Can't capture and echo PID by $!

Question

I am trying to start tcpdump as a background process, and get its PID. The script will monitor the system for disk space, cpu and memory, and kill the tcpdump process if any goes beyond the threshold. But I cant get the PID using $!. The output is blank.

#!/bin/bash
`nohup tcpdump port 389 -i bond0 -s 0 -w /tmp/capture.pcap  > nohup2.out 2>&1&`
my_pid=$!                                                                      
echo $my_pid                                                                   

flag=1                                                                         
exit 1                                                                         

and the output is

bash # sh t.sh

bash #     

Answer 1

Change this:

`nohup tcpdump port 389 -i bond0 -s 0 -w /tmp/capture.pcap  > nohup2.out 2>&1&`

to this:

nohup tcpdump port 389 -i bond0 -s 0 -w /tmp/capture.pcap  > nohup2.out 2>&1 &

The backticks mean that the command should be run in a subshell and its output captured and substituted into a new command. For example, this command:

`echo foo bar`

is equivalent to this command:

foo bar

because it runs echo foo bar, captures the output foo bar, and then runs foo bar. In your case you have absolutely no need of that (if only because you're redirecting the output to a file), so you can just drop them.