Is there a way to create a bash script that will only run for X hours? I'm currently setting up a cron job to initiate a script every night. This script essentially runs until a certain condition is met, exporting it's status to a holding variable to keep track of 'where it is' after each iteration. The intention is to start-up the process every night, run for a few hours, and then stop, holding the status until the process starts up the next night.
Short of somehow collecting the start time, and checking it against the current time in each iteration of the loop, is there an easier way to do this? Bash scripting is not my forte (I know enough to get things done and be dangerous) and I have not done something like this before. Any help would be appreciated. Thanks.
Use GNU Coreutils
GNU coreutils contains an actual timeout binary, usually invoked like this:
# timeout after 5 seconds when sleeping for 30
/usr/bin/timeout 5s /bin/sleep 30
In your case, you'd want to specify hours instead of seconds, so to timeout in 2 hours use something like 2h instead of 5s. See timeout(1) or info coreutils 'timeout invocation' for additional options.
Hacks and Workarounds
Native timeouts or the GNU timeout command are really the best options. However, see the following for some ideas if you decide to roll your own:
How do I run a command, and have it abort (timeout) after N seconds?
The TMOUT variable using read and process or command substitution.
Do it as you described - it is the cleanest way.
But if for some strange reason want kill the process after a time, can use the next
./long_runner &
(sleep 5; kill $!; wait; exit 0) &
will kill the long_runner after 5 secs.
By using the SIGALRM facility you can rig a signal to be sent after a certain time, but traditionally, this was not easily accessible from shell scripts (people would write small custom C or Perl programs for this). These days, GNU coreutils ships with a timeout command which does this by wrapping your command:
timeout 4h yourprogram
Related
The problem I want to tackle is as follows. I have a long(1 to 2 hours) running task that has to be run everyday. So the goto option was cron. But the catch is that I have to give a 24 hour gap between successive runs. So using cron now would involve rewriting the cron job file after every run. This might be clear after this example.
The long running job 'LR' starts at 6PM on Monday and finishes at 7:30PM sameday.
On Tuesday it's supposed to start at 7:30 PM and not 6PM (like it did on monday). This is because there has to be a 24hr gap between successive runs.
The obvious option here was to have a process running an infinite loop. start the LR job. Then sleep for 24hr and continue with the loop. This works perfectly too. In my setup there is a bash script which is running this loop.
while [ 1 == 1 ]; do
/bin/jobs/long_run.py
/bin/jobs/cleanup.sh
sleep 86400
done
So my question is what is the total amount of CPU resource spent and what is the RAM usage.
Not sure if this affects the answer in anyway; I'm running this on termux on an android phone.
Also please recommend other light weight options.
There is nothing to worry about resources, while a script executes sleep, it really sleeps. You should worry for if anything happens between two executions, like restart, downtime etc. This structure:
while true; do
sh script.sh
sleep 86400
done
does not resume and you don't save the time for the next execution anywhere. Similar to this structure is to have a wrapper, suppose f() is your jobs
f() {
echo working
}
wrapper() {
f
echo sleeping
sleep 86400
wrapper
}
wrapper
so now you call the wrapper, which works, sleeps and calls itself. You could use just this, if you are ok with what could go wrong, at least print the datetime somewhere.
You can replace the internal sleep and wrapper call with job scheduling with cron or at. Probably at is not a standard packet for all distributions (at least not for mine) while cron is. You could install it. For at the wrapper would be like this:
wrapper() {
f
at now +1 day wrapper
}
With cron, you could edit the crontab, like this but better use a crontab file like this, what you have to do is to parse date command, create the date prefix, update crontab.
Note: There may be other cron jobs for user, existing or added after that, this is considered in the last link.
How do I start multiple processes in bash and time how long they take?
From this question I know how to start multiple processes in a bash script but using time script.sh doesn't work because the processes spawned end after the script ends.
I tried using wait but that didn't change anything.
Here is the script in its entirety:
for i in `seq $1`
do
( ./client & )
done
wait # This doesn't seem to change anything
I'm trying to get the total time for all the processes to finish and not the time for each process.
Why the parentheses around client invocation? That's going to run the command in a subshell. Since the background job isn't in the top level shell, that's why the wait is ineffective (there's no jobs in this shell to wait for).
Then you can add time back inside the for loop and it should work.
I have a simple bash script that runs some tasks which can take varying amounts of time to complete (from 15 mins to 5 hours). The script loops using a for loop, so that I can run it an arbitrary number of times, normally back-to-back.
However, I have been requested to have each iteration of the script start at the top of the hour. Normally, I would use cron and kick it off that way, every hour, but since the runtime of the script is highly variable, that becomes trickier.
It is not allowable for multiple instances of the script to be running at once.
So, I'd like to include the logic to wait for 'top of the hour' within the script, but I'm not sure of the best way to do that, or if there's some way to (ab)use 'at' or something more elegant like that. Any ideas?
You can still use cron. Just make your script use a lock file. With the flock utility you can do:
#!/bin/bash
exec 42> /tmp/myscriptname.lock
flock -n 42 || { echo "Previous instance still running"; exit 1; }
rest of your script here
Now, simply schedule your job every hour in cron, and the new instance will simply exit if the old one's still running. There is no need to clean up any lock files.
I have written a script to initiate multi-processing
for i in `seq 1 $1`
do
/usr/bin/php index.php name&
done
wait
A cron run every min - myscript.sh 3 now three background process get initiated and after some time I see list of process via ps command. I see all the processes are together in "Sleep" or "Running" mode...Now I wanted to achieve that when one goes to sleep other processes must process..how can I achieve it?. Or this is normal.
This is normal. A program that can run will be given time by the operating system... when possible. If all three are sleeping, then the system is most likely busy and time is being given to other processes.
I'm not a scripting expert and I was wondering what was an acceptable way to run a script for at most x milliseconds (and yet finish before x milliseconds if the script is done before the timeout).
I solved that problem using Bash in a way that I think is very hacky and I wonder if there's a better way to do it.
Basically I've got one shell script called sleep_kill.sh that takes a PID as the first argument and a timeout as its second argument and that does this:
sleep $2
kill -9 $1 2> /dev/null 1> /dev/null
So if the PID corresponds to a script that finishes before timing out, nothing is going to be killed (I take it that the OS shall not have the time to be reusing this PID for another [unrelated] process seen that it's 'cycling' through all the process IDs before starting to reuse them).
Anyway, then I call my script that may "hang" or timeout:
command_that_may_hang.sh
PID=$!
sleep_kill.sh $PID .3
wait $PID > /dev/null 2>&1
And I'll be waiting at most 300 ms for command_that_may_hang.sh. Yet if command_that_may_hang.sh took only 10 ms to execute, I won't be "stuck" for 300 ms.
It would be great if some shell expert could explain the drawbacks of this approach and what should be done instead.
Have a look at this script: http://www.pixelbeat.org/scripts/timeout
Note timeouts of less that one second are pretty much nonsensical on most systems due to scheduling delays etc. Note also that newer coreutils has the timeout command included and it has a resolution of 1 second.