Question

This has been bugging me slightly.

I know you can do

php foo.php

or

php -f foo.php

Is there no way to just launch a script with the execute bit set

./foo.php

Given the folowing:

#!/usr/bin/php

<?php 
exit('hello');

I get “Could not open input file” or ” bad interpreter: No such file or directory” depending on if there’s whitespace after “bin/php”.

Answer 1

The example code you’ve given seems OK and works for me (I added the the closing ?> though)

If you can run the file with /usr/bin/php foo.php there’s something weird going on. Just a wild guess, but maybe you text editor leaves a BOM (byte order mark) at the beginning of the file, so that the #! aren’t the very first two bytes in that file. (you can find out by doing hexdump -C foo.php | head)