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…
Related
I have a requirement where I need to trigger 10 shell scripts at a time. I may have 200+ shell scripts to be executed.
e.g. if I trigger 10 jobs and two jobs completed, I need to trigger another 2 jobs which will make number of jobs currently executing to 10.
I need your help and suggestion to cater this requirement.
Yes with GNU Parallel like this:
parallel -j 10 < ListOfJobs.txt
Or, if your jobs are called job_1.sh to job_200.sh:
parallel -j 10 job_{}.sh ::: {1..200}
Or. if your jobs are named with discontiguous, random names but are all shell scripts named with .sh suffix in one directory:
parallel -j 10 ::: *.sh
There is a very good overview here. There are lots of questions and answers on Stack Overflow here.
Simply run them as background jobs:
for i in {1..10}; { ./script.sh & }
Adding more jobs if less than 10 are running:
while true; do
pids=($(jobs -pr))
((${#pids[#]}<10)) && ./script.sh &
done &> /dev/null
There are different ways to handle this:
Launch them together as background tasks (1)
Launch them in parallel (1)
Use the crontab (2)
Use at (3)
Explanations:
(1) You can launch the processes exactly when you like (by launching a command, click a button or whatever event you choose)
(2) The processes will be launched at the same time, every (working) day, periodically.
(3) You choose a time when the processes will be launched together once.
I have used below to trigger 10 jobs a time.
max_jobs_trigger=10
while mapfile -t -n ${max_jobs_trigger} ary && ((${#ary[#]})); do
jobs_to_trigger=`printf '%s\n' "${ary[#]}"`
#Trigger script in background
done
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:
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"
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'm running two tests on a remote server, here is the command I used several hours ago:
% ./test1.sh; ./test2.sh
The two tests are supposed to run one by one.If the second runs before the first completes, everything will be in ruin, and I'll have to restart the whole procedure.
The dilemma is, these two tasks cost too many hours to complete, and when I prepare to logout the server and wait for the result. I don't know how to switch both of them to background... If I use Ctrl+Z, only the first task will be suspended, while the second starts doing nothing useful while wiping out current data.
Is it possible to switch both of them to background, preserving their orders? Actually I should make these two tasks in the same process group like (./test1.sh; ./test2.sh) &, but sadly, the first test have run several hours, and it's quite a pity to restart the tests.
An option is to kill the second test before it starts, but is there any mechanism to cope with this?
First rename the ./test2.sh to ./test3.sh. Then do [CTRL+Z], followed by bg and disown -h. Then save this script (test4.sh):
while :; do
sleep 5;
pgrep -f test1.sh &> /dev/null
if [ $? -ne 0 ]; then
nohup ./test3.sh &
break
fi
done
then do: nohup ./test4.sh &.
and you can logout.
First, screen or tmux are your friends here, if you don't already work with them (they make remote machine work an order of magnitude easier).
To use conditional consecutive execution you can write:
./test1.sh && ./test2.sh
which will only execute test2.sh if test1.sh returns with 0 (conventionally meaning: no error). Example:
$ true && echo "first command was successful"
first command was successful
$ ! true && echo "ain't gonna happen"
More on control operators: http://www.humbug.in/docs/the-linux-training-book/ch08s01.html