How to check if another instance of my shell script is running - bash

Question

GNU bash, version 1.14.7(1)
I have a script is called "abc.sh"
I have to check this from abc.sh script only...
inside it I have written following statement
status=`ps -efww | grep -w "abc.sh" | grep -v grep | grep -v $$ | awk '{ print $2 }'`
if [ ! -z "$status" ]; then
echo "[`date`] : abc.sh : Process is already running"
exit 1;
fi
I know it's wrong because every time it exits as it found its own process in 'ps'
how to solve it?
how can I check that script is already running or not from that script only ?

Answer 1

An easier way to check for a process already executing is the pidof command.
if pidof -x "abc.sh" >/dev/null; then
echo "Process already running"
fi
Alternatively, have your script create a PID file when it executes. It's then a simple exercise of checking for the presence of the PID file to determine if the process is already running.
#!/bin/bash
# abc.sh
mypidfile=/var/run/abc.sh.pid
# Could add check for existence of mypidfile here if interlock is
# needed in the shell script itself.
# Ensure PID file is removed on program exit.
trap "rm -f -- '$mypidfile'" EXIT
# Create a file with current PID to indicate that process is running.
echo $$ > "$mypidfile"
...
Update:
The question has now changed to check from the script itself. In this case, we would expect to always see at least one abc.sh running. If there is more than one abc.sh, then we know that process is still running. I'd still suggest use of the pidof command which would return 2 PIDs if the process was already running. You could use grep to filter out the current PID, loop in the shell or even revert to just counting PIDs with wc to detect multiple processes.
Here's an example:
#!/bin/bash
for pid in $(pidof -x abc.sh); do
if [ $pid != $$ ]; then
echo "[$(date)] : abc.sh : Process is already running with PID $pid"
exit 1
fi
done

Answer 2

I you want the "pidof" method, here is the trick:
if pidof -o %PPID -x "abc.sh">/dev/null; then
echo "Process already running"
fi
Where the -o %PPID parameter tells to omit the pid of the calling shell or shell script. More info in the pidof man page.

Answer 3

Here's one trick you'll see in various places:
status=`ps -efww | grep -w "[a]bc.sh" | awk -vpid=$$ '$2 != pid { print $2 }'`
if [ ! -z "$status" ]; then
echo "[`date`] : abc.sh : Process is already running"
exit 1;
fi
The brackets around the [a] (or pick a different letter) prevent grep from finding itself. This makes the grep -v grep bit unnecessary. I also removed the grep -v $$ and fixed the awk part to accomplish the same thing.

Answer 4

Working solution:
if [[ `pgrep -f $0` != "$$" ]]; then
echo "Another instance of shell already exist! Exiting"
exit
fi
Edit: I checked out some comments lately, so I tried attempting same with some debugging. I will also will explain it.
Explanation:
$0 gives filename of your running script.
$$ gives PID of your running script.
pgrep searches for process by name and returns PID.
pgrep -f $0 searches by filename, $0 being the current bash script filename and returns its PID.
So, pgrep checks if your script PID ($0) is equal to current running script ($$). If yes, then the script runs normally. If no, that means there's another PID with same filename running, so it exits. The reason I used pgrep -f $0 instead of pgrep bash is that you could have multiple instances of bash running and thus returns multiple PIDs. By filename, its returns only single PID.
Exceptions:
Use bash script.sh not ./script.sh as it doesn't work unless you have shebang.
Fix: Use #!/bin/bash shebang at beginning.
The reason sudo doesn't work is that it returns pgrep returns PID of both bash and sudo, instead of returning of of bash.
Fix:
#!/bin/bash
pseudopid="`pgrep -f $0 -l`"
actualpid="$(echo "$pseudopid" | grep -v 'sudo' | awk -F ' ' '{print $1}')"
if [[ `echo $actualpid` != "$$" ]]; then
echo "Another instance of shell already exist! Exiting"
exit
fi
while true
do
echo "Running"
sleep 100
done
The script exits even if the script isn't running. That is because there's another process having that same filename. Try doing vim script.sh then running bash script.sh, it'll fail because of vim being opened with same filename
Fix: Use unique filename.

Answer 5

Someone please shoot me down if I'm wrong here
I understand that the mkdir operation is atomic, so you could create a lock directory
#!/bin/sh
lockdir=/tmp/AXgqg0lsoeykp9L9NZjIuaqvu7ANILL4foeqzpJcTs3YkwtiJ0
mkdir $lockdir || {
echo "lock directory exists. exiting"
exit 1
}
# take pains to remove lock directory when script terminates
trap "rmdir $lockdir" EXIT INT KILL TERM
# rest of script here

Answer 6

Here's how I do it in a bash script:
if ps ax | grep $0 | grep -v $$ | grep bash | grep -v grep
then
echo "The script is already running."
exit 1
fi
This allows me to use this snippet for any bash script. I needed to grep bash because when using with cron, it creates another process that executes it using /bin/sh.

Answer 7

I find the answer from #Austin Phillips is spot on. One small improvement I'd do is to add -o (to ignore the pid of the script itself) and match for the script with basename (ie same code can be put into any script):
if pidof -x "`basename $0`" -o $$ >/dev/null; then
echo "Process already running"
fi

Answer 8

pidof wasn't working for me so I searched some more and came across pgrep
for pid in $(pgrep -f my_script.sh); do
if [ $pid != $$ ]; then
echo "[$(date)] : my_script.sh : Process is already running with PID $pid"
exit 1
else
echo "Running with PID $pid"
fi
done
Taken in part from answers above and https://askubuntu.com/a/803106/802276

Answer 9

Use the PS command in a little different way to ignore child process as well:
ps -eaf | grep -v grep | grep $PROCESS | grep -v $$

Answer 10

I create a temporary file during execution.
This is how I do it:
#!/bin/sh
# check if lock file exists
if [ -e /tmp/script.lock ]; then
echo "script is already running"
else
# create a lock file
touch /tmp/script.lock
echo "run script..."
#remove lock file
rm /tmp/script.lock
fi

Answer 11

I have found that using backticks to capture command output into a variable, adversly, yeilds one too many ps aux results, e.g. for a single running instance of abc.sh:
ps aux | grep -w "abc.sh" | grep -v grep | wc -l
returns "1". However,
count=`ps aux | grep -w "abc.sh" | grep -v grep | wc -l`
echo $count
returns "2"
Seems like using the backtick construction somehow temporarily creates another process. Could be the reason why the topicstarter could not make this work. Just need to decrement the $count var.

Answer 12

I didn't want to hardcode abc.sh in the check, so I used the following:
MY_SCRIPT_NAME=`basename "$0"`
if pidof -o %PPID -x $MY_SCRIPT_NAME > /dev/null; then
echo "$MY_SCRIPT_NAME already running; exiting"
exit 1
fi

Answer 13

This is compact and universal
# exit if another instance of this script is running
for pid in $(pidof -x `basename $0`); do
[ $pid != $$ ] && { exit 1; }
done

Answer 14

The cleanest fastest way:
processAlreadyRunning () {
process="$(basename "${0}")"
pidof -x "${process}" -o $$ &>/dev/null
}

Answer 15

For other variants (like AIX) that don't have pidof or pgrep. Reliability is greatly improved by getting a "static" view of the process table as opposed to piping it directly to grep. Setting IFS to null will preserve the carriage returns when the ps output is assigned to a variable.
#!/bin/ksh93
IFS=""
script_name=$(basename $0)
PSOUT="$(ps ax)"
ANY_TEXT=$(echo $PSOUT | grep $script_name | grep -vw $$ | grep $(basename $SHELL))
if [[ $ANY_TEXT ]]; then
echo "Process is already running"
echo "$ANY_TEXT"
exit
fi

Answer 16

[ "$(pidof -x $(basename $0))" != $$ ] && exit
https://github.com/x-zhao/exit-if-bash-script-already-running/blob/master/script.sh