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Find the age of the oldest file in one line or return zero


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







4















I want to find the age of the oldest file in a certain directory or return 0 if there aren't any files in this directory. I also need a one-line command doing it. So far this is my command for finding the age in seconds of the oldest file in the directory:



expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))


The problem is that if there are no files it is returning the following error:



$ expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))
stat: cannot stat ‘0’: No such file or directory
-bash: 1554373460 - : syntax error: operand expected (error token is "- ")


So in this case I want the command to return just 0 and to suppress the error printout.










share|improve this question


















  • 2





    Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.

    – Jeff Schaller
    16 hours ago











  • I am passing this line to specialized software. Then according to the output of the command, I can trigger an alarm and if I make it on more than a single line, I need to write more complex logic. The idea is to check a specific directory where there should not be any files for more than 20 seconds, I want to trigger an alarm if the age of the oldest file is more than 30 seconds.

    – Georgе Stoyanov
    16 hours ago













  • also I would be very happy if you have any ideas, how I can simplify my command for finding the age of the oldest file

    – Georgе Stoyanov
    16 hours ago











  • ls -lt | tail -1 will give you the oldest file; you can parse out the date or go through the stat stuff without having to do a single line shell loop

    – mpez0
    7 hours ago




















4















I want to find the age of the oldest file in a certain directory or return 0 if there aren't any files in this directory. I also need a one-line command doing it. So far this is my command for finding the age in seconds of the oldest file in the directory:



expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))


The problem is that if there are no files it is returning the following error:



$ expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))
stat: cannot stat ‘0’: No such file or directory
-bash: 1554373460 - : syntax error: operand expected (error token is "- ")


So in this case I want the command to return just 0 and to suppress the error printout.










share|improve this question


















  • 2





    Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.

    – Jeff Schaller
    16 hours ago











  • I am passing this line to specialized software. Then according to the output of the command, I can trigger an alarm and if I make it on more than a single line, I need to write more complex logic. The idea is to check a specific directory where there should not be any files for more than 20 seconds, I want to trigger an alarm if the age of the oldest file is more than 30 seconds.

    – Georgе Stoyanov
    16 hours ago













  • also I would be very happy if you have any ideas, how I can simplify my command for finding the age of the oldest file

    – Georgе Stoyanov
    16 hours ago











  • ls -lt | tail -1 will give you the oldest file; you can parse out the date or go through the stat stuff without having to do a single line shell loop

    – mpez0
    7 hours ago
















4












4








4


1






I want to find the age of the oldest file in a certain directory or return 0 if there aren't any files in this directory. I also need a one-line command doing it. So far this is my command for finding the age in seconds of the oldest file in the directory:



expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))


The problem is that if there are no files it is returning the following error:



$ expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))
stat: cannot stat ‘0’: No such file or directory
-bash: 1554373460 - : syntax error: operand expected (error token is "- ")


So in this case I want the command to return just 0 and to suppress the error printout.










share|improve this question














I want to find the age of the oldest file in a certain directory or return 0 if there aren't any files in this directory. I also need a one-line command doing it. So far this is my command for finding the age in seconds of the oldest file in the directory:



expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))


The problem is that if there are no files it is returning the following error:



$ expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))
stat: cannot stat ‘0’: No such file or directory
-bash: 1554373460 - : syntax error: operand expected (error token is "- ")


So in this case I want the command to return just 0 and to suppress the error printout.







shell-script files directory






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 16 hours ago









Georgе StoyanovGeorgе Stoyanov

164421




164421








  • 2





    Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.

    – Jeff Schaller
    16 hours ago











  • I am passing this line to specialized software. Then according to the output of the command, I can trigger an alarm and if I make it on more than a single line, I need to write more complex logic. The idea is to check a specific directory where there should not be any files for more than 20 seconds, I want to trigger an alarm if the age of the oldest file is more than 30 seconds.

    – Georgе Stoyanov
    16 hours ago













  • also I would be very happy if you have any ideas, how I can simplify my command for finding the age of the oldest file

    – Georgе Stoyanov
    16 hours ago











  • ls -lt | tail -1 will give you the oldest file; you can parse out the date or go through the stat stuff without having to do a single line shell loop

    – mpez0
    7 hours ago
















  • 2





    Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.

    – Jeff Schaller
    16 hours ago











  • I am passing this line to specialized software. Then according to the output of the command, I can trigger an alarm and if I make it on more than a single line, I need to write more complex logic. The idea is to check a specific directory where there should not be any files for more than 20 seconds, I want to trigger an alarm if the age of the oldest file is more than 30 seconds.

    – Georgе Stoyanov
    16 hours ago













  • also I would be very happy if you have any ideas, how I can simplify my command for finding the age of the oldest file

    – Georgе Stoyanov
    16 hours ago











  • ls -lt | tail -1 will give you the oldest file; you can parse out the date or go through the stat stuff without having to do a single line shell loop

    – mpez0
    7 hours ago










2




2





Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.

– Jeff Schaller
16 hours ago





Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.

– Jeff Schaller
16 hours ago













I am passing this line to specialized software. Then according to the output of the command, I can trigger an alarm and if I make it on more than a single line, I need to write more complex logic. The idea is to check a specific directory where there should not be any files for more than 20 seconds, I want to trigger an alarm if the age of the oldest file is more than 30 seconds.

– Georgе Stoyanov
16 hours ago







I am passing this line to specialized software. Then according to the output of the command, I can trigger an alarm and if I make it on more than a single line, I need to write more complex logic. The idea is to check a specific directory where there should not be any files for more than 20 seconds, I want to trigger an alarm if the age of the oldest file is more than 30 seconds.

– Georgе Stoyanov
16 hours ago















also I would be very happy if you have any ideas, how I can simplify my command for finding the age of the oldest file

– Georgе Stoyanov
16 hours ago





also I would be very happy if you have any ideas, how I can simplify my command for finding the age of the oldest file

– Georgе Stoyanov
16 hours ago













ls -lt | tail -1 will give you the oldest file; you can parse out the date or go through the stat stuff without having to do a single line shell loop

– mpez0
7 hours ago







ls -lt | tail -1 will give you the oldest file; you can parse out the date or go through the stat stuff without having to do a single line shell loop

– mpez0
7 hours ago












2 Answers
2






active

oldest

votes


















4














If it must be one line:



stat -c %Y ./* 2>/dev/null | awk -v d="$(date +%s)" 'BEGIN {m=d} $0 < m {m = $0} END {print d - m}'




  • stat -c %Y ./* 2>/dev/null print the timestamp of all files, ignoring errors (so no files results in no output)


  • With awk:





    • -v d="$(date +%s)" save the current timestamp in a variable d


    • BEGIN {m=d} initialize m to d


    • $0 < m {m = $0} keeping track of the minimum in m


    • END {print d - m} print the difference.








share|improve this answer


























  • unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

    – Georgе Stoyanov
    16 hours ago











  • @George ah, oops, I inverted the check for min

    – muru
    16 hours ago



















5














With zsh and perl:



perl -le 'print 0+-M $ARGV[0]' /path/to/dir/*(N-Om[1])


(add the D glob qualifier if you also want to consider hidden files (but not . nor ..)).



Note that for symlinks, that considers the modification time of the file it resolves to. Remove the - in the glob qualifiers to consider the modification time of the symlink instead (and use (lstat$ARGV[0] && -M _) in perl to get the age of the symlink).



That gives the age in days. Multiply by 86400 to get a number of seconds:



perl -le 'print 86400*-M $ARGV[0]' /path/to/dir/*(N-Om[1])




  • (N-Om[1]): glob qualifier:



    • N: turns on nullglob for that glob. So if there's no file in the directory, expands to nothing causing perl's -M to return undef.


    • -: causes next glob qualifiers to apply on the target of symlinks


    • Om: reverse (capital) order by modification time (so from oldest to newest like ls -rt)


    • [1]: select first matching file only




  • -M file: gets the age of the content of the file.


  • 0+ or 86400* force a conversion to number (for the undef case).






share|improve this answer


























  • I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

    – Georgе Stoyanov
    12 hours ago








  • 1





    @GeorgеStoyanov, the syntax is for zsh, not bash.

    – Stéphane Chazelas
    11 hours ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














If it must be one line:



stat -c %Y ./* 2>/dev/null | awk -v d="$(date +%s)" 'BEGIN {m=d} $0 < m {m = $0} END {print d - m}'




  • stat -c %Y ./* 2>/dev/null print the timestamp of all files, ignoring errors (so no files results in no output)


  • With awk:





    • -v d="$(date +%s)" save the current timestamp in a variable d


    • BEGIN {m=d} initialize m to d


    • $0 < m {m = $0} keeping track of the minimum in m


    • END {print d - m} print the difference.








share|improve this answer


























  • unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

    – Georgе Stoyanov
    16 hours ago











  • @George ah, oops, I inverted the check for min

    – muru
    16 hours ago
















4














If it must be one line:



stat -c %Y ./* 2>/dev/null | awk -v d="$(date +%s)" 'BEGIN {m=d} $0 < m {m = $0} END {print d - m}'




  • stat -c %Y ./* 2>/dev/null print the timestamp of all files, ignoring errors (so no files results in no output)


  • With awk:





    • -v d="$(date +%s)" save the current timestamp in a variable d


    • BEGIN {m=d} initialize m to d


    • $0 < m {m = $0} keeping track of the minimum in m


    • END {print d - m} print the difference.








share|improve this answer


























  • unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

    – Georgе Stoyanov
    16 hours ago











  • @George ah, oops, I inverted the check for min

    – muru
    16 hours ago














4












4








4







If it must be one line:



stat -c %Y ./* 2>/dev/null | awk -v d="$(date +%s)" 'BEGIN {m=d} $0 < m {m = $0} END {print d - m}'




  • stat -c %Y ./* 2>/dev/null print the timestamp of all files, ignoring errors (so no files results in no output)


  • With awk:





    • -v d="$(date +%s)" save the current timestamp in a variable d


    • BEGIN {m=d} initialize m to d


    • $0 < m {m = $0} keeping track of the minimum in m


    • END {print d - m} print the difference.








share|improve this answer















If it must be one line:



stat -c %Y ./* 2>/dev/null | awk -v d="$(date +%s)" 'BEGIN {m=d} $0 < m {m = $0} END {print d - m}'




  • stat -c %Y ./* 2>/dev/null print the timestamp of all files, ignoring errors (so no files results in no output)


  • With awk:





    • -v d="$(date +%s)" save the current timestamp in a variable d


    • BEGIN {m=d} initialize m to d


    • $0 < m {m = $0} keeping track of the minimum in m


    • END {print d - m} print the difference.









share|improve this answer














share|improve this answer



share|improve this answer








edited 14 hours ago









Stéphane Chazelas

313k57592948




313k57592948










answered 16 hours ago









murumuru

37.2k589164




37.2k589164













  • unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

    – Georgе Stoyanov
    16 hours ago











  • @George ah, oops, I inverted the check for min

    – muru
    16 hours ago



















  • unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

    – Georgе Stoyanov
    16 hours ago











  • @George ah, oops, I inverted the check for min

    – muru
    16 hours ago

















unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

– Georgе Stoyanov
16 hours ago





unfortunately, it does return 0 no matter if the directory is empty or it has more than one file

– Georgе Stoyanov
16 hours ago













@George ah, oops, I inverted the check for min

– muru
16 hours ago





@George ah, oops, I inverted the check for min

– muru
16 hours ago













5














With zsh and perl:



perl -le 'print 0+-M $ARGV[0]' /path/to/dir/*(N-Om[1])


(add the D glob qualifier if you also want to consider hidden files (but not . nor ..)).



Note that for symlinks, that considers the modification time of the file it resolves to. Remove the - in the glob qualifiers to consider the modification time of the symlink instead (and use (lstat$ARGV[0] && -M _) in perl to get the age of the symlink).



That gives the age in days. Multiply by 86400 to get a number of seconds:



perl -le 'print 86400*-M $ARGV[0]' /path/to/dir/*(N-Om[1])




  • (N-Om[1]): glob qualifier:



    • N: turns on nullglob for that glob. So if there's no file in the directory, expands to nothing causing perl's -M to return undef.


    • -: causes next glob qualifiers to apply on the target of symlinks


    • Om: reverse (capital) order by modification time (so from oldest to newest like ls -rt)


    • [1]: select first matching file only




  • -M file: gets the age of the content of the file.


  • 0+ or 86400* force a conversion to number (for the undef case).






share|improve this answer


























  • I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

    – Georgе Stoyanov
    12 hours ago








  • 1





    @GeorgеStoyanov, the syntax is for zsh, not bash.

    – Stéphane Chazelas
    11 hours ago
















5














With zsh and perl:



perl -le 'print 0+-M $ARGV[0]' /path/to/dir/*(N-Om[1])


(add the D glob qualifier if you also want to consider hidden files (but not . nor ..)).



Note that for symlinks, that considers the modification time of the file it resolves to. Remove the - in the glob qualifiers to consider the modification time of the symlink instead (and use (lstat$ARGV[0] && -M _) in perl to get the age of the symlink).



That gives the age in days. Multiply by 86400 to get a number of seconds:



perl -le 'print 86400*-M $ARGV[0]' /path/to/dir/*(N-Om[1])




  • (N-Om[1]): glob qualifier:



    • N: turns on nullglob for that glob. So if there's no file in the directory, expands to nothing causing perl's -M to return undef.


    • -: causes next glob qualifiers to apply on the target of symlinks


    • Om: reverse (capital) order by modification time (so from oldest to newest like ls -rt)


    • [1]: select first matching file only




  • -M file: gets the age of the content of the file.


  • 0+ or 86400* force a conversion to number (for the undef case).






share|improve this answer


























  • I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

    – Georgе Stoyanov
    12 hours ago








  • 1





    @GeorgеStoyanov, the syntax is for zsh, not bash.

    – Stéphane Chazelas
    11 hours ago














5












5








5







With zsh and perl:



perl -le 'print 0+-M $ARGV[0]' /path/to/dir/*(N-Om[1])


(add the D glob qualifier if you also want to consider hidden files (but not . nor ..)).



Note that for symlinks, that considers the modification time of the file it resolves to. Remove the - in the glob qualifiers to consider the modification time of the symlink instead (and use (lstat$ARGV[0] && -M _) in perl to get the age of the symlink).



That gives the age in days. Multiply by 86400 to get a number of seconds:



perl -le 'print 86400*-M $ARGV[0]' /path/to/dir/*(N-Om[1])




  • (N-Om[1]): glob qualifier:



    • N: turns on nullglob for that glob. So if there's no file in the directory, expands to nothing causing perl's -M to return undef.


    • -: causes next glob qualifiers to apply on the target of symlinks


    • Om: reverse (capital) order by modification time (so from oldest to newest like ls -rt)


    • [1]: select first matching file only




  • -M file: gets the age of the content of the file.


  • 0+ or 86400* force a conversion to number (for the undef case).






share|improve this answer















With zsh and perl:



perl -le 'print 0+-M $ARGV[0]' /path/to/dir/*(N-Om[1])


(add the D glob qualifier if you also want to consider hidden files (but not . nor ..)).



Note that for symlinks, that considers the modification time of the file it resolves to. Remove the - in the glob qualifiers to consider the modification time of the symlink instead (and use (lstat$ARGV[0] && -M _) in perl to get the age of the symlink).



That gives the age in days. Multiply by 86400 to get a number of seconds:



perl -le 'print 86400*-M $ARGV[0]' /path/to/dir/*(N-Om[1])




  • (N-Om[1]): glob qualifier:



    • N: turns on nullglob for that glob. So if there's no file in the directory, expands to nothing causing perl's -M to return undef.


    • -: causes next glob qualifiers to apply on the target of symlinks


    • Om: reverse (capital) order by modification time (so from oldest to newest like ls -rt)


    • [1]: select first matching file only




  • -M file: gets the age of the content of the file.


  • 0+ or 86400* force a conversion to number (for the undef case).







share|improve this answer














share|improve this answer



share|improve this answer








edited 11 hours ago

























answered 14 hours ago









Stéphane ChazelasStéphane Chazelas

313k57592948




313k57592948













  • I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

    – Georgе Stoyanov
    12 hours ago








  • 1





    @GeorgеStoyanov, the syntax is for zsh, not bash.

    – Stéphane Chazelas
    11 hours ago



















  • I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

    – Georgе Stoyanov
    12 hours ago








  • 1





    @GeorgеStoyanov, the syntax is for zsh, not bash.

    – Stéphane Chazelas
    11 hours ago

















I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

– Georgе Stoyanov
12 hours ago







I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error: -bash: syntax error near unexpected token '(', the one giving me an error is running a rather old version of perl: v5.16.3

– Georgе Stoyanov
12 hours ago






1




1





@GeorgеStoyanov, the syntax is for zsh, not bash.

– Stéphane Chazelas
11 hours ago





@GeorgеStoyanov, the syntax is for zsh, not bash.

– Stéphane Chazelas
11 hours ago


















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