Quick-and-dirty way to ensure only one instance of a shell script is running at a time

Question

What\'s a quick-and-dirty way to make sure that only one instance of a shell script is running at a given time?

Answer 1

All approaches that test the existence of "lock files" are flawed.

Why? Because there is no way to check whether a file exists and create it in a single atomic action. Because of this; there is a race condition that WILL make your attempts at mutual exclusion break.

Instead, you need to use mkdir. mkdir creates a directory if it doesn't exist yet, and if it does, it sets an exit code. More importantly, it does all this in a single atomic action making it perfect for this scenario.

if ! mkdir /tmp/myscript.lock 2>/dev/null; then
    echo "Myscript is already running." >&2
    exit 1
fi

For all details, see the excellent BashFAQ: http://mywiki.wooledge.org/BashFAQ/045

If you want to take care of stale locks, fuser(1) comes in handy. The only downside here is that the operation takes about a second, so it isn't instant.

Here's a function I wrote once that solves the problem using fuser:

#       mutex file
#
# Open a mutual exclusion lock on the file, unless another process already owns one.
#
# If the file is already locked by another process, the operation fails.
# This function defines a lock on a file as having a file descriptor open to the file.
# This function uses FD 9 to open a lock on the file.  To release the lock, close FD 9:
# exec 9>&-
#
mutex() {
    local file=$1 pid pids 

    exec 9>>"$file"
    { pids=$(fuser -f "$file"); } 2>&- 9>&- 
    for pid in $pids; do
        [[ $pid = $$ ]] && continue

        exec 9>&- 
        return 1 # Locked by a pid.
    done 
}

You can use it in a script like so:

mutex /var/run/myscript.lock || { echo "Already running." >&2; exit 1; }

If you don't care about portability (these solutions should work on pretty much any UNIX box), Linux' fuser(1) offers some additional options and there is also flock(1).