CODEWITHSUNDEEP
phpformsinputipgeocode
user990175
user990175

Reputation: 311

IP Geocode using PHP echo

I'm trying to display the users country (based on their IP) in a form input value.

Here's my code so far ..

<input style="display:block" type="text" name="country" id="country" value="<?php               
$pageContent = file_get_contents('http://freegeoip.net/json/echo $_SERVER['REMOTE_ADDR']');
$parsedJson  = json_decode($pageContent);
echo $parsedJson->country_name; ?>" />

I'm using PHP JSON-decode to get the data from "http://freegeoip.net/json/(IP ADDRESS)".

That website geocodes the IP address and returns a country name.

What I want to do is to be able to substitute in a users IP address into that web address, which would then return the country name of the user. The best way I could think of was by using 

<?php echo $_SERVER['REMOTE_ADDR']; ?>

but when I put it in I get a Server Error.

How would I go about doing this?

Upvotes: 2

Views: 6426

Answers (4)

Samuel Dauzon
Samuel Dauzon

Reputation: 11334

Since Freegeoip will became a paid service (https://ipstack.com/), if you want a free simple service you can use ip-api.com instead of freegeoip.net : 

function getDataOfIp($ip) {
    try {
        $pageContent = file_get_contents('http://ip-api.com/json/' . $ip);
        $parsedJson  = json_decode($pageContent);
        return [
            "country_name" => $parsedJson->country,
            "country_code" => $parsedJson->countryCode,
            "time_zone" => $parsedJson->timezone
        ];
    } 
    catch (Exception $e) {
        return null;
    }
}

You should adapt return value by the data you need.

Upvotes: 0

daan.desmedt
daan.desmedt

Reputation: 3820

I've been needing the GeoIPservices myself in some project lately. I've created a ** PHP wrapper**, in case others are looking for a PHP helper for this.

Can be found at https://github.com/DaanDeSmedt/FreeGeoIp-PHP-Wrapper

Upvotes: 1

Jeroen
Jeroen

Reputation: 13257

$pageContent = file_get_contents('http://freegeoip.net/json/' . $_SERVER['REMOTE_ADDR']);

You can't use echo within a function. Instead, you should use the concatenating operator.

Also, better code would be something like this:

<?php               
$pageContent = file_get_contents('http://freegeoip.net/json/' . $_SERVER['REMOTE_ADDR']);
$parsedJson  = json_decode($pageContent);
?>
<input style="display:block" type="text" name="country" id="country" value="<?php echo htmlspecialchars($parsedJson->country_name); ?>" />

Upvotes: 11

Brad
Brad

Reputation: 163488

$pageContent = file_get_contents('http://freegeoip.net/json/' . $_SERVER['REMOTE_ADDR']);

No need to echo into a string parameter... just concatenate with ..

Upvotes: 2

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