I am trying to write a script that displays an image for a time, then returns back to the shell script. I tried:
feh outfile.jpeg | sleep 10 | exit
but this doesn't work. Im not interested in exiting the script, only in getting back to the other functions of the script.
I just want to show the image for a short time, then return to what the script is doing.
How might I achieve this?
You could do:
feh outfile.jpeg & # run process in background
pid=$! # obtain PID of last backgrounded process
sleep 10
kill $pid # kill feh
Here is a working solution:
(feh outfile.jpeg&) && (sleep 10 && pkill feh)
It will run feh in backgroud and then count to 10 before killing feh. A better way would be to kill the right pid though...
Related
So I want to start a docker image, then a Django back-end and finally an angular front-end, let them run as long as I need to do tests/develop and then kill them when I'm done. To do this I first tried starting them all in a script and have them run in a background, and have a second script do kill %n for both processes. This doesn't work because the background processes are in another context, so the second script cannot reference them.
Then I tried this:
#!/bin/bash
# Exit Angular, Django and kill docker_img
function clean_up()
{
echo "Exiting..."
kill %2
kill %1
docker stop docker_img
reset
exit
}
# Trigger cleanup on CTRL + C
trap clean_up SIGINT
# Start docker database
docker start docker_img
# Start django backend
cd ~/Projects/DjangoBackend
source venv/bin/activate
python src/manage.py runserver &
sleep 3
echo 'Done starting django, starting angular'
sleep 1
# Start angular front end
cd ~/Projects/AngularFront
npm start &
However, after npm start & runs, the trap stops working, so it effectively becomes useless. I'm guessing it could be because once my script is done running the trap is no longer active, but I don't know how to fix this. What can I do?
If you are looking to kill a process in unix/linux, one way of doing it is you can record their PID in a file using ps -ef command.
And then use kill -9 to kill the process.
Example:
$ ps -ef | grep <process_name> | awk -F ' ' '{print $2}' > pid.txt
$ kill -9 `cat pid.txt`
ps -ef command will give all the running processes, using grep and process name, you can get PID of the particular process
awk is used to extract only PID from above command
kill -9 will forcefully kill the process
The answer seems to have been pretty easy, all I had to do was add wait to the end of the script, which allows the script to wait until the processes are done executing. Since two of the processes are servers, they don't stop unless prompted, so it'll just wait until SIGINT is received, at that point it'll run the clean_up function and exit gracefully.
Additionally, one could use the same trap but with the EXIT trigger instead of SIGINT to clean up when the script exits on it's own due to the processes closing.
I have written a bash script to carry out some tests on my system. The tests run in the background and in parallel. The tests can take a long time and sometimes I may wish to abort the tests part way through.
If I Control+C then it aborts the parent script, but leaves the various children running. I wish to make it so that I can hit Control+C or otherwise to quit and then kill all child processes running in the background. I have a bit of code that does the job if I'm running running the background jobs directly from the terminal, but it doesn't work in my script.
I have a minimal working example.
I have tried using trap in combination with pgrep -P $$.
#!/bin/bash
trap 'kill -n 2 $(pgrep -P $$)' 2
sleep 10 &
wait
I was hoping that on hitting control+c (SIGINT) would kill everything that the script started but it actually says:
./breakTest.sh: line 1: kill: (3220) - No such process
This number changes, but doesn't seem to apply to any running processes, so I don't know where it is coming from.
I guess if the contents of the trap command get evaluated where the trap command occurs then it might explain the outcome. The 3220 pid might be for pgrep itself.
I'd appreciate some insight here
Thanks
I have found a solution using pkill. This example also deals with many child processes.
#!/bin/bash
trap 'pkill -P $$' SIGINT SIGTERM
for i in {1..10}; do
sleep 10 &
done
wait
This appears to kill all the child processes elegantly. Though I don't properly understand what the issue was with my original code, apart from sending the correct signal.
in bash whenever you you use & after a command it places that command as a background job ( this background jobs are called job_spec ) which is incremented by one until you exit that terminal session. You can use the jobs command to get the list of the background jobs running. To work with this jobs you have to use the % with the job id. The jobs command also accept other options such as jobs -p to see the proces sids of all jobs , jobs -p %JOB_SPEC to see the process of id of that particular job.
#!/usr/bin/env bash
trap 'kill -9 %1' 2
sleep 10 &
wait
or
#!/usr/bin/env bash
trap 'kill -9 $(jobs -p %1)' 2
sleep 10 &
wait
I implemented something like this few years back, you can take a look at it async bash
You can try something like the following:
pkill -TERM -P <your_parent_id_here>
How do I repeatedly start and kill a bash script that takes a long time. I have a analyze_realtime.sh that runs indefinitely, but I only want to run it for X second bursts (just say 15s for now).
while true; do analyze_realtime.sh; sleep 15; done
The problem with this is that analyze_realtime.sh never finishes, so this logic doesn't work. Is there a way to kill the process after 15 seconds, then start it again?
I was thinking something with analyze_realtime.sh&, ps, and kill may work. Is there anything simpler?
while true;
do
analyze_realtime.sh &
jobpid=$! # This gets the pid of the bg job
sleep 15
kill $jobpid
if ps -p $jobpid &>/dev/null; then
echo "$jobpid didn't get killed. Moving on..."
fi
done
You can do more under the if-statement, sending other SIGNALs if SIGHUP didn't work.
Try this out
while true; do
analyze_realtime.sh & # put script execution in background
sleep 15
kill %1
done
Explanation
%1 refer to the latest process ran in background
You can use timeout utility from coreutils:
while true; do
timeout 15 analyze_realtime.sh
done
(Inspired by this answer)
Is there a way I can write a simple script to run a program, close that program about 5 seconds later, and then repeat?
I just want to be able to run a program that I wrote over and over again but to do so Id have to close it like 5 seconds after running it.
Thanks!
If your command is non-interactive (requires no user interaction):
Launch your program in the background with control operator &, which gives you access to its PID (process ID) via $!, by which you can kill the running program instance after sleeping for 5 seconds:
#!/bin/bash
# Start an infinite loop.
# Use ^C to abort.
while :; do
# Launch the program in the background.
/path/to/your/program &
# Wait 5 seconds, then kill the program (if still alive).
sleep 5 && { kill $! && wait $!; } 2>/dev/null
done
If your command is interactive:
More work is needed if your command must run in the foreground to allow user interaction: then it is the command to kill the program after 5 seconds that must run in the background:
#!/bin/bash
# Turn on job control, so we can bring a background job back to the
# foreground with `fg`.
set -m
# Start an infinite loop.
# CAVEAT: The only way to exit this loop is to kill the current shell.
# Setting up an INT (^C) trap doesn't help.
while :; do
# Launch program in background *initially*, so we can reliably
# determine its PID.
# Note: The command line being set to the bakground is invariably printed
# to stderr. I don't know how to suppress it (the usual tricks
# involving subshells and group commands do not work).
/path/to/your/program &
pid=$! # Save the PID of the background job.
# Launch the kill-after-5-seconds command in the background.
# Note: A status message is invariably printed to stderr when the
# command is killed. I don't know how to suppress it (the usual tricks
# involving subshells and group commands do not work).
{ (sleep 5 && kill $pid &) } 2>/dev/null
# Bring the program back to the foreground, where you can interact with it.
# Execution blocks until the program terminates - whether by itself or
# by the background kill command.
fg
done
Check out the watch command. It will let you run a program repeatedly monitoring the output. Might have to get a little fancy if you need to kill that program manually after 5 seconds.
https://linux.die.net/man/1/watch
A simple example:
watch -n 5 foo.sh
To literally answer your question:
Run 10 times with sleep 5:
#!/bin/bash
COUNTER=0
while [ $COUNTER -lt 10 ]; do
# your script
sleep 5
let COUNTER=COUNTER+1
done
Run continuously:
#!/bin/bash
while [ 1 ]; do
# your script
sleep 5
done
If there is no input on the code, you can simply do
#!/bin/bash
while [ 1 ]
do
./exec_name
if [ $? == 0 ]
then
sleep 5
fi
done
In my bash script I am writing, I am trying to start a process (sleep) in the background and then suspend it. Finally, the process with be finished. For some reason through, when I send the kill command with the stop signal, it just keeps running as if it received nothing. I can do this from the command line, but the bash script is not working as intended.
sleep 15&
pid=$!
kill -s STOP $pid
jobs
kill -s CONT $pid
You can make it work by enabling 'monitor mode' in your script: set -m
Please see why-cant-i-use-job-control-in-a-bash-script for further information