Parsing JSON generated by PHP using jquery returns 'undefined'

Question

I am creating a simple module for our senior project that checks whether an email already exists in the database. This is the important segment of my php file which displays the formated JSON:

die(json_encode(array(2 => "Email already taken!")));

For that example, that should return = {"2","Email already taken!"}

And here is the javascript code (that's embedded):

$(function() {
   $("#signin").submit(function() {
     // validate and process form here
     $.ajax({
        type: "GET",
        url: "mailcheck.php",
        data: "email="+document.getElementById('email').value,
        success: function(data) {
        var parsed = $.parseJSON(data);
        //display message back to user here
          document.getElementById('msg').innerHTML = parsed.Msg;
          $("input#email").val("");
        },  
     });
   return false;
   });
});

The problem is, I always get 'undefined'. What's the problem?

Wouldn't die() also echo out the contents? codepad.viper-7.com/w3uR9C — asprin
– asprin, Commented Jun 24, 2014 at 4:03
the die() function echoes out its contents then it will exit, thus leaving the succeeding echo() or die() statements not displayed. — mharz14
– mharz14, Commented Jun 24, 2014 at 6:01
if I could not use die(), could I use exit; instead? I tried exit and it also worked... — mharz14
– mharz14, Commented Jun 24, 2014 at 6:10

PeterKA · Accepted Answer · 2014-06-24 03:43:40Z

1

In your PHP file echo json_encode as @pc-shooter has mentioned. And you do not need to parse JSON if you set dataType: 'json', in you ajax call.

You may also need to set to content type in you PHP file.

header('Content-Type: application/json');
echo json_encode(array(2 => "Email already taken!"));

answered Jun 24, 2014 at 3:43

PeterKA

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