I am this code to get the IP address on my locolhost and I get
::1
result
code is:
function ip()
{
if (!empty($_SERVER['HTTP_CLIENT_IP']))
{
$ip = $_SERVER['HTTP_CLIENT_IP'];
} elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR']))
{
$ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
}
else
{
$ip = $_SERVER['REMOTE_ADDR'];
}
return $ip;
}
echo ip();
where is error?
Actually, I want to get country & city name after I get IP... Any other solution?
Thanks
::1 is shorthand for localhost, using IPv6 terminology, so it is working. Try connecting using your hostname, and you should see the address change to be more obviously an address...
As #Roland writes it is an IPv6 address. If you don't intend on using IPv6 in your application then you should tell your web server to stop listening on IPv6 ports. When you send a request the client will try IPv6 first if it doesn't work then it will fall back to IPv4 and you should get the much more familiar looking 127.0.0.1.
You could use the following:
$server_address = $_SERVER['SERVER_ADDR'];
$port_used = $_SERVER['SERVER_PORT'];
$ip_address = $_SERVER['REMOTE_ADDR'];
// on my test machine this gives the following results:
$server_address = 127.0.0.1
$port_used = 80
$ip_address = 127.0.0.1
Edited: to include the geo-location aspect of the question, that I hadn't noticed 'til after I submitted the original answer.
Rather than repeat answers found elsewhere, I'll first link to this (well-answered) SO question:
google-geolocation-api-library, and then to the Google results page for the search terms geolocation php site:stackoverflow.com, which links to many other -probably relevant- answers that might better address your questions than I'm able.
If you print_r($_SERVER) you will see what information is available, what it looks like, and what fields might be empty.
Related
I want to browse some sites using a proxy, so I want to change the IP address of my client to the specified IP.
I am also testing it on localhost, so my goal is to visit my localhost site with PHP script (also localhost). The problem is how to test this (maybe some simple test by writing out changed IP to file).
I'm not entirely sure what exactly you mean, but you can check your clients IP address with this:
if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
$ip = $_SERVER['HTTP_CLIENT_IP'];
} elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
$ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
} else {
$ip = $_SERVER['REMOTE_ADDR'];
}
echo $ip;
Ofcourse this will not change your clients IP address. You need an actual proxy server for that.
I have PHP code that is supposed to detect a user's IP address. The code below returns an IP address, but the IP address is sometimes a local network IP (e.g. 10.0.0.1) and not a public IP. How can I ensure that I always get the public IP? Thanks. BTW, this code is from another StackOverflow post. Also, this code is used in a website that is being accessed over the internet from a completely separate network than that of my Apache web server.
if (isset($_SERVER["HTTP_CLIENT_IP"])){
$ip = $_SERVER["HTTP_CLIENT_IP"];
} elseif (isset($_SERVER["HTTP_X_FORWARDED_FOR"])){
$ip = $_SERVER["HTTP_X_FORWARDED_FOR"];
} elseif (isset($_SERVER["HTTP_X_FORWARDED"])){
$ip = $_SERVER["HTTP_X_FORWARDED"];
} elseif (isset($_SERVER["HTTP_FORWARDED_FOR"])){
$ip = $_SERVER["HTTP_FORWARDED_FOR"];
} elseif (isset($_SERVER["HTTP_FORWARDED"])){
$ip = $_SERVER["HTTP_FORWARDED"];
} else {
$ip = $_SERVER["REMOTE_ADDR"];
}
Eliminate all the if(){} else{} code, and just request the REMOTE_ADDR:
$ip = $_SERVER["REMOTE_ADDR"];
It's the only reliable source of a user's remote IP address as all the other _SERVER keys can be masked by the client.
If the IP address is still local (and youre SURE that you're dealing with clients not in the networks or on a local VPN) the you may be dealing with either a server caching system (Squid Proxy eg). Have a look at http://www.nineteenlabs.com/2007/08/24/high-anonymous-proxy-squid-25/
here i am using $_SERVER['remote_addr'] i am getting the IP address INTERNET providers. please help me out.
function getRealIpAddr()
{
if (!empty($_SERVER['HTTP_CLIENT_IP'])) //check ip from share internet
{
$ip=$_SERVER['HTTP_CLIENT_IP'];
}
elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) //to check ip is pass from proxy
{
$ip=$_SERVER['HTTP_X_FORWARDED_FOR'];
}
else
{
$ip=$_SERVER['REMOTE_ADDR'];
}
return $ip;
}
It would seem that the question is looking for the server's IP address, and the problem is that the supplied code is returning the remote IP. If you want the IP address of the server, try $_SERVER['SERVER_ADDR'] . See $_SERVER for more details, or clearly define what you mean by "system IP".
You will not be able to get the client's local IP address from PHP like this, you will only ever get the
common gateway IP address
as you put it. This is because the request will appear to be coming from the router or whatever is between your client and the web server
My server is behind a proxy and I'm trying to get a IP of a user I'm using in PHP
$IP = $_SERVER['HTTP_X_FORWARDED_FOR'];
The problem I see lately some users Ip's are like this below example Ip's.
173.0.0.000, 190.000.00.01
173.0.0.000, 190.000.00.02
173.0.0.000, 190.000.00.03
173.0.0.000, 190.000.00.04
173.0.0.000, 190.000.00.05
How would I go about just getting the correct IP like below. It doesn't happen on all the Ip's.
The format for X-Forwarded-For is: client1, proxy1, proxy2. So you will want to grab the first address in the list:
$IP = current(explode(',', $_SERVER['HTTP_X_FORWARDED_FOR']));
echo $IP; // 173.0.0.000
That will grab the first IP address whether there are multiple addresses or just one.
i have following php code to get vistor ip
function VisitorIP()
{
if(isset($_SERVER['HTTP_X_FORWARDED_FOR']))
$TheIp=$_SERVER['HTTP_X_FORWARDED_FOR'];
else $TheIp=$_SERVER['REMOTE_ADDR'];
return trim($TheIp);
}
$Users_IP_address = VisitorIP();
but every time i execute this script on my localhost i got ::1 as IP
how can i get exact internet IP & system IP address
because my script will run on LAN nwteork and i want to record IP of lan pcs & internet IP and save it in mysql database.
i need PHP code which privide local system IP and system internet IP.
Thanks
Technically, you are getting the right IP. ::1 is your loopback (localhost) IPv6 address.
First, throw this stupid function away, because HTTP_X_FORWARDED_FOR is not an IP address but merely an HTTP header, and make it just
$Users_IP_address = $_SERVER['REMOTE_ADDR'];
Next, you have to set up your server to work with ipv4, not ipv6.
Though it is not really a PHP question and should be asked on the serverfault, provided with full system setup: OS, version, your rights etc.
When you run above script on localhost you can get ::1 this result.
It's very common.
If you want to get your system IP address with localhost use this code.
1) Script for localhost
<?PHP
$ip = #file_get_contents("http://www.apps-tube.com/apps/ip-address/ip.php");
echo $ip;
?>
2) Script for Servers
<?PHP
function getUserIP()
{
$client = #$_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]
?>