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Answer 1

+4 A:

There are two problems. Halfway down the PHP you have <php not <?php. But really you should be using json_encode. Just make a PHP array with your data, json_encode it, and it'll do all the clever JSON stuff for you. It'll be much easier to maintain that way.

Another problem: JSON keys should be strings.

Try something like this:

$result = array();
if ($calcular <= 9) {
    $result['mensagem'] = 1;
}
//... etc.

echo json_encode($result);

Edit: I've just done a check, and the following works for me:

$(document).ready(function () {
    $.ajax({
        type: "POST",
        url: 'validar.php',
        dataType: 'json',
        success: function(msg) {
            alert('Mensagem '+ msg.mensagem);
        },
        error : function () {
            alert('Ocorreu um erro');
        }
    });
});

Skilldrick 2010-06-15 09:57:18

Hey, thank you for your answer, i remade the code, but firebug is giving the same problem:msg is null[Break on this error] alert('Mensagem ' + msg.mensagem );

Davidslv 2010-06-15 10:10:15

Oh man... thank you very much!!!!

Davidslv 2010-06-15 10:32:37

Happy to help :)

Skilldrick 2010-06-15 10:36:58

Answer 2

+1 A:

I'm guessing it's because your JSON syntax isn't correct (look it up and see what quotes you missed :). You could use json_encode in PHP to save you from building it manually.

Also note that you should be serving an appropriate content type from the PHP responder (application/json). I don't know if that affects jQuery's built in parsing, but it feels cleaner.

Alex Ciminian 2010-06-15 09:57:44

Thank you, i will check that... i had a similar code in Perl and didn't had any problem... i really don't know what is happening here..Thanks

Davidslv 2010-06-15 10:11:28

Answer 3

A:

JSON needs to be in a strict format, which your PHP is not giving. Use json_encode to make sure that your output is valid.

Something like:

<?php
    if(!empty($_POST)) { // ensure _POST exists, and is not empty
        foreach($_POST as $key => $val) {
            // if the string 'pergunta' is contained within that _POST key, add to a total
            if(strpos($key, "pergunta") !== false) { 
                $calcular += $val;
            }
        }

        // return '1' for 9 and under, '2' for 10 or 11, '3' for 12 and over
        $mensagem = ($calcular <= 9) ? 1 : ($calcular > 9 && $calcular <= 11) ? 2 : 3;

        // use json_encode to ensure that what we echo back is valid JSON, eg. {"mensagem":1}
        echo json_encode(array('mensagem' => $mensagem));
    }
?>

Adam Hepton 2010-06-15 10:09:21

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Problem with Json in Jquery

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