Question

I’ve searched around but haven’t quite found what I’m looking for. In a nutshell I have created a bash script to run in a infinite while loop, sleeping and checking if a process is running. The only problem is even if the process is running, it says it is not and opens another instance.

I know I should check by process name and not process id, since another process could jump in and take the id. However all perl programs are named Perl5.10.0 on my system, and I intend on having multiple instances of the same perl program open.

The following “if” always returns false, what am I doing wrong here???

while true; do

 if [ ps -p $pid ]; then
  echo "Program running fine"
  sleep 10

 else
  echo "Program being restarted\n"
  perl program_name.pl &
  sleep 5
  read -r pid < "${filename}_pid.txt"
 fi

done

Answer 1

Get rid of the square brackets. It should be:

if ps -p $pid; then

The square brackets are syntactic sugar for the test command. This is an entirely different beast and does not invoke ps at all:

if test ps -p $pid; then

In fact that yields “-bash: [: -p: binary operator expected” when I run it.