Bash script to count number of executions instances does not work.

jose_spain · 05-30-2019, 08:01 AM

I am writing a script that detects if there is any instance of it already running and shows on screen the number of instances.

The content of the "detect_itself.sh" script is:

#!/bin/sh

INSTANCES_NUMBER=`ps -ef | grep detect_itself.sh | grep -v -i grep | wc -l`
echo "Number of detect_itself.sh instances running now =" $INSTANCES_NUMBER

echo "Second method:"
ps -ef | grep detect_itself.sh | grep -v -i grep | wc -l

echo "Third method:"
echo `ps -ef | grep detect_itself.sh | grep -v -i grep | wc -l`

echo "Please, press a key"
read -r key

When executing the script it shows by screen:

Number of detect_itself.sh instances running now = 2
Second method:
1
Third method:
2
Please, press a key

But I expected it to show:

Number of detect_itself.sh instances running now = 1
Second method:
1
Third method:
1
Please, press a key

I do not understand why if I execute ps -ef | grep detect_itself.sh | grep -v -i grep | wc -l it returns the value 1, but if I save this value in a variable and show it with echo it shows 2.

pan64 · 05-30-2019, 10:11 AM

you need to use pgrep instead of that ps | grep | grep | wc chain.
Also please use code tags to keep formatting of your original script.

The test you made shows clearly the method you use (that pipe chain) is not reliable.

michaelk · 05-30-2019, 10:59 AM

Command substitution i.e $(command) or `command` starts a child process.

This is why method 1 and 3 return 2 and not 1.

ondoho · 05-31-2019, 05:21 AM

seems too complex, too many pipes.

questions:
how do you start the script?
is it in your $PATH?
is it always started with the same command?

maybe

Code:

pidof -x "$0"

is enough?

BW-userx · 05-31-2019, 07:34 AM

Looking at this and thinking what good, or real world usage could this possibly be used for, a script that checks itself to see if it has another one running or not, if yes how many?

then I remember something I wrote in a script long long ago to see if more than one (child) was running. this is a snippet of it. Maybe (hopefully) you can figure out how to do something with it to get what you're looking for. It actually goes in line with what pan64 is telling you, you need to use.

Code:

[ $(pgrep -u $USER $appname | wc -l) -ge 2 ]

chrism01 · 06-07-2019, 12:55 AM

A classic usage example is a cron job that runs so frequently that successive runs may overlap if eg the amt of incoming data grows enough (& it is designed to single task).
There are a few ways of checking this situation, but checking the process table is more reliable/accurate than lock files - I've seen that fail a few times.

MadeInGermany · 06-07-2019, 12:32 PM

The backticks spawn a sub shell. Dependent on your shell interpreter, it can put another process in the proctable that can match your grep.
As suggested already, pgrep is the way out.

Code:

pgrep -x detect_itself.sh

Or

Code:

pgrep -f detect_itself.sh

Furthermore,
| wc -l is a builtin: pgrep -c ...

ondoho · 06-12-2019, 04:12 AM

Quote:

Originally Posted by MadeInGermany

Furthermore,
| wc -l is a builtin: pgrep -c ...

in recent bash versions, neither wc nor pgrep are builtins.

OP seems to have gone AWOL.

in any case, i use this in one of my scripts:

Code:

function only_me_or_exit {
# argument: pidfile
# make sure only 1 instance is running
touch "$1"
read lastPID < "$1"
# if lastPID is not null and a process with that pid exists , exit
[ ! -z "$lastPID" -a -d /proc/$lastPID ] && { echo "An instance of $me is already running with pid $lastPID." ; exit 1 ; }
# else - save my pid in the lock file, and continue
echo $$ > "$1"
}

been a while since i wrote it, code might be inferior.