Logging IP in MYSQL Database

Beginners - Learning PHP
#1

Help! I don’t know why this isn’t working. I’m trying to log an ip in a database and retrieve it along with a integer. All it has been doing is logging 0 as the ip and 0 as the integer.

[php] $data = mysql_query(“SELECT ip,points FROM information”)
or die(mysql_error());

if (isset($_SERVER[“HTTP_X_FORWARDED_FOR”])) {
$ip = $_SERVER[“HTTP_X_FORWARDED_FOR”];
} else {
$ip = $_SERVER[“REMOTE_ADDR”];
}

while($info = mysql_fetch_array( $data ))
{
if($info[‘ip’] != null && $info[‘ip’] == ‘ip2long($ip)’){
Print “You have $info[points] points!”;
break;
}
if($info[‘ip’] != null && $info[‘ip’] != ‘ip2long($ip)’){
mysql_query(“INSERT INTO information VALUES (‘INET_ATON($ip)’,‘0’)”);
Print “You have 0 points!”;
break;
}
}[/php]

#2

if (isset($_SERVER[“HTTP_X_FORWARDED_FOR”])) { will always be truthy as that specific variable is always set. Try using !empty() instead of isset() and let me know if it works.

#3

It still does not work. Nothing has changed. When I do
Print $ip;

It prints out this.
::1

#4

IPv6.

Use the MySQL INET6_ATON for it.

#5

So would this work?

[php] $data = mysql_query(“SELECT ip,points FROM information”)
or die(mysql_error());

if (!empty($_SERVER[“HTTP_X_FORWARDED_FOR”])) {
$ip = $_SERVER[“HTTP_X_FORWARDED_FOR”];
} else {
$ip = $_SERVER[“REMOTE_ADDR”];
}

while($info = mysql_fetch_array( $data ))
{
if($info[‘ip’] != null && $info[‘ip’] == ip2long($ip)){
Print “You have $info[points] points!”;
break;
}
if($info[‘ip’] != null && $info[‘ip’] != ip2long($ip)){
mysql_query(“INSERT INTO information VALUES (‘INET6_ATON($ip)’,‘0’)”);
Print “You have 0 points!”;
break;
}
}[/php]

#6

Edit: It still doesn’t work. Im currently using the code above. Also when I refresh the page it just sends multiple queries to MYSQL.

#7

Hmm now Im just having trouble logging the IP. I keep getting 0 in MYSQL.

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