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${count}
else
echo "Not full yet"
fi
I've got the following code in my shell script:
SERVER=`ps -ef | grep -v grep | grep -c sof2ded`
if ["$SERVER" != "0"]; then
echo "Already Running, exiting"
exit
else
echo "Starting up the server..."
cd /home/sof2/
/home/sof2/crons/start.sh > /dev/null 2>&1
fi
I did chmod a+x status.sh
Now I try to run the script but it's returning this error:
./status.sh: line 5: [1: command not found
Starting up the server...
Any help would be greatly appreciated.
Could you please try changing a few things in your script as follows and let me know if that helps you?(changed back-tick to $ and changed [ to [[ in code)
SERVER=$(ps -ef | grep -v grep | grep -c sof2ded)
if [[ "$SERVER" -eq 0 ]]; then
echo "Already Running, exiting"
exit
else
echo "Starting up the server..."
cd /home/sof2/
/home/sof2/crons/start.sh > /dev/null 2>&1
fi
The problem is with the test command. "But", I hear you say, "I am not using the test command". Yes you are, it is also known as [.
if statement syntax is if command. The brackets are not part of if syntax.
Commands have arguments separated (tokenized) by whitespace, so:
[ "$SERVER" != "0" ]
The whitespace is needed because the command is [ and then there are 4 arguments passed to it (the last one must be ]).
A more robust way of comparing numerics is to use double parentheses,
(( SERVER == 0 ))
Notice that you don't need the $ or the quotes around SERVER. Also the spacing is less important, but useful for readability.
[[ is used for comparing text patterns.
As a comment, backticks ` ` are considered deprecated because they are difficult to read, they are replaced with $( ... ).
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))" != $$ ] && exit
https://github.com/x-zhao/exit-if-bash-script-already-running/blob/master/script.sh
I need to develop a shell script that would not be started if another instance of them self is running.
If I build a test.sh that monitors itself I need to know if it is already running and then abort, otherwise (if it not previously running) I can run
#!/bin/bash
loop() {
while [ 1 ]; do
echo "run";
#-- (... omissis ...)
sleep 30
done
}
daemon="`/bin/basename $0`"
pidlist=`/usr/bin/pgrep $daemon | grep -v $$`
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=$$
pidlist=`/usr/bin/pgrep $daemon | grep -v $PID`
echo "1:[ $pidlist ]"
Etc. Since the value of $PID is established before the sub-shell is run, it should be the same for all of the commands.
Is this process going to be popular enough that other people want to run the same daemon on the machine? Maybe you never have multiple users on the machine, but remember that someone else might be wanting to run the command too.
I have the following bash script:
tail -F -n0 /private/var/log/system.log | while read line
do
if [ ! `echo $line | grep -c 'launchd'` -eq 0 ]; then
echo 'launchd message'
exit 0
fi
done
For some reason, it is echoing launchd message, waiting for a full 5 seconds, and then exiting.
Why is this happening and how do I make it exit immediately after it echos launchd message?
Since you're using a pipe, the while loop is being run in a subshell. Run it in the main shell instead.
#!/bin/bash
while ...
do
...
done < <(tail ...)
As indicated by Ignacio, your tail | while creates a subshell. The delay is because it's waiting for the next line to be written to the log file before everything closes.
You can add this line immediately before your exit command if you'd prefer not using process substitution:
kill -SIGPIPE $$
Unfortunately, I don't know of any way to control the exit code using this method. It will be 141 which is 128 + 13 (the signal number of SIGPIPE).
If you're trying to make the startup of a daemon dependent on another one having started, there's probably a better way to do that.
By the way, if you're really writing a Bash script (which you'd have to be to use <() process substitution), you can write your if like this: if [[ $line == *launchd* ]].
You can also exit the subshell with a tell-tale exit code and then test the value of "$?" to get the same effect you're looking for:
tail -F -n0 /private/var/log/system.log | while read line
do
if [ ! `echo $line | grep -c 'launchd'` -eq 0 ]; then
echo 'launchd message'
exit 10
fi
done
if [ $? -eq 10 ]; then exit 0; fi