Pass JavaScript variable to php script

Question

I have 2 files "index.php" and "userip.php". I want to pass the variable varip to the file "userip.php" with ajax. If this is successful I want to share POST['name']; in a session. I thought that the session would be set but when I reload the index.php page the echo shows nothing. Can someone help me out?

index.php (jQuery section):

    <script type="text/javascript">   
    $.getJSON("http://ip.jsontest.com/", function(data) {
        var varip = "";
        $.each(data, function(k, v) {
            varip += v;
            $.ajax({
                type: "POST",
                url: "userip.php",
                data: "name="+varip,
                success: function(data){
                    alert("ok");
                }
            });
        });            
    });  
    </script>

index.php (php section):

<?php
echo $_SESSION['userip'];
?>

userip.php:

session_start();
if(!empty($_POST['name'])){ 
    $variable = $_POST['name'];
    $_SESSION['userip'] = $variable;
}

Add session_start() to index.php

Greg
– Greg

2014-04-09 23:00:53 +00:00
Commented Apr 9, 2014 at 23:00 — Greg
– Greg, Commented Apr 9, 2014 at 23:00
possible duplicate of javascript variable to php script

Guillermo Gutiérrez
– Guillermo Gutiérrez

2014-06-18 21:06:37 +00:00
Commented Jun 18, 2014 at 21:06 — Guillermo Gutiérrez
– Guillermo Gutiérrez, Commented Jun 18, 2014 at 21:06

Niet the Dark Absol · Accepted Answer · 2014-04-09 23:07:49Z

1

The problem is that you are missing session_start() in your index.php file, so at that point $_SESSION hasn't been loaded.

But it looks like you're getting the user's IP address?

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

answered Apr 9, 2014 at 23:07

Niet the Dark Absol

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1 Comment

Rene Over a year ago

omg thats it-,- Thanks man. Im so stupid sometimes. $_SERVER['REMOTE_ADDR']; isnt working because its a proxy and all clients have the same IP. I tried this: stackoverflow.com/a/14985633/1516246 but it wasnt working. Every $_SERVER opinion output was the proxy IP

Joseph118 · Accepted Answer · 2014-04-09 23:01:25Z

-1

change

data: "name="+varip,

to

data: { name: varip  },

answered Apr 9, 2014 at 23:01

Joseph118

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