Re:fopen error

First of all... thanks for trying to help... I'll show the full piece of code I made : $file = "http://www.ivan.ivao.org/whazzup.txt"; $fp = fsockopen($file, 80); if (!$fp) { echo "does not open"; } else { echo $fp; ---------> Here is what I say it does't return anything. This line is just to figure out what fsockopen() returns. while (!feof($fp)) { echo fgets ($fp,500); } fclose ($fp); } Thanks again Alfonso ====================================== What do you mean "it returns nothing?" fsockopen() returns a file handle. Was the file handle valid, i.e., not 0? If true, then you must READ from the opened file handle. Please, be more specific in your error discussion. ---- Original Message ----- From: "Alfonso Ballesteros" <alfonsoballesteros@wanadoo.es> To: <php-general@lists.php.net> Sent: Thursday, October 17, 2002 6:58 AM Subject: Re: [PHP] fopen error I'm still stuck with this... Using fsockopen() I'm not able to open the remote file. I use $file = "http://www.xxx.xxx/file.txt"; $fp = fsockopen($file, 80); and $fp returns nothing... Any idea? Alfonso "Marek Kilimajer" <kilimajer@webglobe.sk> escribió en el mensaje news:3DAE87AE.1000307@webglobe.sk... > if allow_url_fopen is not on, you can still use fsockopen(), and no it > doesn't affect security, if you don't open files based on unchecked user > input. > > Alfonso Ballesteros wrote: > > >Hello > > > >I'm a newcomer to PHP and I wish to know how to get the contents of a file > >located in an URL if the parameter allow_url_fopen is "no value" in the > >php.ini file. > > > >Does this parameter, if is "on", affect the security of the server? > > > >Thanks in advance > > > >Alfonso > > > > > > > > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php