Simple bash script count running processes by name

Question

i'm working on a small bash script which counts how often a script with a certain name is running.

ps -ef | grep -v grep | grep scrape_data.php | wc -l

is the code i use, via ssh it outputs the number of times scrape_data.php is running. Currently the output is 3 for example. So this works fine.

Now I'm trying to make a little script which does something when the count is smaller than 1.

#!/bin/sh


if [ ps -ef | grep -v grep | grep scrape_data.php | wc -l ] -lt 1; then
        exit 0

 #HERE PUT CODE TO START NEW PROCESS

else

        exit 0
fi

The script above is what I have so far, but it does not work. I'm getting this error:

[root@s1 crons]# ./check_data.sh
./check_data.sh: line 4: [: missing `]'
wc: invalid option -- e

What am I doing wrong in the if statement?

Answer 1

Your test syntax is not correct, the lt should be within the test bracket:

if [ $(ps -ef | grep -v grep | grep scrape_data.php | wc -l) -lt 1 ]; then

  echo launch

else
  echo no launch

  exit 0
fi

or you can test the return value of pgrep:

pgrep scrape_data.php &> /dev/null

if [ $? ]; then
  echo no launch
fi

Answer 2

if you're using Bash then drop [ and -lt and use (( for arithmetic comparisons.

ps provides the -C switch, which accepts the process name to look for.
grep -v trickery are just hacks.

#!/usr/bin/env bash

proc="scrape_data.php"
limit=1

numproc="$(ps hf -opid,cmd -C "$proc" | awk '$2 !~ /^[|\\]/ { ++n } END { print n }')"

if (( numproc < limit ))
then
    # code when less than 'limit' processes run
    printf "running processes: '%d' less than limit: '%d'.\n" "$numproc" "$limit"
else
    # code when more than 'limit' processes run
    printf "running processes: '%d' more than limit: '%d'.\n" "$numproc" "$limit"
fi