CODEWITHSUNDEEP
phpjqueryjson
ngplayground
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Reputation: 21657

Struggling with a small json and php to get IP

PHP

<?php
header('Content-type: application/json');
$return['ip'] = $_SERVER['REMOTE_ADDR'];  
$results[] = array(
      'ip' => $return['ip']
   );
echo json_encode($results);
?>

jQuery

$.getJSON("http://domain.com/json/",
        function(data){
            console.log(data.ip);
        });
    });

But when I run the jQuery I've checked Fire bug and it says the following

GET http://domain.com/json/ 200 OK 81ms

And doesn't respond with the IP that I requested for. Have I missed something?

UPDATED CODE

PHP

<?php
header('Content-type: application/json');
$return['ip'] = $_SERVER['REMOTE_ADDR'];  
$results = array(
      'ip' => $return['ip']
   );
echo json_encode($results);
?>

jQuery

$.getJSON("http://domain.com/json/", function(data){
            console.log(data.ip);
        });

Firebug Error

SyntaxError: invalid label {"ip":"XXX.XXX.XXX.X"}

An arrow points at the first quotation mark just before the word ip.

Upvotes: 1

Views: 330

Answers (1)

Dan
Dan

Reputation: 548

You are returning:

[{'ip': 'XXX.XXX.XXX.XXX'}]

But you are treating it as if you are returning:

{'ip': 'XXX.XXX.XXX.XXX'}

You either need to change your JavaScript to console.log(data[0].ip) or change your PHP to: $results = array( ... ); rather than $results[] = array( ... );

Either will fix your problem. :)

Upvotes: 5

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