How to run a program for a fixed period of time?
I am looking for a method built-in to ubuntu that will allow me to run a script or program or whatever for a fixed period of time.
I found a program that does this in a way I like, but the package is unavailable for Ubuntu. In any case, I was hoping for something built-in.
The only thing i can think of is to get the current time and set a cron job 30 minutes from 'now' that will kill the program. I was hoping there was a way to do this without setting up a script, but if I need to - it wont be the end of the world. After the 30 minute interval I would like to put my laptop in a sleep mode, but this can be separate from the timer thing.
Thanks in advance.
command-line scripts timeout
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
asked May 26 '11 at 16:50
Teque5Teque5
207139
add a comment |
I am looking for a method built-in to ubuntu that will allow me to run a script or program or whatever for a fixed period of time.
I found a program that does this in a way I like, but the package is unavailable for Ubuntu. In any case, I was hoping for something built-in.
The only thing i can think of is to get the current time and set a cron job 30 minutes from 'now' that will kill the program. I was hoping there was a way to do this without setting up a script, but if I need to - it wont be the end of the world. After the 30 minute interval I would like to put my laptop in a sleep mode, but this can be separate from the timer thing.
Thanks in advance.
command-line scripts timeout
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
asked May 26 '11 at 16:50
Teque5Teque5
207139
add a comment |
I am looking for a method built-in to ubuntu that will allow me to run a script or program or whatever for a fixed period of time.
I found a program that does this in a way I like, but the package is unavailable for Ubuntu. In any case, I was hoping for something built-in.
The only thing i can think of is to get the current time and set a cron job 30 minutes from 'now' that will kill the program. I was hoping there was a way to do this without setting up a script, but if I need to - it wont be the end of the world. After the 30 minute interval I would like to put my laptop in a sleep mode, but this can be separate from the timer thing.
Thanks in advance.
command-line scripts timeout
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
asked May 26 '11 at 16:50
Teque5Teque5
207139
I am looking for a method built-in to ubuntu that will allow me to run a script or program or whatever for a fixed period of time.
I found a program that does this in a way I like, but the package is unavailable for Ubuntu. In any case, I was hoping for something built-in.
The only thing i can think of is to get the current time and set a cron job 30 minutes from 'now' that will kill the program. I was hoping there was a way to do this without setting up a script, but if I need to - it wont be the end of the world. After the 30 minute interval I would like to put my laptop in a sleep mode, but this can be separate from the timer thing.
Thanks in advance.
command-line scripts timeout
command-line scripts timeout
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
asked May 26 '11 at 16:50
Teque5Teque5
207139
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
asked May 26 '11 at 16:50
Teque5Teque5
207139
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
edited Sep 7 '13 at 20:48
Braiam
52.2k20137222
52.2k20137222
asked May 26 '11 at 16:50
Teque5Teque5
207139
asked May 26 '11 at 16:50
Teque5Teque5
207139
asked May 26 '11 at 16:50
Teque5Teque5
207139
207139
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
I've just wrote the following and it seems to work:
ping google.com& PID=$!; sleep 3; kill $PID
Of course you should substitute ping with the command you want to run and 3 with a timeout in seconds. Let me know if you need a more detailed explanation on how it works.
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Just remove the backslash before!. Once more sorry for the confusion.
– Adam Byrtek
May 26 '11 at 19:09
add a comment |
Why not use /usr/bin/timeout?
$ timeout --help
Usage: timeout [OPTION] DURATION COMMAND [ARG]...
or: timeout [OPTION]
Start COMMAND, and kill it if still running after DURATION.
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
add a comment |
A simple (and not much tested) version of hatimerun:
#!/bin/sh
help(){
echo "Usage" >&2
echo " $0 -t TIME COMMAND" >&2
exit 1
}
TEMP=`getopt -o t: -n "$0" -- "$@"`
if [ $? != 0 ] ; then
help
fi
eval set -- "$TEMP"
while true ; do
case "$1" in
-t) timeout=$2; shift 2;;
--) shift; break;;
esac
done
if [ -z "$timeout" ]; then
help
fi
cmd="$*"
$cmd&
echo "kill $!" | at now+$timeout
See the manpage of at for how to specify the time. Like with at the minimum time resolution is 1 minute.
Another way is to use upstart instead of this script.
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
add a comment |
Using Perl:
perl -e "alarm 5; exec @ARGV" systemctl list-timers
Using timelimit command:
sudo apt install timelimit
timelimit -t5 node app.js
Using timeout command:
timeout -sHUP time command
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
2
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
1
@Olorin now the post is edited by OP including another solutions as well as a sample of thetimeoutcommand.
– αғsнιη
Feb 11 at 5:48
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.
– Rakib Fiha
Feb 11 at 7:00
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
I've just wrote the following and it seems to work:
ping google.com& PID=$!; sleep 3; kill $PID
Of course you should substitute ping with the command you want to run and 3 with a timeout in seconds. Let me know if you need a more detailed explanation on how it works.
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Just remove the backslash before!. Once more sorry for the confusion.
– Adam Byrtek
May 26 '11 at 19:09
add a comment |
I've just wrote the following and it seems to work:
ping google.com& PID=$!; sleep 3; kill $PID
Of course you should substitute ping with the command you want to run and 3 with a timeout in seconds. Let me know if you need a more detailed explanation on how it works.
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Just remove the backslash before!. Once more sorry for the confusion.
– Adam Byrtek
May 26 '11 at 19:09
add a comment |
I've just wrote the following and it seems to work:
ping google.com& PID=$!; sleep 3; kill $PID
Of course you should substitute ping with the command you want to run and 3 with a timeout in seconds. Let me know if you need a more detailed explanation on how it works.
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
I've just wrote the following and it seems to work:
ping google.com& PID=$!; sleep 3; kill $PID
Of course you should substitute ping with the command you want to run and 3 with a timeout in seconds. Let me know if you need a more detailed explanation on how it works.
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
edited Sep 7 '13 at 21:09
Eliah Kagan
82.5k22227369
82.5k22227369
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
answered May 26 '11 at 17:18
Adam ByrtekAdam Byrtek
8,32712526
8,32712526
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Just remove the backslash before!. Once more sorry for the confusion.
– Adam Byrtek
May 26 '11 at 19:09
add a comment |
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Just remove the backslash before!. Once more sorry for the confusion.
– Adam Byrtek
May 26 '11 at 19:09
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
This exits in a syntax error after '&;'. For example just running 'vlc &;' gets the same error; but I see what you are trying to do...
– Teque5
May 26 '11 at 18:54
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Sorry about that, I tested it on Zsh instead of Bash. It works when you remote the semicolon, corrected.
– Adam Byrtek
May 26 '11 at 18:57
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Well when i remove the semicolon it doesn't get the PID. I get: 'bash: kill: $!: arguments must be process or job IDs'. It does TRY to kill it after 10 seconds though...
– Teque5
May 26 '11 at 19:00
Just remove the backslash before
!. Once more sorry for the confusion.– Adam Byrtek
May 26 '11 at 19:09
Just remove the backslash before
!. Once more sorry for the confusion.– Adam Byrtek
May 26 '11 at 19:09
add a comment |
Why not use /usr/bin/timeout?
$ timeout --help
Usage: timeout [OPTION] DURATION COMMAND [ARG]...
or: timeout [OPTION]
Start COMMAND, and kill it if still running after DURATION.
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
add a comment |
Why not use /usr/bin/timeout?
$ timeout --help
Usage: timeout [OPTION] DURATION COMMAND [ARG]...
or: timeout [OPTION]
Start COMMAND, and kill it if still running after DURATION.
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
add a comment |
Why not use /usr/bin/timeout?
$ timeout --help
Usage: timeout [OPTION] DURATION COMMAND [ARG]...
or: timeout [OPTION]
Start COMMAND, and kill it if still running after DURATION.
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
Why not use /usr/bin/timeout?
$ timeout --help
Usage: timeout [OPTION] DURATION COMMAND [ARG]...
or: timeout [OPTION]
Start COMMAND, and kill it if still running after DURATION.
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
answered Feb 7 '12 at 21:24
waltinatorwaltinator
22.6k74169
22.6k74169
add a comment |
add a comment |
A simple (and not much tested) version of hatimerun:
#!/bin/sh
help(){
echo "Usage" >&2
echo " $0 -t TIME COMMAND" >&2
exit 1
}
TEMP=`getopt -o t: -n "$0" -- "$@"`
if [ $? != 0 ] ; then
help
fi
eval set -- "$TEMP"
while true ; do
case "$1" in
-t) timeout=$2; shift 2;;
--) shift; break;;
esac
done
if [ -z "$timeout" ]; then
help
fi
cmd="$*"
$cmd&
echo "kill $!" | at now+$timeout
See the manpage of at for how to specify the time. Like with at the minimum time resolution is 1 minute.
Another way is to use upstart instead of this script.
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
add a comment |
A simple (and not much tested) version of hatimerun:
#!/bin/sh
help(){
echo "Usage" >&2
echo " $0 -t TIME COMMAND" >&2
exit 1
}
TEMP=`getopt -o t: -n "$0" -- "$@"`
if [ $? != 0 ] ; then
help
fi
eval set -- "$TEMP"
while true ; do
case "$1" in
-t) timeout=$2; shift 2;;
--) shift; break;;
esac
done
if [ -z "$timeout" ]; then
help
fi
cmd="$*"
$cmd&
echo "kill $!" | at now+$timeout
See the manpage of at for how to specify the time. Like with at the minimum time resolution is 1 minute.
Another way is to use upstart instead of this script.
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
add a comment |
A simple (and not much tested) version of hatimerun:
#!/bin/sh
help(){
echo "Usage" >&2
echo " $0 -t TIME COMMAND" >&2
exit 1
}
TEMP=`getopt -o t: -n "$0" -- "$@"`
if [ $? != 0 ] ; then
help
fi
eval set -- "$TEMP"
while true ; do
case "$1" in
-t) timeout=$2; shift 2;;
--) shift; break;;
esac
done
if [ -z "$timeout" ]; then
help
fi
cmd="$*"
$cmd&
echo "kill $!" | at now+$timeout
See the manpage of at for how to specify the time. Like with at the minimum time resolution is 1 minute.
Another way is to use upstart instead of this script.
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
A simple (and not much tested) version of hatimerun:
#!/bin/sh
help(){
echo "Usage" >&2
echo " $0 -t TIME COMMAND" >&2
exit 1
}
TEMP=`getopt -o t: -n "$0" -- "$@"`
if [ $? != 0 ] ; then
help
fi
eval set -- "$TEMP"
while true ; do
case "$1" in
-t) timeout=$2; shift 2;;
--) shift; break;;
esac
done
if [ -z "$timeout" ]; then
help
fi
cmd="$*"
$cmd&
echo "kill $!" | at now+$timeout
See the manpage of at for how to specify the time. Like with at the minimum time resolution is 1 minute.
Another way is to use upstart instead of this script.
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
edited Jul 20 '18 at 21:45
David Foerster
28.4k1366111
28.4k1366111
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
answered May 26 '11 at 17:50
Florian DieschFlorian Diesch
65.5k16164181
65.5k16164181
add a comment |
add a comment |
Using Perl:
perl -e "alarm 5; exec @ARGV" systemctl list-timers
Using timelimit command:
sudo apt install timelimit
timelimit -t5 node app.js
Using timeout command:
timeout -sHUP time command
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
2
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
1
@Olorin now the post is edited by OP including another solutions as well as a sample of thetimeoutcommand.
– αғsнιη
Feb 11 at 5:48
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.
– Rakib Fiha
Feb 11 at 7:00
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
add a comment |
Using Perl:
perl -e "alarm 5; exec @ARGV" systemctl list-timers
Using timelimit command:
sudo apt install timelimit
timelimit -t5 node app.js
Using timeout command:
timeout -sHUP time command
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
2
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
1
@Olorin now the post is edited by OP including another solutions as well as a sample of thetimeoutcommand.
– αғsнιη
Feb 11 at 5:48
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.
– Rakib Fiha
Feb 11 at 7:00
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
add a comment |
Using Perl:
perl -e "alarm 5; exec @ARGV" systemctl list-timers
Using timelimit command:
sudo apt install timelimit
timelimit -t5 node app.js
Using timeout command:
timeout -sHUP time command
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
Using Perl:
perl -e "alarm 5; exec @ARGV" systemctl list-timers
Using timelimit command:
sudo apt install timelimit
timelimit -t5 node app.js
Using timeout command:
timeout -sHUP time command
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
edited Feb 11 at 3:52
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
answered Feb 7 at 5:27
Rakib FihaRakib Fiha
195
195
2
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
1
@Olorin now the post is edited by OP including another solutions as well as a sample of thetimeoutcommand.
– αғsнιη
Feb 11 at 5:48
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.
– Rakib Fiha
Feb 11 at 7:00
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
add a comment |
2
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
1
@Olorin now the post is edited by OP including another solutions as well as a sample of thetimeoutcommand.
– αғsнιη
Feb 11 at 5:48
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.
– Rakib Fiha
Feb 11 at 7:00
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
2
2
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
That's already in askubuntu.com/a/102328/760903
– Olorin
Feb 7 at 5:27
1
1
@Olorin now the post is edited by OP including another solutions as well as a sample of the
timeout command.– αғsнιη
Feb 11 at 5:48
@Olorin now the post is edited by OP including another solutions as well as a sample of the
timeout command.– αғsнιη
Feb 11 at 5:48
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.
myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$ For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.– Rakib Fiha
Feb 11 at 7:00
@αғsнιη Do you mind answering a short query about timeout and timelimit command? When I used timeout and timelimit command, it gave me error like this after it exited from that timeout period.
myname@myname:~$ myname@myname:~$ myname@myname:~$ myname@myname:~$ For each enter, it kept making in the same line instead of going to another line. Then I had to exit from the session and login again. But, perl command worked fine without giving any error.– Rakib Fiha
Feb 11 at 7:00
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
sorry, I have no idea what's happening there
– αғsнιη
Feb 12 at 5:22
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