Passing array argument to bash from php

Question

I am trying to pass array from shell_exec and want to access the array in bash file. I tried to research from many websites but did not find the proper solution for it. Plea

Answer 1

If your php variable $array is a php Array, you need to convert it to a string to do the call, eg:

 shell_exec('sh get_countries.sh '.join(' ',$array));

Then your countries.sh should be exactly what you gave as an example:

#!/bin/bash
declare -a myArray
myArray=("$@") 
for arg in "${myArray[@]}"; do
 echo "$arg"
done

Make sure you permit the file for execution with chmod a+rx countries.sh.