bash script to execute php file

Question

I want to write something detecting where PHP is installed and then execute php after that eg:

./find_php.sh /path/to/php/file.php

So I can get it to detect the php I want to use and stick it in a variable, but I don't know how to get it to execute the following php file after. Here's a skeleton of what I've got:

#!/bin/bash
if [ -e /usr/local/zend/bin/php ]; then
   PHP=/usr/local/zend/bin/php
else
   PHP=/usr/bin/php
fi
$PHP <execute php>

Edit - thanks got it, just needed exec $PHP "$@"

Answer 1

Make your last line:

exec $PHP "$@"

That will run the php binary with the arguments you gave to the script and have it take over the shell script process as well.

If you would like to have your script do other things after executing the php script, drop the exec.