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

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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.

shell-script files directory

share|improve this question

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.
  
  – Jeff Schaller♦
  13 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
  13 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
  13 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
  4 hours ago
  
  

  
  
  
  

add a comment | 

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.

shell-script files directory

share|improve this question

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.
  
  – Jeff Schaller♦
  13 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
  13 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
  13 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
  4 hours ago
  
  

  
  
  
  

add a comment | 

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

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

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

$ 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 "- ")

share|improve this question

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

share|improve this question

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

asked 13 hours ago

Georgе StoyanovGeorgе Stoyanov

164421

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

edited 11 hours ago

Stéphane Chazelas

313k57592948

answered 13 hours ago

murumuru

37.2k589164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
  
  – Georgе Stoyanov
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @George ah, oops, I inverted the check for min
  
  – muru
  13 hours ago
  

  
  
  
  

add a comment | 

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

edited 8 hours ago

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @GeorgеStoyanov, the syntax is for zsh, not bash.
  
  – Stéphane Chazelas
  8 hours ago
  

  
  
  
  

add a comment | 

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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

edited 11 hours ago

Stéphane Chazelas

313k57592948

answered 13 hours ago

murumuru

37.2k589164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
  
  – Georgе Stoyanov
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @George ah, oops, I inverted the check for min
  
  – muru
  13 hours ago
  

  
  
  
  

add a comment | 

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

edited 11 hours ago

Stéphane Chazelas

313k57592948

answered 13 hours ago

murumuru

37.2k589164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
  
  – Georgе Stoyanov
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @George ah, oops, I inverted the check for min
  
  – muru
  13 hours ago
  

  
  
  
  

add a comment | 

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

edited 11 hours ago

Stéphane Chazelas

313k57592948

answered 13 hours ago

murumuru

37.2k589164

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

share|improve this answer

edited 11 hours ago

Stéphane Chazelas

313k57592948

answered 13 hours ago

murumuru

37.2k589164

edited 11 hours ago

Stéphane Chazelas

313k57592948

edited 11 hours ago

Stéphane Chazelas

313k57592948

answered 13 hours ago

murumuru

37.2k589164

answered 13 hours ago

murumuru

37.2k589164

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

edited 8 hours ago

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @GeorgеStoyanov, the syntax is for zsh, not bash.
  
  – Stéphane Chazelas
  8 hours ago
  

  
  
  
  

add a comment | 

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

edited 8 hours ago

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @GeorgеStoyanov, the syntax is for zsh, not bash.
  
  – Stéphane Chazelas
  8 hours ago
  

  
  
  
  

add a comment | 

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

edited 8 hours ago

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

313k57592948

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

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

share|improve this answer

edited 8 hours ago

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

313k57592948

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

313k57592948

answered 11 hours ago

Stéphane ChazelasStéphane Chazelas

313k57592948