Re: Variable won't work in function, even when I global it?
| From: | Leif K-Brooks | Date: | Mon, 20 May 2002 06:01:02 +0000 |
| Subject: | Re: Variable won't work in function, even when I global it? | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-98426@lists.php.net to get a copy of this message | ||
Ok, I guess I'll send it. Thanks for trying to help. Here it is:
$ip = $REMOTE_ADDR;
$query = mysql_query("select COUNT(*) as rowexists from ips where ip
= '$ip'");
$result = mysql_fetch_array($query);
if(($result['rowexists'] == 0) && ($loggedin)){
mysql_query("INSERT INTO
ips (ip,username) VALUES
('$ip','$userinfo[username]')") or print "Error: ".mysql_error();
}
$time_end = getmicrotime();
$time_taken = $time_end - $time_start;
$query = mysql_query("select COUNT(*) as cnt from ips");
$result = mysql_fetch_array($query);
$visits = $result['cnt'] + 2800;
eval($getlayout['footer']);
Miguel Cruz wrote:
Rather than continuing to shroud this in mystery, could you at least give us a taste of what is in $footertext?