我使用下面的PHP代码来获取访问者的IP地址:
<?php echo $_SERVER['REMOTE_ADDR']; ?>
但是,当访问者使用代理时,我无法从他们那里获得真实的IP地址。在这种情况下,有没有办法获得访问者的IP地址?
发布于 2012-11-30 22:01:14
试试这段php代码。
<?PHP
function getUserIP()
{
// Get real visitor IP behind CloudFlare network
if (isset($_SERVER["HTTP_CF_CONNECTING_IP"])) {
$_SERVER['REMOTE_ADDR'] = $_SERVER["HTTP_CF_CONNECTING_IP"];
$_SERVER['HTTP_CLIENT_IP'] = $_SERVER["HTTP_CF_CONNECTING_IP"];
}
$client = @$_SERVER['HTTP_CLIENT_IP'];
$forward = @$_SERVER['HTTP_X_FORWARDED_FOR'];
$remote = $_SERVER['REMOTE_ADDR'];
if(filter_var($client, FILTER_VALIDATE_IP))
{
$ip = $client;
}
elseif(filter_var($forward, FILTER_VALIDATE_IP))
{
$ip = $forward;
}
else
{
$ip = $remote;
}
return $ip;
}
$user_ip = getUserIP();
echo $user_ip; // Output IP address [Ex: 177.87.193.134]
?>
发布于 2012-11-30 22:08:55
这是我见过的最常见的技术:
function getUserIP() {
if( array_key_exists('HTTP_X_FORWARDED_FOR', $_SERVER) && !empty($_SERVER['HTTP_X_FORWARDED_FOR']) ) {
if (strpos($_SERVER['HTTP_X_FORWARDED_FOR'], ',')>0) {
$addr = explode(",",$_SERVER['HTTP_X_FORWARDED_FOR']);
return trim($addr[0]);
} else {
return $_SERVER['HTTP_X_FORWARDED_FOR'];
}
}
else {
return $_SERVER['REMOTE_ADDR'];
}
}
请注意,这并不能保证您将始终获得正确的用户IP,因为有许多方法可以隐藏它。
发布于 2015-03-04 07:36:54
这是我的方法:
function getRealUserIp(){
switch(true){
case (!empty($_SERVER['HTTP_X_REAL_IP'])) : return $_SERVER['HTTP_X_REAL_IP'];
case (!empty($_SERVER['HTTP_CLIENT_IP'])) : return $_SERVER['HTTP_CLIENT_IP'];
case (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) : return $_SERVER['HTTP_X_FORWARDED_FOR'];
default : return $_SERVER['REMOTE_ADDR'];
}
}
使用方法:
$ip = getRealUserIp();
https://stackoverflow.com/questions/13646690
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