tailf output as script parameter

Question

Why is it when I leave the following tailf running:

tailf /var/log/z-way-server.log | grep --line-buffered device-info | gawk 'BEGIN { FS = "\"" } ; {print $4,"is",$8}'

I get my desired output:

Den Window Sensor is off

However, when I want this passed as a parameter to a script (which uses $1 within the script):

./message.sh $(tailf /var/log/z-way-server.log | grep --line-buffered device-info | gawk 'BEGIN { FS = "\"" } ; {print $4,"is",$8}')

It never passes the message as desired? For clarification, message.sh Test would post the word "Test" to a Slack channel.

Answer 1

The problem is that your message.sh command is never started. The shell first needs to evaluate the $(...) command in order to pass the result as arguments, but you are using tailf which never ends. Compare with this:

ls $(echo hello;sleep 10) & sleep 1; ps

The ps will show you that ls has not been started, and will only get the hello arg when the sleep 10 finishes and the $() part has finished.

Answer 2

try to pass with double quotes. and if your OS supports timeout command then try with timeout (refer the below post)

https://stackoverflow.com/questions/10430126/how-to-stop-tail-f-command-executed-in-sub-shell

if you don't want timeout, then try like this

#!/bin/bash

tailf /var/log/z-way-server.log | grep --line-buffered device-info | gawk 'BEGIN { FS = "\"" } ; {print $4,"is",$8}' | while read line
do
    /absolute/path/of/message.sh "${line}"
done