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.p hp
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 ;-)
--
/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.p hp
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 ;-)
--