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

Answer 1

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

Answer 2

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

Answer 3

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..."

Answer 4

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"

Answer 5

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

Answer 6

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.