Quick-and-dirty way to ensure only one instance of

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

Take a look to FLOM (Free LOck Manager) http://sourceforge.net/projects/flom/: you can synchronize commands and/or scripts using abstract resources that does not need lock files in a filesystem. You can synchronize commands running in different systems without a NAS (Network Attached Storage) like an NFS (Network File System) server.

Using the simplest use case, serializing "command1" and "command2" may be as easy as executing:

flom -- command1

and

flom -- command2

from two different shell scripts.

Answer 2

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).

Answer 3

PID and lockfiles are definitely the most reliable. When you attempt to run the program, it can check for the lockfile which and if it exists, it can use ps to see if the process is still running. If it's not, the script can start, updating the PID in the lockfile to its own.

Answer 4

Actually although the answer of bmdhacks is almost good, there is a slight chance the second script to run after first checked the lockfile and before it wrote it. So they both will write the lock file and they will both be running. Here is how to make it work for sure:

lockfile=/var/lock/myscript.lock

if ( set -o noclobber; echo "$$" > "$lockfile") 2> /dev/null ; then
  trap 'rm -f "$lockfile"; exit $?' INT TERM EXIT
else
  # or you can decide to skip the "else" part if you want
  echo "Another instance is already running!"
fi

The noclobber option will make sure that redirect command will fail if file already exists. So the redirect command is actually atomic - you write and check the file with one command. You don't need to remove the lockfile at the end of file - it'll be removed by the trap. I hope this helps to people that will read it later.

P.S. I didn't see that Mikel already answered the question correctly, although he didn't include the trap command to reduce the chance the lock file will be left over after stopping the script with Ctrl-C for example. So this is the complete solution

Answer 5

Really quick and really dirty? This one-liner on the top of your script will work:

[[ $(pgrep -c "`basename \"$0\"`") -gt 1 ]] && exit

Of course, just make sure that your script name is unique. :)

Answer 6

This example is explained in the man flock, but it needs some impovements, because we should manage bugs and exit codes:

   #!/bin/bash
   #set -e this is useful only for very stupid scripts because script fails when anything command exits with status more than 0 !! without possibility for capture exit codes. not all commands exits >0 are failed.

( #start subprocess
  # Wait for lock on /var/lock/.myscript.exclusivelock (fd 200) for 10 seconds
  flock -x -w 10 200
  if [ "$?" != "0" ]; then echo Cannot lock!; exit 1; fi
  echo $$>>/var/lock/.myscript.exclusivelock #for backward lockdir compatibility, notice this command is executed AFTER command bottom  ) 200>/var/lock/.myscript.exclusivelock.
  # Do stuff
  # you can properly manage exit codes with multiple command and process algorithm.
  # I suggest throw this all to external procedure than can properly handle exit X commands

) 200>/var/lock/.myscript.exclusivelock   #exit subprocess

FLOCKEXIT=$?  #save exitcode status
    #do some finish commands

exit $FLOCKEXIT   #return properly exitcode, may be usefull inside external scripts

You can use another method, list processes that I used in the past. But this is more complicated that method above. You should list processes by ps, filter by its name, additional filter grep -v grep for remove parasite nad finally count it by grep -c . and compare with number. Its complicated and uncertain