Run PHP script from bash

[Alvaro G Vicario]

I'm trying to run a PHP file as a shell script (with the regular
/usr/bin/php interpreter). It looks like this:

#!/usr/bin/php -q -d html_errors=0
<? echo "This is a test\n"; ?>

This is the result of the execution:

$ ./cibervivienda.php
Error in argument 1, char 3: option not found
Error in argument 1, char 4: option not found -
Error in argument 1, char 3: option not found

I can solve it removing the "-d html_errors=0" part but then I get
HTML-formated errors. I can't find the logic because the option is
definitively accepted if I type it myself:

$ /usr/bin/php -q -d html_errors=0
<? echo "This is a test\n"; ?>
This is a test

I'm not sure whether it's a PHP or a bash issue. Thank you in advance,
--
-- Álvaro G. Vicario - Burgos, Spain
-- Questions sent to my mailbox will be billed ;-)
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