Php tracking own location gives a randomized IP - php

Question

-- Please scroll down to where I marked the PHP --
To explain in better detail.
I made a Leaflet map and in that map I want to load my own location.
Here's my code for that in Javascript, but this is out of question like #Pocketsand and I already discussed. So then scroll down to the PHP code and see if you can get the IP address through the browser.
$part_content = "<div id=\"mapid\"></div>";
//.setView([".$longitude.", ".$latitude."], ".$zoom_factor.");
$part_content .= "<script>
var map = L.map('mapid').fitWorld();
L.tileLayer('https://api.tiles.mapbox.com/v4/{id}/{z}/{x}/{y}.png?access_token=pk.eyJ1IjoibWFwYm94IiwiYSI6ImNpejY4NXVycTA2emYycXBndHRqcmZ3N3gifQ.rJcFIG214AriISLbB6B5aw', {
maxZoom: 18,
attribution: 'Map data © OpenStreetMap contributors, ' +
'CC-BY-SA, ' +
'Imagery © Mapbox',
id: 'mapbox.streets'
}).addTo(map);
function onLocationFound(e) {
var radius = e.accuracy / 2;
L.marker(e.latlng).addTo(map)
.bindPopup(\"You are within \" + radius + \" meters from this point\").openPopup();
L.circle(e.latlng, radius).addTo(map);
}
function onLocationError(e) {
alert(e.message);
}
map.on('locationfound', onLocationFound);
map.on('locationerror', onLocationError);
map.locate({setView: true, maxZoom: 16});
</script>";
It's the same code as in this maps source code.
When I go to Firefox, it partly works and on someone else's computer it works fine, when on my computer I get the error from the image I showed you.
So I can't locate on my computer to my own location, as in google maps it works perfectly and my extensions also don't seem to block that part.
PHP:
Basically this:
$ip = !empty($_SERVER['HTTP_X_FORWARDED_FOR']) ? $_SERVER['HTTP_X_FORWARDED_FOR'] : $_SERVER['REMOTE_ADDR'];
Gives me a random IP, because of HTTP_X_FORWARDED_FOR and REMOTE_ADDR gives me the correct IP, but not when I use this from a different IP address then the local one... thats why I check if the proxy is not empty.
This is the full php code for the tracker:
$ip = !empty($_SERVER['HTTP_X_FORWARDED_FOR']) ? $_SERVER['HTTP_X_FORWARDED_FOR'] : $_SERVER['REMOTE_ADDR'];
$url = "http://freegeoip.net/json/$ip";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 5);
$data = curl_exec($ch);
curl_close($ch);
if ($data) {
$location = json_decode($data);
$longitude = $location->longitude;
$latitude = $location->latitude;
$longitude = str_replace(",", ".", $longitude);
$latitude = str_replace(",", ".", $latitude);
}
My current problem is:
$ip = $_SERVER['HTTP_CLIENT_IP'] ? $_SERVER['HTTP_CLIENT_IP'] : ($_SERVER['HTTP_X_FORWARDE‌​D_FOR'] ? $_SERVER['HTTP_X_FORWARDED_FOR'] : $_SERVER['REMOTE_ADDR']);
Which gets the correct IP address, but it almost seems like it gets the IP hosts location. Which is not my intention, I intend to get the clients location through the IP address, which currently isn't working.
I tried and tried, but couldn't seem to figure this out and hope you guys know more about this.
Thanks in advance!

Answer 1

I am guessing you didn't visit the URL in the error message? It's a Chrome security thing, try running off a local webserver or getting a https certificate on your remote server.
The Chrome Security team and I propose that, for new and particularly
powerful web platform features, browser vendors tend to prefer to make
the the feature available only to secure origins by default.
[...]
Definitions:
“Secure origins” are origins that match at least one of the following
(scheme, host, port) patterns:
(https, *, *)
(wss, *, *)
(*, localhost, ) (, 127/8, *)
(*, ::1/128, *)
(file, *, —)
(chrome-extension, *, —)
This list may be incomplete, and may need to be changed.
Source: https://www.chromium.org/Home/chromium-security/prefer-secure-origins-for-powerful-new-features

Answer 2

To get a client public ip address, in PHP 5.3 or greater use:
<?php
$ip = getenv('HTTP_CLIENT_IP')?:
getenv('HTTP_X_FORWARDED_FOR')?:
getenv('HTTP_X_FORWARDED')?:
getenv('HTTP_FORWARDED_FOR')?:
getenv('HTTP_FORWARDED')?:
getenv('REMOTE_ADDR');
?>
And if that is not working, use an external provider like https://geolocation-db.com
A JSON-P callback example:
<?php
$jsonp = file_get_contents('https://geolocation-db.com/jsonp');
$data = jsonp_decode($jsonp);
print $data->IPv4 . '<br>';
print $data->country_code . '<br>';
print $data->country_name . '<br>';
print $data->state . '<br>';
print $data->city . '<br>';
print $data->postal . '<br>';
print $data->latitude . '<br>';
print $data->longitude . '<br>';
// Strip callback function name and parenthesis
function jsonp_decode($jsonp) {
if($jsonp[0] !== '[' && $jsonp[0] !== '{') {
$jsonp = substr($jsonp, strpos($jsonp, '('));
}
return json_decode(trim($jsonp,'();'));
}
?>
And a JSON example:
<?php
$json = file_get_contents('https://geolocation-db.com/json');
$data = json_decode($json);
print $data->country_code . '<br>';
print $data->country_name . '<br>';
print $data->state . '<br>';
print $data->city . '<br>';
print $data->postal . '<br>';
print $data->latitude . '<br>';
print $data->longitude . '<br>';
print $data->IPv4 . '<br>';
?>

Answer 3

Instead of going with a PHP function I did a leaflet function instead, which tracks the user through the browser.
Using the library: leaflet.locate
Where then I could use the:
// create control and add to map
var lc = L.control.locate().addTo(map);
// request location update and set location
lc.start();
But this isn't the answer to the PHP part of my code, even though this leaflet function also works for now.
Sadly the PHP part can't find the exact address through the IP if the server is at another place, but the leaflet way of doing it through the browser does work. Hope you guys find use in these answers.