i m sorry im newbie here,.. so i want to ask about, how to get the IP from the server,.
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first i user PEAR2 to get an api from Mikrotik, then i want to get an IP for making a log, to log what an IP that Connect or Dissconnect from the network itself... the reason why i make the log from PEAR2, i want to make a notification android based, that if the client from the mikrotik's is disconnect or connect, it will send a notif to an android apps..
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in those case PHP i use this script,
echo $server_ip = gethostbyname($_SERVER['SERVER_NAME']);
?> is this a correct..?
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many thanks for you all. im sorry for My bad English
You want echo $server_ip = $_SERVER['SERVER_ADDR']
What you have gives you the host name.
Like this for the server ip:
If you mean getting the user's IP address, you can do something like :
<?php
if(!empty($_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'];
}
?>
<?php echo "<br />".$ip;?>
Related
I am using software called gns3 to build networks.
I have the need to find out devices external ip from within an gns3 internal network. This would be like a pc behind a router doing NAT.
Please could someone tell me if there is a way that you can echo out the clients external ip in php so I can add a web server to one of my virtual networks within gns3 so I can visit it on some devices and find their external ip’s for testing?
Thanks in advance
You need to make a request to an external website that could tell you the ip address you have. You may use checkip.dyndns.org - click the link to see your own ip addres.
if you want to do that from php here is an example code:
<?php
//get website content as a string
$ipCheckUrl = 'http://checkip.dyndns.org';
$subject = file_get_contents($ipCheckUrl);
//extract from string just ip address
$pattern = '/Current IP Address\:\s*(\d+\.\d+\.\d+\.\d+)/U';
$ip = preg_match($pattern,$subject,$result) ? $result[1] : 'Error';
//print ip address.
echo $ip;
i'm running an hidden service as the final exam of my school, but i've a problem: i have to show that the IP of the same computer which connects to my website can change in a bit , but when i try functions such as $ip=$_SERVER['REMOTE_ADDR']; it always gives me 127.0.0.1 instead of the last tor node which connects to me.
How can i solve this ? Is it normal on the Tor network? Cheers in advance!
There is a no direct way , but you can try using a third party server to get actual ip even from localhost server.
<?php
$content = file_get_contents('http://checkip.dyndns.com/');
preg_match('/Current IP Address: \[?([:.0-9a-fA-F]+)\]?/', $content, $ip);
echo $externalIp = $ip[1];
?>
Here is a working PHPFiddle . Hope this will help, cheers :)
I have am using websocket and trying to merge with my custom PHP app and Mysql database. In Database I store IP, ClientID and username when user connects to server. These infromation are to be used when user disconnect from websocket server or sends message. I run server.php with php server.php
The server page is https://github.com/Flynsarmy/PHPWebSocket-Chat/blob/master/server.php
As mentioned above I stored necessary information in Mysql database to identify which users sends message, disconnect and connect.
So for that I need to identify my own mechine IP for further identification with various clients. So when I try to use
<?php
echo $_SERVER["REMOTE_ADDR"];
in server.php. It gives an error saying undefined.
You don't.
WebSockets does not deal with web requests. The $_SERVER superglobal does not get populated because it does not make sense to populate it.
Deal with the socket connections directly.
You most likely want to play around with socket_getsockname().
Please try this,
var_dump($_SERVER);
and check if it prints... [REMOTE_ADDR] => .......
But if you are after clients IP,
as a PHP developer I use following code
$ip=NULL;
if (!empty($_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'];
I am not a php expert. I develop android apps. In my app i am getting the user's ip address from this url http://ip2country.sourceforge.net/ip2c.php?format=JSON. As you can see when some open this url it returns some info including the IP address. I only want to get the IP and (if possible) country. Most of the time this url is busy and doesn't returns ip and gives max active user connections error. Can you please give me any php file which i can put in my own webhost and call the url to get ip. The data returned should be in json so i can parse it easily.
Thanks
<?php
$json = file_get_contents("http://ip2country.sourceforge.net/ip2c.php?format=JSON");
//this $json will have the response that the website sends.
echo json_encode($json);
?>
You can have this object wherever you call this php file and do the needful
Run this php file to check the output
Another way: EDIT
<?php
$visitor_ip = $_SERVER['REMOTE_ADDR'];
echo $visitor_ip;
$data1 = file_get_contents("http://api.hostip.info/?ip=$visitor_ip");
echo "<br> $data1";
?>
You can use PHP to get the users IP address via $_SERVER['REMOTE_ADDR']. You can then use an IP to Location lookup website to translate that into a country and merge the results:
$ip = $_SERVER['REMOTE_ADDR'];
$loc = json_decode(file_get_contents('http://ipinfo.io/'.$ip.'/json'), true);
$country = isset($loc['country']) ? $loc['country'] : 'Unknown';
$result = array('ip'=>$ip, 'country'=>$country);
header('Content-Type: application/json');
echo json_encode($result);
You get the IP address in PHP you have to use $_SERVER['REMOTE_ADDR'] then use http://ipinfo.io to pass that IP to this website. You can get the Location through JSON data, just follow the first answer on this question Getting the location from an IP address.
I have successfully implemented it by following the answer.
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