RE: [PHP] require() problem

this sounds like a problem with your path. is the required file that you are specifying in $configfile. in a different directory? This example assumes that the $configfile is in the same directory. also, unless your $configfile is supposed to run/insert PHP/HTML code, you should be using include. for example: $configfile = "config.inc.php3"; include("$configfile"); -----Original Message----- From: Andre Fadel [mailto:fadel@fee.unicamp.br] Sent: Monday, August 07, 2000 7:13 PM To: php-general@lists.php.net Subject: [PHP] require() problem I was just wondering if require will take variables instead of file names. I have this on my script: require($configfile); But I'm getting a error like this: Fatal error: Failed opening required 'config.inc.php3' (include_path='') Although the $configfile variable does not contain that, if i tell it to print $configfile right before the require statement, i get something different (the right stuff!!). So, what's the deal with that? What could be wrong? Thanks a lot! -- André Fadel fadel@fee.unicamp.br Hoje E Sempre GUARANI! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net