bash script to monitor myself - bash

I need to develop a shell script that would not be started if another instance of them self is running.
If I build a test.sh that monitors itself I need to know if it is already running and then abort, otherwise (if it not previously running) I can run
#!/bin/bash
loop() {
while [ 1 ]; do
echo "run";
#-- (... omissis ...)
sleep 30
done
}
daemon="`/bin/basename $0`"
pidlist=`/usr/bin/pgrep $daemon | grep -v $$`
echo "1:[ $pidlist ]"
pidlist=$(/usr/bin/pgrep $daemon | grep -v $$)
echo "2:[ $pidlist ]"
echo "3:[ `/usr/bin/pgrep $daemon | grep -v $$` ]"
echo "4:["
/usr/bin/pgrep $daemon | grep -v $$
echo "]"
if [ -z "$pidlist" ]; then
loop &
else
echo "Process $daemon is already running with pid [ $pidlist ]"
fi
exit 0;
When I run the above script for the first time (no previous instances running) I get this output:
1:[ 20341 ]
2:[ 20344 ]
3:[ 20347 ]
4:[
]
I cannot understand why only 4th attempt does not return anything (as expected). What's wrong in my script?
Do I have to redirect output of 4th command on a temporary file and then query that file in order to decide if I can run (or not) the loop function?
Thanks anyone would help me!

Sub-shells...the first three are run in sub-shells and hence $$ has changed to the PID of the sub-shell.
Try using:
PID=$$
pidlist=`/usr/bin/pgrep $daemon | grep -v $PID`
echo "1:[ $pidlist ]"
Etc. Since the value of $PID is established before the sub-shell is run, it should be the same for all of the commands.
Is this process going to be popular enough that other people want to run the same daemon on the machine? Maybe you never have multiple users on the machine, but remember that someone else might be wanting to run the command too.

Related

How to avoid the same bash script from running more than once when its called from another script?

I have a script called "upcall" which calls 4 different scripts. In upcall I call them in the way show. The first part of the script works when I run the script directly (bash upload_cloud1), but does not when its called from the script below. Im sure there is a way to fix this, but just not sure what it is. I have it currently setup in crontab to run every 15 mins to check for used space.
#!/bin/bash
if [[ "`pidof -x $(basename $0) -o %PPID`" ]]; then
echo "This script is already running with PID `pidof -x $(basename $0) -o %PPID`"
exit; fi
count=$(</opt/rclone/scripts/upcount)
size=$(df -k /dev/sda2 | tail -1 | awk '{print $3}')
if [ "$size" -gt "234003200" ]; then
bash /opt/rclone/scripts/upload_cloud${count}
else
echo "Not full yet"
fi

Check if bash script already running except itself with arguments

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.

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

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 $$ | awk '{ print $2 }'`
if [ ! -z "$status" ]; then
echo "[`date`] : abc.sh : Process is already running"
exit 1;
fi
I know it's wrong because every time it exits as it found its own process in 'ps'
how to solve it?
how can I check that script is already running or not from that script only ?
An easier way to check for a process already executing is the pidof command.
if pidof -x "abc.sh" >/dev/null; then
echo "Process already running"
fi
Alternatively, have your script create a PID file when it executes. It's then a simple exercise of checking for the presence of the PID file to determine if the process is already running.
#!/bin/bash
# abc.sh
mypidfile=/var/run/abc.sh.pid
# Could add check for existence of mypidfile here if interlock is
# needed in the shell script itself.
# Ensure PID file is removed on program exit.
trap "rm -f -- '$mypidfile'" EXIT
# Create a file with current PID to indicate that process is running.
echo $$ > "$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 != $$ ]; then
echo "[$(date)] : abc.sh : Process is already running with PID $pid"
exit 1
fi
done
I you want the "pidof" method, here is the trick:
if pidof -o %PPID -x "abc.sh">/dev/null; then
echo "Process already running"
fi
Where the -o %PPID parameter tells to omit the pid of the calling shell or shell script. More info in the pidof man page.
Here's one trick you'll see in various places:
status=`ps -efww | grep -w "[a]bc.sh" | awk -vpid=$$ '$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 $$ and fixed the awk part to accomplish the same thing.
Working solution:
if [[ `pgrep -f $0` != "$$" ]]; 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` != "$$" ]]; then
echo "Another instance of shell already exist! Exiting"
exit
fi
while true
do
echo "Running"
sleep 100
done
The script exits even if the script isn't running. That is because there's another process having that same filename. Try doing vim script.sh then running bash script.sh, it'll fail because of vim being opened with same filename
Fix: Use unique filename.
Someone please shoot me down if I'm wrong here
I understand that the mkdir operation is atomic, so you could create a lock directory
#!/bin/sh
lockdir=/tmp/AXgqg0lsoeykp9L9NZjIuaqvu7ANILL4foeqzpJcTs3YkwtiJ0
mkdir $lockdir || {
echo "lock directory exists. exiting"
exit 1
}
# take pains to remove lock directory when script terminates
trap "rmdir $lockdir" EXIT INT KILL TERM
# rest of script here
Here's how I do it in a bash script:
if ps ax | grep $0 | grep -v $$ | 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.
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 $$ >/dev/null; then
echo "Process already running"
fi
pidof wasn't working for me so I searched some more and came across pgrep
for pid in $(pgrep -f my_script.sh); do
if [ $pid != $$ ]; 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
Use the PS command in a little different way to ignore child process as well:
ps -eaf | grep -v grep | grep $PROCESS | grep -v $$
I create a temporary file during execution.
This is how I do it:
#!/bin/sh
# check if lock file exists
if [ -e /tmp/script.lock ]; then
echo "script is already running"
else
# create a lock file
touch /tmp/script.lock
echo "run script..."
#remove lock file
rm /tmp/script.lock
fi
I have found that using backticks to capture command output into a variable, adversly, yeilds one too many ps aux results, e.g. for a single running instance of abc.sh:
ps aux | grep -w "abc.sh" | grep -v grep | wc -l
returns "1". However,
count=`ps aux | grep -w "abc.sh" | grep -v grep | wc -l`
echo $count
returns "2"
Seems like using the backtick construction somehow temporarily creates another process. Could be the reason why the topicstarter could not make this work. Just need to decrement the $count var.
I didn't want to hardcode abc.sh in the check, so I used the following:
MY_SCRIPT_NAME=`basename "$0"`
if pidof -o %PPID -x $MY_SCRIPT_NAME > /dev/null; then
echo "$MY_SCRIPT_NAME already running; exiting"
exit 1
fi
This is compact and universal
# exit if another instance of this script is running
for pid in $(pidof -x `basename $0`); do
[ $pid != $$ ] && { exit 1; }
done
The cleanest fastest way:
processAlreadyRunning () {
process="$(basename "${0}")"
pidof -x "${process}" -o $$ &>/dev/null
}
For other variants (like AIX) that don't have pidof or pgrep. Reliability is greatly improved by getting a "static" view of the process table as opposed to piping it directly to grep. Setting IFS to null will preserve the carriage returns when the ps output is assigned to a variable.
#!/bin/ksh93
IFS=""
script_name=$(basename $0)
PSOUT="$(ps ax)"
ANY_TEXT=$(echo $PSOUT | grep $script_name | grep -vw $$ | grep $(basename $SHELL))
if [[ $ANY_TEXT ]]; then
echo "Process is already running"
echo "$ANY_TEXT"
exit
fi
[ "$(pidof -x $(basename $0))" != $$ ] && exit
https://github.com/x-zhao/exit-if-bash-script-already-running/blob/master/script.sh

Continue script if only one instance is running? [duplicate]

This question already has answers here:
Quick-and-dirty way to ensure only one instance of a shell script is running at a time
(43 answers)
Closed 5 years ago.
now this is embarrassing. I'm writing quick script and I can't figure out why this statement don't work.
if [ $(pidof -x test.sh | wc -w) -eq 1 ]; then echo Passed; fi
I also tried using back-ticks instead of $() but it still wouldn't work.
Can you see what is wrong with it? pidof -x test.sh | wc -w returns 1 if I run it inside of script, so I don't see any reason why basically if [ 1 -eq 1 ] wouldn't pass.
Thanks a lot!
Jefromi is correct; here is the logic I think you want:
#!/bin/bash
# this is "test.sh"
if [ $(pidof -x test.sh| wc -w) -gt 2 ]; then
echo "More than 1"
exit
fi
echo "Only one; doing whatever..."
Ah, the real answer: when you use a pipeline, you force the creation of a subshell. This will always cause you to get an increased number:
#!/bin/bash
echo "subshell:"
np=$(pidof -x foo.bash | wc -w)
echo "$np processes" # two processes
echo "no subshell:"
np=$(pidof -x foo.bash)
np=$(echo $np | wc -w)
echo "$np processes" # one process
I'm honestly not sure what the shortest way is to do what you really want to. You could avoid it all by creating a lockfile - otherwise you probably have to trace back via ppid to all the top-level processes and count them.
you don't have to pass the result of pidof to wc to count how many there are..use the shell
r=$(pidof -x -o $$ test.sh)
set -- $r
if [ "${##}" -eq 1 ];then
echo "passed"
else
echo "no"
fi
If you use the -o option to omit the PID of the script ($$), then only the PID of the subshell and any other instances of the script (and any subshells they might spawn) will be considered, so the test will pass when there's only one instance:
if [ $(pidof -x -o $$ test.sh | wc -w) -eq 1 ]; then echo Passed; fi
Here's how I would do it:
if [ "`pgrep -c someprocess`" -gt "1" ]; then
echo "More than one process running"
else
echo "Multiple processes not running"
fi
If you don't want to use a lockfile ... you can try this:
#!/bin/bash
if [[ "$(ps -N -p $$ -o comm,pid)" =~ $'\n'"${0##*/}"[[:space:]] ]]; then
echo "aready running!"
exit 1
fi
PS: it might need adjustment for a weird ${0##*/}
Just check for the existence of any one (or more) process identified as test.sh, the return code will be 1 if none are found:
pidof -x test.sh >/dev/null && echo "Passed"

Kill other bash daemons from the same script

I am having a hell of a time trying to write a "kill all other daemon processes" function for use within a bash daemon. I do not ever want more than one daemon running at once. Any suggestions? This is what I have:
#!/bin/bash
doService(){
while
do
something
sleep 15
done
}
killOthers(){
otherprocess=`ps ux | awk '/BashScriptName/ && !/awk/ {print $2}'| grep -Ev $$`
WriteLogLine "Checking for running daemons."
if [ "$otherprocess" != "" ]; then
WriteLogLine "There are other daemons running, killing all others."
VAR=`echo "$otherprocess" |grep -Ev $$| sed 's/^/kill /'`
`$VAR`
else
WriteLogLine "There are no daemons running."
fi
}
killOthers
doService
It works some of the time, it doesn't others. There is almost nothing consistent.
You've already eliminated the current process ID using grep -v so there's no reason to do it again when you issue the kill. There's also no reason to build the kill in a variable. Just do:
kill $otherprocess
But why not just use:
pkill -v $$ BashScriptName
or
pkill -v $$ $0
without any grep.
Then you can do:
if [[ $? ]]
then
WriteLogLine "Other daemons killed."
else
WriteLogLine "There are no daemons running."
fi
Could you try the old 'lock file' trick here? Test for a file: if it doesn't exists, create it and then startup; otherwise exit.
Like:
#!/bin/bash
LOCKFILE=/TMP/lockfile
if [ -f "$LOCKFILE" ]; then
echo "Lockfile detected, exiting..."
exit 1
fi
touch $LOCKFILE
while :
do
sleep 30
done
rm $LOCKFILE # assuming an exit point here, probably want a 'trap'-based thing here.
The downside is you have to clean-up lock-files from time to time, if an orphan is left behind.
Can you convert this to a 'rc' (or S*/K* script ?) so you can specify 'once' in the inittab (or equivalent method - not sure on MacOS) ?
Like what is described here:
http://aplawrence.com/Unixart/startup.html
EDIT:
Possibly this Apple Doc might help here:
http://developer.apple.com/mac/library/DOCUMENTATION/MacOSX/Conceptual/BPSystemStartup/Articles/StartupItems.html
If you run your service under runit — the service mustn't fork into the background — you'll have a guarantee there is exactly one instance of it running. runit starts the service if it isn't running or if it quit or crashed, stops it if you ask, keeps a pidfile around.

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