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Find the age of the oldest file in one line or return zero
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}
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♦
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 thestat
stuff without having to do a single line shell loop
– mpez0
7 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 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 thestat
stuff without having to do a single line shell loop
– mpez0
7 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♦
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 thestat
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
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
16 hours ago
@George ah, oops, I inverted the check for min
– muru
16 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 $ARGV[0]' /path/to/dir/*(N-Om[1])
(N-Om[1])
: glob qualifier:
N
: turns onnullglob
for that glob. So if there's no file in the directory, expands to nothing causingperl
's-M
to returnundef
.
-
: causes next glob qualifiers to apply on the target of symlinks
Om
: reverse (capital) order by modification time (so from oldest to newest likels -rt
)
[1]
: select first matching file only
-M file
: gets the age of the content of the file.
0+
or86400*
force a conversion to number (for theundef
case).
I am getting just 0 as an output even though it should show me a couple of thousands of seconds, both commands actually gives me the same output and on another machine I am getting an error:-bash: syntax error near unexpected token '('
, the one giving me an error is running a rather old version of perl:v5.16.3
– Georgе Stoyanov
12 hours ago
1
@GeorgеStoyanov, the syntax is forzsh
, notbash
.
– Stéphane Chazelas
11 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 = $0}
keeping track of the minimum inm
END {print d - m}
print the difference.
unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
– Georgе Stoyanov
16 hours ago
@George ah, oops, I inverted the check for min
– muru
16 hours ago
add a comment |
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 = $0}
keeping track of the minimum inm
END {print d - m}
print the difference.
unfortunately, it does return 0 no matter if the directory is empty or it has more than one file
– Georgе Stoyanov
16 hours ago
@George ah, oops, I inverted the check for min
– muru
16 hours ago
add a comment |
If it must be one line:
stat -c %Y ./* 2>/dev/null | awk -v d="$(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 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
add a comment |
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
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 symlinks
Om
: 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
12 hours ago
1
@GeorgеStoyanov, the syntax is forzsh
, notbash
.
– Stéphane Chazelas
11 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 symlinks
Om
: 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
12 hours ago
1
@GeorgеStoyanov, the syntax is forzsh
, notbash
.
– Stéphane Chazelas
11 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 symlinks
Om
: 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 symlinks
Om
: 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 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 forzsh
, notbash
.
– Stéphane Chazelas
11 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
12 hours ago
1
@GeorgеStoyanov, the syntax is forzsh
, notbash
.
– 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
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♦
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 thestat
stuff without having to do a single line shell loop– mpez0
7 hours ago