Question

I have been facing a very peculiar issue with shell scripts.

Here is the scenario

Script1 (spawns in background)–> Script2

Script2 has the following code

function check_log()
{
    logfile=$1
    tail -5f ${logfile} | while read line
    do
      echo $line
      if echo $line|grep "${triggerword}";then
        echo "Logout completion detected"
        start_leaks_detection
        triggerwordfound=true
        echo "Leaks detection complete"
      fi
      if $triggerwordfound;then
        echo "Trigger word found and processing complete.Exiting"
        break
      fi

    done
        echo "Outside loop"
        exit 0

}

check_log "/tmp/somefile.log" "Logout detected"

Now the break in while loop does not help here. I can see “Logout completion detected” as well as “Leaks detection complete” being echoed on the stdout, but not the string “outside loop”

I am assuming this has to do something with tail -f creating a subshell. What I want to do is, exit that subshell as well as exit Script2 to get control back to Script1.

Can someone please shed some light on how to do this?

Answer 1

Try this, although it’s not quite the same (it doesn’t skip the beginning of the log file at startup):

triggerwordfound=
while [ -z "$triggerwordfound" ]; do
    while read line; do
        echo $line
        if echo $line|grep "${triggerword}";then
            echo "Logout completion detected"
            start_leaks_detection
            triggerwordfound=true
            echo "Leaks detection complete"
        fi
    done
done < "$logfile"
echo "Outside loop"

The double loop effectively does the same thing as tail -f.