This question already has answers here:
Timeout a command in bash without unnecessary delay
(24 answers)
Closed 6 years ago.
This is a CentOS 6.x box, on it I have two things that I need to run one right after the other - a shell script and a .sql script.
I want to write a shell script that calls the first script, lets it run and then terminates it after a certain number of hours, and then calls the .sql script (they can't run simultaneously).
I'm unsure how to do the middle part, that is terminating the first script after a certain time limit, any suggestions?
script.sh &
sleep 4h && kill $!
script.sql
This will wait 4 hours then kill the first script and run the second. It always waits 4 hours, even if the script exits early.
If you want to move on immediately, that's a little trickier.
script.sh &
pid=$!
sleep 4h && kill "$pid" 2> /dev/null &
wait "$pid"
Related
This question already has answers here:
Timeout a command in bash without unnecessary delay
(24 answers)
Closed 1 year ago.
In my bash script I run a command that activates a script. I repeat this command many times in a for loop and as such want to wait until the script is finished before running it again. My bash script is as follows
for k in $(seq 1 5)
do
sed_param='s/mu = .*/mu = '${mu}';/'
sed -i "$sed_param" brusselator.c
make brusselator.tst &
done
As far as I know the & at the end lets the script know to wait until the command is finished, but this isn't working. Is there some other way?
Furthermore, sometimes the command can take very very long, in this case I would maximally want to wait 5 seconds. But if the command is done earlier I would't want to wait 5 seconds. Is there some way to achieve this?
There is the timeout command. You would use it like
timeout -k 5 make brusselator.tst
Maybe you would like to see also if it exited successfully, failed or was killed because it timed out.
timeout -k 5 make brusselator.tst && echo OK || echo Failed, status $?
If the command times out, and --preserve-status is not set, then command exits with status 124. Different status would mean that make failed for different reason before timing out.
I am trying to launch multiple instances of imagesnap simultaneously from a single bash script on a Mac. Also, it would be great to give (some of) the arguments by user input when running the script.
I have 4 webcams connected, and want to take series of images from each camera with a given interval. Being an absolute beginner with bash scripts, I don't know where to start searching. I have tested that 4 instances of imagesnap works nicely when running them manually from Terminal, but that's about it.
To summarise I'm looking to make a bash script that:
run multiple instances of imagesnap.
has user input for some of the arguments for imagesnap.
ideally start all the imagesnap instances at (almost) the same time.
--EDIT--
After thinking about this I have a vague idea of how this script could be organised using the ability to take interval images with imagesnap -t x.xx:
Run multiple scripts from within the main script
or
Use subshells to run multiple instances of imagesnap
Start each sub script or subshell in parallel if possible.
Since each instance of imagesnap will run until terminated it would be great if they could all be stopped with a single command
the following quick hack (saved as run-periodically.sh) might do the right thing:
#!/bin/bash
interval=5
start=$(date +%s)
while true; do
# run jobs in the background
for i in 1 2 3 4; do
"$#" &
done
# wait for all background jobs to finish
wait
# figure out how long we have to sleep
end=$(date +%s)
delta=$((start + interval - end))
# if it's positive sleep for this amount of time
if [ $delta -gt 0 ]; then
sleep $delta || exit
fi
start=$((start + interval))
done
if you put this script somewhere appropriate and make it executable, you can run it like:
run-periodically.sh imagesnap arg1 arg2
but while testing, I ran with:
sh run-periodically.sh sh -c "date; sleep 2"
which will cause four copies of "start a shell that displays the date then waits a couple of seconds" to be run in parallel every interval seconds. if you want to run different things in the different jobs, then you might want to put them into this script explicitly or maybe another script which this one calls…
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 $1`
do
( ./client & )
done
wait # This doesn't seem to change anything
I'm trying to get the total time for all the processes to finish and not the time for each process.
Why the parentheses around client invocation? That's going to run the command in a subshell. Since the background job isn't in the top level shell, that's why the wait is ineffective (there's no jobs in this shell to wait for).
Then you can add time back inside the for loop and it should work.
This question already has answers here:
How would I get a cron job to run every 30 minutes?
(6 answers)
Closed 7 years ago.
The community reviewed whether to reopen this question 2 months ago and left it closed:
Original close reason(s) were not resolved
I want to schedule a command like ./example every 6 minutes and when 6 minutes is done it exits the process and runs it again. How would I do that in Bash? I run CentOS.
I would make a cronjob running every sixth minutes and using the timeout command to kill it after, say, 5 minutes and 50 seconds.
This is a sample crontab rule:
*/6 * * * * cd /path/to/your/file && timeout -s9 290s ./example
It changes working directory to where you have your script and then executes the script. Note that I send it signal 9 (SIGKILL) using the -s9 flag which means "terminate immediately". In most cases you might want to consider sending SIGTERM instead, which tells the script to "exit gracefully". If that is the case you can consider giving the script a little bit more time to exit by decreasing the timeout value even more. To send SIGTERM instead of SIGKILL, just remove the -s9 flag.
You edit your crontab by running crontab -e
Replace mycommand in the script below...
#! /bin/bash
## create an example command to launch for demonstration purposes
function mycommand { D=$(date) ; while true ; do echo $D; sleep 1s ; done; }
while true
do
mycommand & PID=$!
sleep 6m
kill $PID ; wait $PID 2>/dev/null
done
Every six minutes, this kills the command then restarts it.
Use Ctrl-C as one way to terminate this sequence.
This question already has answers here:
Timeout a command in bash without unnecessary delay
(24 answers)
Closed 9 years ago.
I'm writing a script and would like to know how to ask one of the commands to exit after few seconds. For eg. let's suppose my script runs 2 application commands in it.
#!/bin/bash
for i in `cat servers`
do
<command 1> $i >> Output_file #Consistency command
<command 2> $i >> Output_file #Communication check
done
These commands are to check consistency & communication to/from application. I want to know how do I make sure that command 1 & 2 runs for only few seconds and if there is no response from particular host, move on to next command.
bash coreutils has got 'timeout` command.
From manual:
DESCRIPTION
Start COMMAND, and kill it if still running after NUMBER seconds. SUFFIX may be "s" for seconds (the default), "m" for
minutes, "h" for hours or "d" for days.
for example:
timeout 5 sleep 6