I know I can run my bash script in the background by using bash script.sh & disown or alternatively, by using nohup. However, I want to run my script in the background by default, so when I run bash script.sh or after making it executable, by running ./script.sh it should run in the background by default. How can I achieve this?
Self-contained solution:
#!/bin/sh
# Re-spawn as a background process, if we haven't already.
if [[ "$1" != "-n" ]]; then
nohup "$0" -n &
exit $?
fi
# Rest of the script follows. This is just an example.
for i in {0..10}; do
sleep 2
echo $i
done
The if statement checks if the -n flag has been passed. If not, it calls itself with nohup (to disassociate the calling terminal so closing it doesn't close the script) and & (to put the process in the background and return to the prompt). The parent then exits to leave the background version to run. The background version is explicitly called with the -n flag, so wont cause an infinite loop (which is hell to debug!).
The for loop is just an example. Use tail -f nohup.out to see the script's progress.
Note that I pieced this answer together with this and this but neither were succinct or complete enough to be a duplicate.
Simply write a wrapper that calls your actual script with nohup actualScript.sh &.
Wrapper script wrapper.sh
#! /bin/bash
nohup ./actualScript.sh &
Actual script in actualScript.sh
#! /bin/bash
for i in {0..10}
do
sleep 10 #script is running, test with ps -eaf|grep actualScript
echo $i
done
tail -f 10 nohup.out
0
1
2
3
4
...
Adding to Heath Raftery's answer, what worked for me is a variation of what he suggested such as this:
if [[ "$1" != "-n" ]]; then
$0 -n & disown
exit $?
fi
Related
I have two bash scripts:
a.sh:
echo "running"
doit=true
if [ $doit = true ];then
./b.sh &
fi
some-long-operation-binary
echo "done"
b.sh:
for i in {0..50}; do
echo "counting";
sleep 1;
done
I get this output:
> ./a.sh
running
counting
Why do I only see the first "counting" from b.sh and then nothing anymore? (Currently some-long-operation-binary just sleep 5 for this example). I first thought that due to setting b.sh in the background, its STDOUT is lost, but why do I see the first output? More importantly: is b.sh still running and doing its thing (its iteration)?
For context:
b.sh is going to poll a service provided by some-long-operation-binary, which is only available after some time the latter has run, and when ready, would write its content to a file.
Apologies if this is just rubbish, it's a bit late...
You should add #!/bin/bash or the like to b.sh that uses a Bash-like expansion, to make sure Bash is actually running the script. Otherwise there may be (indeed) only one loop iteration happening.
When you start a background process, it is usually a good practice to kill it and wait for it, no matter which way the script exits.
#!/bin/bash
set -e -o pipefail
declare -i show_counter=1
counter() {
local -i i
for ((i = 0;; ++i)); do
echo "counting $((i))"
sleep 1
done
}
echo starting
if ((show_counter)); then
counter &
declare -i counter_pid="${!}"
trap 'kill "${counter_pid}"
wait -n "${counter_pid}" || :
echo terminating' EXIT
fi
sleep 10 # long-running process
So I've looked up other questions and answers for this and as you can imagine, there are lots of ways to find this. However, my situation is kind of different.
I'm able to check whether a bash script is already running or not and I want to kill the script if it's already running.
The problem is that with the below code, -since I'm running this within the same script- the script kills itself too because it sees a script already running.
result=`ps aux | grep -i "myscript.sh" | grep -v "grep" | wc -l`
if [ $result -ge 1 ]
then
echo "script is running"
else
echo "script is not running"
fi
So how can I check if a script is already running besides it's own self and kill itself if there's another instance of the same script is running, else, continue without killing itself.
I thought I could combine the above code with $$ command to find the script's own PID and differentiate them this way but I'm not sure how to do that.
Also a side note, my script can be run multiple times at the same time within the same machine but with different arguments and that's fine. I only need to identify if script is already running with the same arguments.
pid=$(pgrep myscript.sh | grep -x -v $$)
# filter non-existent pids
pid=$(<<<"$pid" xargs -n1 sh -c 'kill -0 "$1" 2>/dev/null && echo "$1"' --)
if [ -n "$pid" ]; then
echo "Other script is running with pid $pid"
echo "Killing him!"
kill $pid
fi
pgrep lists the pids that match the name myscript.sh. From the list we filter current $$ shell with grep -v. It the result is non-empty, then you could kill the other pid.
Without the xargs, it would work, but the pgrep myscript.sh will pick up the temporary pid created for command substitution or the pipe. So the pid will never be empty and the kill will always execute complaining about the non-existent process. To do that, for each pid in pids, I check if the pid exists with kill -0. If it does, then it is outputted, effectively filtering all nonexistent pids.
You could also use a normal for loop to filter the pids:
# filter non-existent pids
pid=$(
for i in $pid; do
if kill -0 "$i" 2>/dev/null; then
echo "$i"
fi
done
)
Alternatively, you could use flock to lock the file and use lsof to list current open files with filtering the current one. As it is now, I think it will kill also editors that are editing the file and such. I believe the lsof output could be better filtered to accommodate this.
if [ "${FLOCKER}" != "$0" ]; then
pids=$(lsof -p "^$$" -- ./myscript.sh | awk 'NR>1{print $2}')
if [ -n "$pids" ]; then
echo "Other processes with $(echo $pids) found. Killing them"
kill $pids
fi
exec env FLOCKER="$0" flock -en "$0" "$0" "$#"
fi
I would go with either of 2 ways to solve this problem.
1st solution: Create a watchdog file lets say a .lck file kind of on a location before starting the script's execution(Make sure we use trap etc commands in case script is aborted so that .lck file should be removed) AND remove it once execution of script is completed successfully.
Example script for 1st solution: This is just an example a test one. We need to take care of interruptions in the script, lets say script got interrupted by a command or etc then we could use trap in it too, since at that time it would have not been completed but you may need to kick it off again(since last time it was not completed).
cat file.ksh
#!/bin/bash
PWD=`pwd`
watchdog_file="$PWD/script.lck"
if [[ -f "$watchdog_file" ]]
then
echo "Please wait script is still running, exiting from script now.."
exit 1;
else
touch $watchdog_file
fi
while true
do
echo "singh" > test1
done
if [[ -f "$watchdog_file" ]]
then
rm "$watchdog_file"
fi
2nd solution: Take pid of current running shell using $$ save it in a file. Then check if that process is still running come out of script if NOT running then move on to run statements in script.
I have a bash script (this_script.sh) that invokes multiple instances of another TCL script.
set -m
for vars in $( cat vars.txt );
do
exec tclsh8.5 the_script.tcl "$vars" &
done
while [ 1 ]; do fg 2> /dev/null; [ $? == 1 ] && break; done
The multi threading portion was taken from Aleksandr's answer on: Forking / Multi-Threaded Processes | Bash.
The script works perfectly (still trying to figure out the last line). However, this line is always displaed: exec tclsh8.5 the_script.tcl "$vars"
How do I hide that line? I tried running the script as :
bash this_script.sh > /dev/null
But this hides the output of the invoked tcl scripts too (I need the output of the TCL scripts).
I tried adding the /dev/null to the end of the statement within the for statement, but that too did not work either. Basically, I am trying to hide the command but not the output.
You should use $! to get the PID of the background process just started, accumulate those in a variable, and then wait for each of those in turn in a second for loop.
set -m
pids=""
for vars in $( cat vars.txt ); do
tclsh8.5 the_script.tcl "$vars" &
pids="$pids $!"
done
for pid in $pids; do
wait $pid
# Ought to look at $? for failures, but there's no point in not reaping them all
done
Suppose I have test.sh as below. The intent is to run some background task(s) by this script, that continuously updates some file. If the background task is terminated for some reason, it should be started again.
#!/bin/sh
if [ -f pidfile ] && kill -0 $(cat pidfile); then
cat somewhere
exit
fi
while true; do
echo "something" >> somewhere
sleep 1
done &
echo $! > pidfile
and want to call it like ./test.sh | otherprogram, e. g. ./test.sh | cat.
The pipe is not being closed as the background process still exists and might produce some output. How can I tell the pipe to close at the end of test.sh? Is there a better way than checking for existence of pidfile before calling the pipe command?
As a variant I tried using #!/bin/bash and disown at the end of test.sh, but it is still waiting for the pipe to be closed.
What I actually try to achieve: I have a "status" script which collects the output of various scripts (uptime, free, date, get-xy-from-dbus, etc.), similar to this test.sh here. The output of the script is passed to my window manager, which displays it. It's also used in my GNU screen bottom line.
Since some of the scripts that are used might take some time to create output, I want to detach them from output collection. So I put them in a while true; do script; sleep 1; done loop, which is started if it is not running yet.
The problem here is now that I don't know how to tell the calling script to "really" detach the daemon process.
See if this serves your purpose:
(I am assuming that you are not interested in any stderr of commands in while loop. You would adjust the code, if you are. :-) )
#!/bin/bash
if [ -f pidfile ] && kill -0 $(cat pidfile); then
cat somewhere
exit
fi
while true; do
echo "something" >> somewhere
sleep 1
done >/dev/null 2>&1 &
echo $! > pidfile
If you want to explicitly close a file descriptor, like for example 1 which is standard output, you can do it with:
exec 1<&-
This is valid for POSIX shells, see: here
When you put the while loop in an explicit subshell and run the subshell in the background it will give the desired behaviour.
(while true; do
echo "something" >> somewhere
sleep 1
done)&
I have a large script called mandacalc which I want to always run with the nohup command. If I call it from the command line as:
nohup mandacalc &
everything runs swiftly. But, if I try to include nohup inside my command, so I don't need to type it everytime I execute it, I get an error message.
So far I tried these options:
nohup (
command1
....
commandn
exit 0
)
and also:
nohup bash -c "
command1
....
commandn
exit 0
" # and also with single quotes.
So far I only get error messages complaining about the implementation of the nohup command, or about other quotes used inside the script.
cheers.
Try putting this at the beginning of your script:
#!/bin/bash
case "$1" in
-d|--daemon)
$0 < /dev/null &> /dev/null & disown
exit 0
;;
*)
;;
esac
# do stuff here
If you now start your script with --daemon as an argument, it will restart itself detached from your current shell.
You can still run your script "in the foreground" by starting it without this option.
Just put trap '' HUP on the beggining of your script.
Also if it creates child process someCommand& you will have to change them to nohup someCommand& to work properly... I have been researching this for a long time and only the combination of these two (the trap and nohup) works on my specific script where xterm closes too fast.
Create an alias of the same name in your bash (or preferred shell) startup file:
alias mandacalc="nohup mandacalc &"
Why don't you just make a script containing nohup ./original_script ?
There is a nice answer here: http://compgroups.net/comp.unix.shell/can-a-script-nohup-itself/498135
#!/bin/bash
### make sure that the script is called with `nohup nice ...`
if [ "$1" != "calling_myself" ]
then
# this script has *not* been called recursively by itself
datestamp=$(date +%F | tr -d -)
nohup_out=nohup-$datestamp.out
nohup nice "$0" "calling_myself" "$#" > $nohup_out &
sleep 1
tail -f $nohup_out
exit
else
# this script has been called recursively by itself
shift # remove the termination condition flag in $1
fi
### the rest of the script goes here
. . . . .
the best way to handle this is to use $()
nohup $( command1, command2 ...) &
nohup is expecting one command and in that way You're able to execute multiple commands with one nohup