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 have a scenario to copy file from one server to another, for that i need to check any existing scp is in progress, have wrote a sample shell script but the condition is not being met even though syntax is correct, the main problem here is the output of ps command will gets stored in variable scpstat and the same compared for matching string in if statement, here I'm getting the output of the variable is different from executing outside of the script. can see it is formatted different in script execution when executing sh -x scpsamp.sh, why there is "sh" appended to the output, but while comparing without ps and assigning as scpstat='scp' i can able to get the condition correct, am i doing anything wrong while getting output in to the variable. please help
#!/bin/sh
scpstat=`ps -ef | grep scp | egrep -v 'grep|ssh' | awk '{print $8}')`
if [ "$scpstat" = "scp" ];
then
echo "SCP is in progress"
else
echo "No SCP in progress"
fi
sh -x output
It's notoriously difficult to extract information from the output of ps. If your system has pgrep, it's much easier:
if pgrep scp >/dev/null
then
echo "SCP is in progress"
else
echo "No SCP in progress"
fi
Trying to output a list of cronjobs, not a list of users. The raw output of crontab -l is way too dirty and I can't seem to clean it up. I run this with sudo script.sh or su and then run it. I've tried invoking it sudo script.sh | grep -v no also. I'm mystified why this doesn't work:
#!/bin/bash
#Trying to show all cronjobs but no extraneous info
#
# This shows "no crontab for USER" for every USER without
# a crontab - I only want to see actual cronjobs, not a long
# list of users without crontabs
echo "Here is the basic output that needs manipulation:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
crontab -l -u $USER
done
#
# grep -v fails me
# (grep'ing the output of the script as a whole fails also)
echo "
trying with grep -v no on each line:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
crontab -l -u $USER | grep -v no
done
echo "
maybe with quotes around the no:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
crontab -l -u $USER | grep -v "no"
done
# string manipulation - I can't even get started
echo "
And here I try to put the commmand output into a string so I can manipulate it further, and use an if/then/fi on the product:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
STRING="$(crontab -l -u $USER | grep -v no)"
echo "STRING: $STRING"
done
BTW, is there an easier way to get code to format correctly here than pasting in 4 spaces at the beginning of each line? I must have experimented for 40 minutes. Not complaining, just asking.
crontab is wiring the "no crontab for user" to standard error. To get rid of those messages, you can run
crontab -l -u $USER 2>/dev/null
within your loop.
I would also suggest renaming USER into something else. The USER variable name is reserved, and should be set to your user (login) name. Generally, you should use lower case variable names, to avoid this sort of name clash.
I'm writing a unix shell script and need to check if there are currently running processes with "xyz" in their directory. If yes than continue to next command and show text like "Found It".
If not than don't continue and display text like "Process Not Found".
I tried something like this:
if ps -ef | grep xyz
then
echo "XYZ Process Found!"
else
echo "XYZ Process Not Found!"
fi
But it just showing me the processes and display "process found" even if there's no xyz process.
I believe you want to check the output of the command against a value using Command substition, from the linked bash-hackers wiki The command substitution expands to the output of commands. These commands are executed in a subshell, and their stdout data is what the substitution syntax expands to. Also, count the lines and remove grep. Something like,
if [[ $(ps -ef | grep xyz | grep -v grep | wc -l) != 0 ]]; then
echo "XYZ Process Found!"
else
echo "XYZ Process Not Found!"
fi
Edit
Based on the comments below, you should probably use
if [[ $(ps -ef | grep -c xyz) -ne 1 ]]; then
which is a lot easier to read.
When you run grep xyz, that process - grep xyz - is also running & thus shown in the output of ps -ef.
This running process command line contains xyz. Thus grep passes that line to output.
Hence you always get zero exit status - i.e. success.
2 Solutions:
use if ps -ef | grep '[x]yz'; then. (You may want to suppress grep output with -q)
The grep command being run is grep [x]yz. This gets printed in ps -ef output.
Obviously, grep filters out this line. [x]yz could be matched with \[x\]yz, not with [x]yz.
use if pgrep -f xyz >/dev/null; then
Check man pgrep for more details..
You can also use pgrep. From pgrep(1):
pgrep looks through the currently running processes and lists the
process IDs which match the selection criteria to stdout.
[...]
EXIT STATUS
0 One or more processes matched the criteria.
1 No processes matched.
2 Syntax error in the command line.
3 Fatal error: out of memory etc.
Example output:
[~]% pgrep xterm
18231
19070
31727
You can use it in an if statement like so:
if pgrep xterm > /dev/null; then
echo Found xterm
else
echo xterm not found
fi
Note: pgrep is not a standard utility (ie. it's not in POSIX), but widely available on at least Linux and I believe most BSD systems.
is_xyz_running() {
[ "$(pgrep xyz)" ] && echo true || echo false
}
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