

    – Joe
    Apr 2 at 22:08











  • Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

    – RubberStamp
    Apr 3 at 0:41















6














This will help you:



sec2min() printf "%d:%02d" "$((10#$1 / 60))" "$((10#$1 % 60))"; 




$ sec2min 500
8:20
$ sec2min 1
0:01





share|improve this answer























  • Thanks! That's working perfectly :)

    – Joe
    Apr 2 at 22:08











  • Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

    – RubberStamp
    Apr 3 at 0:41













6












6








6







This will help you:



sec2min() printf "%d:%02d" "$((10#$1 / 60))" "$((10#$1 % 60))"; 




$ sec2min 500
8:20
$ sec2min 1
0:01





share|improve this answer













This will help you:



sec2min() printf "%d:%02d" "$((10#$1 / 60))" "$((10#$1 % 60))"; 




$ sec2min 500
8:20
$ sec2min 1
0:01






share|improve this answer












share|improve this answer



share|improve this answer










answered Apr 2 at 20:02









glenn jackmanglenn jackman

53k573114




53k573114












  • Thanks! That's working perfectly :)

    – Joe
    Apr 2 at 22:08











  • Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

    – RubberStamp
    Apr 3 at 0:41

















  • Thanks! That's working perfectly :)

    – Joe
    Apr 2 at 22:08











  • Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

    – RubberStamp
    Apr 3 at 0:41
















Thanks! That's working perfectly :)

– Joe
Apr 2 at 22:08





Thanks! That's working perfectly :)

– Joe
Apr 2 at 22:08













Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

– RubberStamp
Apr 3 at 0:41





Here's the same operation using a second iteration of awk ... awk ' min = ($1 / 60)-(($1 % 60)*1/60) ; sec = $1 % 60 ; print min,":",sec'

– RubberStamp
Apr 3 at 0:41

















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