This question already has an answer here:
Kill background process when another process ends in Linux
(1 answer)
Closed 4 years ago.
So im starting 4 codes at the same time and I want 3 of them to run in a loop until the 4th program finish.
loopProgram1 &
loopProgram2 &
loopProgram3 &
Program4
So I want loopPrograms 1,2 and 3 to execute and then all of them exit once program4 is done. Is there any way I can do this?
Here is one in bash that takes care of the background processes too if you should kill the main script. First the looper:
$ cat loopProgram
while true # loops forever echoing its parameter number set in main
do
echo $1
sleep 1
done
and the main:
$ cat main.bash
function finish {
kill %1 %2 %3
}
trap finish EXIT
bash loopProgram 1 &
bash loopProgram 2 &
bash loopProgram 3 &
sleep 3 # this mimics your Program4
Run it:
$ bash main.bash
2
1
3
1
2
3
2
1
3
$
Since my loopers loop forever, if you'd leave the kill in the bottom of main.bash after the sleep and ^C'ing it would leave the loopers running in the background.
(I got the influensa, I hope the example is clear. I probably forget it after the next nap.)
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.
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"
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
Heres what I'm trying to do. I have 4 shell scripts. Script 1 needs to be run first, then 2, then 3, then 4, and they must be run in that order. Script 1 needs to be running (and waiting in the background) for 2 to function properly, however 1 takes about 3 seconds to get ready for use. I tried doing ./1.sh & ./2.sh & ./3.sh & ./4.sh, but this results in a total mess,since 2 starts requesting things from 1 when 1 is not ready yet. So, my question is, from one shell script, how do I get it to start script 1, wait like 5 seconds, start script 2, wait like 5 seconds, etc. without stopping any previous scripts from running (i.e. they all have to be running in the background for any higher numbered script to work). Any suggestions would be much appreciated!
May I introduce you to the sleep command?
./1.sh & sleep 5
./2.sh & sleep 5
./3.sh & sleep 5
./4.sh
#!/bin/sh
./1.sh &; sleep 5;./2.sh &; sleep 5; ./3.sh &; sleep 5; ./4.sh