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Find the age of the oldest file in one line or return zero
Determining number of files in a directory without counting themCheck for subdirectoryAdd mtime to grep -c output and sort the output by mtimehow to find files of current date in remote server and copy those file to local server using rcp using shell scripting (ksh)?Search for files that contain a string and list their names sorted by the modified daterecursive search and print most recent tar.gz file in each dirFind and rm command deleted the files inside the directory and the directory itselfHow do I change the sorting method of files used by asterisk (*) in bash?Find files, print creation date and file namescp <source>:<filepath> <dest> works from command line, but cp: cannot stat error from script. Why?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
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
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
2
Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.
– Jeff Schaller♦
14 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 thestat
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
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
shell-script files directory
asked 14 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♦
14 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 thestat
stuff without having to do a single line shell loop
– mpez0
4 hours ago
add a comment |
2
Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.
– Jeff Schaller♦
14 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 thestat
stuff without having to do a single line shell loop
– mpez0
4 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♦
14 hours ago
Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.
– Jeff Schaller♦
14 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
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
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
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 |
2 Answers
2
active
oldest
votes
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 variabled
BEGIN m=d
initializem
tod
$0 < m m = $0
keeping track of the minimum inm
END print d - m
print the difference.
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 |
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 $ARGV[0]' /path/to/dir/*(N-Om[1])
(N-Om[1])
: glob qualifier:N
: turns onnullglob
for that glob. So if there's no file in the directory, expands to nothing causingperl
's-M
to returnundef
.-
: causes next glob qualifiers to apply on the target of symlinksOm
: reverse (capital) order by modification time (so from oldest to newest likels -rt
)[1]
: select first matching file only
-M file
: gets the age of the content of the file.0+
or86400*
force a conversion to number (for theundef
case).
I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error:-bash: syntax error near unexpected token '('
, the one giving me an error is running a rather old version of perl:v5.16.3
– Georgе Stoyanov
9 hours ago
1
@GeorgеStoyanov, the syntax is forzsh
, notbash
.
– Stéphane Chazelas
9 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 variabled
BEGIN m=d
initializem
tod
$0 < m m = $0
keeping track of the minimum inm
END print d - m
print the difference.
unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
– Georgе Stoyanov
13 hours ago
@George ah, oops, I inverted the check for min
– muru
13 hours ago
add a comment |
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 variabled
BEGIN m=d
initializem
tod
$0 < m m = $0
keeping track of the minimum inm
END print d - m
print the difference.
unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
– Georgе Stoyanov
13 hours ago
@George ah, oops, I inverted the check for min
– muru
13 hours ago
add a comment |
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 variabled
BEGIN m=d
initializem
tod
$0 < m m = $0
keeping track of the minimum inm
END print d - m
print the difference.
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 variabled
BEGIN m=d
initializem
tod
$0 < m m = $0
keeping track of the minimum inm
END print d - m
print the difference.
edited 12 hours ago
Stéphane Chazelas
313k57592948
313k57592948
answered 13 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
13 hours ago
@George ah, oops, I inverted the check for min
– muru
13 hours ago
add a comment |
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
unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
– Georgе Stoyanov
13 hours ago
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
@George ah, oops, I inverted the check for min
– muru
13 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 onnullglob
for that glob. So if there's no file in the directory, expands to nothing causingperl
's-M
to returnundef
.-
: causes next glob qualifiers to apply on the target of symlinksOm
: reverse (capital) order by modification time (so from oldest to newest likels -rt
)[1]
: select first matching file only
-M file
: gets the age of the content of the file.0+
or86400*
force a conversion to number (for theundef
case).
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 forzsh
, notbash
.
– Stéphane Chazelas
9 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 onnullglob
for that glob. So if there's no file in the directory, expands to nothing causingperl
's-M
to returnundef
.-
: causes next glob qualifiers to apply on the target of symlinksOm
: reverse (capital) order by modification time (so from oldest to newest likels -rt
)[1]
: select first matching file only
-M file
: gets the age of the content of the file.0+
or86400*
force a conversion to number (for theundef
case).
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 forzsh
, notbash
.
– Stéphane Chazelas
9 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 onnullglob
for that glob. So if there's no file in the directory, expands to nothing causingperl
's-M
to returnundef
.-
: causes next glob qualifiers to apply on the target of symlinksOm
: reverse (capital) order by modification time (so from oldest to newest likels -rt
)[1]
: select first matching file only
-M file
: gets the age of the content of the file.0+
or86400*
force a conversion to number (for theundef
case).
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 onnullglob
for that glob. So if there's no file in the directory, expands to nothing causingperl
's-M
to returnundef
.-
: causes next glob qualifiers to apply on the target of symlinksOm
: reverse (capital) order by modification time (so from oldest to newest likels -rt
)[1]
: select first matching file only
-M file
: gets the age of the content of the file.0+
or86400*
force a conversion to number (for theundef
case).
edited 8 hours ago
answered 12 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
9 hours ago
1
@GeorgеStoyanov, the syntax is forzsh
, notbash
.
– Stéphane Chazelas
9 hours ago
add a comment |
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 forzsh
, notbash
.
– Stéphane Chazelas
9 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
9 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
9 hours ago
1
1
@GeorgеStoyanov, the syntax is for
zsh
, not bash
.– Stéphane Chazelas
9 hours ago
@GeorgеStoyanov, the syntax is for
zsh
, not bash
.– Stéphane Chazelas
9 hours ago
add a comment |
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Out of curiosity, why does it have to be in one line? It's much less readable & maintainable that way.
– Jeff Schaller♦
14 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 thestat
stuff without having to do a single line shell loop– mpez0
4 hours ago