Tell me more ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 4 down vote favorite
1

I'm trying to send a form to a PHP using $.ajax in jQuery. I send the whole thing as JSON, but, when I try to get the response, I get the 'parsererror'. What am I doing wrong?

jQuery fragment:

$("form").submit(function(){
    $.ajax({type: "POST",
        url: "inscrever_curso.php",
        data: {cpf : $("input#cpf").val(),nome : $("input#nome").val()},
        dataType: "json",
        success: function(data){
            alert("sucesso: "+data.msg);
        },
        error: function(XMLHttpRequest, textStatus, errorThrown){
            alert ("erro: "+textStatus);
        }
    });
    return false;
});

PHP:

<?php
    $return[msg]="Testing, testing.";
    echo json_encode($return);
?>
share|improve this question
Thanks for the fast reply - I just added the quotes around msg in PHP and still get the error. :( – Bruno Schweller Aug 9 '09 at 21:51
Oh :-( Are you sure it is "parse error" you get from the server ? Is your PHP script no longer than those two lines ? Because those look fine (no parse error, at least) ; you are using PHP >= 5.2, right ? – Pascal MARTIN Aug 9 '09 at 21:54
2  
Shouldn't the answers starting with "Shouldn't" be comments against the original question and not use up answer space? – Question Mark Aug 9 '09 at 21:54
What do you get when you go to inscrever_curso.php in your web browser? – Rob Aug 9 '09 at 21:55
yeah, I was using a larger script before, but since it wasn't working, I tried to make it simple first. IT doesn't work even with this simple script. – Bruno Schweller Aug 9 '09 at 21:55
show 5 more commentsadd comment (requires an account with 50 reputation)

3 Answers

up vote 6 down vote accepted

I don't know if it's the cause of your problem, but here is one thing about your PHP code:

$return[msg]="Testing, testing.";
echo json_encode($return);

Doing this, with E_NOTICE error_reporting level, you get:

Notice: Use of undefined constant msg - assumed 'msg'

If notices are enabled and displayed on your server, is will cause an output that is not valid JSON (the ouput contains the error message before the JSON string).


To avoid that notice, you must use this syntax:

$return['msg']="Testing, testing.";
echo json_encode($return);

Note the quotes around msg: the key of the array-element you are creating is a string constant, so, it is enclosed in quotes.

If you don't put in these quotes, PHP searches for a defined constant called "msg", which probably doesn't exist.

For more information about this, you can take a look at this section of the manual: Why is $foo[bar] wrong?.

(I am not sure it'll solve your problem, but it's good to know anyway.)

share|improve this answer
add comment (requires an account with 50 reputation)
up vote 1 down vote

Your PHP code should probably be:

<?php
    $return = array();
    $return['msg']="Testing, testing.";
    echo json_encode($return);
?>
share|improve this answer
add comment (requires an account with 50 reputation)
up vote 1 down vote

I just found out what was happening: I've got an .htaccess script for friendly URLs, and it was searching for the right PHP file, but in the wrong directory. I just added a '/' before the filename in the URL and then it worked.

share|improve this answer
Ergh :-D Nice job : we probably wouldn't have found this one ^^ Thanks for letting us know what caused the problem ! – Pascal MARTIN Aug 9 '09 at 22:16
add comment (requires an account with 50 reputation)

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.