Thread: PHP Question
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Old 02-28-2004, 11:08 AM  
alex79
Confirmed User
 
Join Date: Jun 2002
Location: france
Posts: 996
Quote:
Originally posted by JDog
if(!$x = fopen($url)) {
echo "Can not access $url";
exit;
}
with this i get next:

Warning: Wrong parameter count for fopen() in /home/****** on line 23
Can not access http://www.********
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