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I am trying to set up a cron job that will keep alive a process that is running on a particular port in a centos server.

Based on the previous discussions I have come up with the following script which will be called by a cron job:

#!/bin/bash
opt="$(lsof -i tcp:9090| awk 'NR!=1 {print $2}')"
echo $opt
if [ -z "$opt" ]
then
    # nohup npm start & 1>/dev/null 2>/dev/null &
    echo "App Restarted" | mail -s "App is restarted"  "myemail"
fi

This gets triggered irrespective of the value in the variable output, not sure what is wrong here.

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1 Answer 1

Reset to default
2 
#assign output of a command to a variable
output="$(command)"

#test variable to see if empty
if [ -z "$output" ]
then
   #do something
   a_command
fi
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8
  • I have created the following script: #!/bin/bash output="$(lsof -i tcp:9090| awk 'NR!=1 {print $2}')" #test variable to see if empty if [[ "!" -z "$output" ]] then #do something echo "reaching here" fi Now I am getting error as line 5: conditional binary operator expected line 5: syntax error near -z' line 5: if [[ "!" -z "$output" ]] ' Commented Feb 26, 2020 at 20:49
  • 2
    Only unquoted ! inside [[ ]] negates. (Quoted ! is good inside [ ]). Why not -n instead of ! -z? Commented Feb 26, 2020 at 20:58
  • I have made minor update to the if and was able to solve my issue: if [ -z "$output" ] Commented Feb 26, 2020 at 21:04
  • @KamilMaciorowski Did I make an error leaving the "!" in? Commented Feb 27, 2020 at 17:51
  • Consult with OP if the negation shall be in place, but in any case, without double quotes. Cheers! Commented Feb 27, 2020 at 18:13

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