Bash calls two bash scripts

Question

I have the following script:

#!/bin/bash

if [ "$EUID" -ne 0 ]
  then
    echo '' 
    echo 'Please run the script as root'
    echo ''
  exit
fi

for run in {1..11}
do
    sudo ./start_ap.sh    
    sleep 10    
    sudo ./tst.sh

done 

The problem is that after executing

sudo ./start_ap.sh

the next lines will not be executed, because the line sudo ./start_ap.sh needs CTRL+C to stop and only then next lines will be executed.

However, I want that the sudo ./start_ap.sh will be terminated after sudo ./tst.sh and at next step this will be repeated 11 times.

So far, after execution of sudo ./start_ap.sh, the next lines will not be executed without killing its process.

How can I realize it?

PS start_ap.sh starts the hostapd and that's why it needs killing for next executions.

Answer 1

You need to run ./start_ap.sh in the background, then kill it after ./tst.sh completes. Note that if you actually run the script as root, there is no need to use sudo inside the script.

for run in {1..11}; do
    ./start_ap.sh & pid=$!
    sleep 10
    ./tst.sh
    kill "$pid"
done