I've another problem. I get an error in FireFox and I don't know what my fault is. I always did it like this way and I never got an error. I already check lower/uppercase mistakes but I can't find anything.

Thanks

$.ajax({type: "POST", url: "ajax/check_username.php", data: {username: username}}).done is not a function

<script type="text/javascript">
$(document).ready(function(){
    $("#username").keyup(function(){
        var username = $("#username").val();
        $(".usernameFeedback").fadeIn("fast");

        $.ajax({
            type: "POST",
            url: "ajax/check_username.php",
            data: { username: username }
        }).done(function( msg ) {
            $("#loadingImage").hide();
            if(msg.status != "error")
                {
                    if(msg.available == "yes")
                    {
                        $(".usernameFeedback span").text(msg.message);
                        $(".usernameFeedback span").removeClass("notok");
                        $(".usernameFeedback span").addClass("ok");
                    }
                    else
                    {
                        $(".usernameFeedback span").text(msg.message);
                        $(".usernameFeedback span").addClass("notok");
                    }
                }
        });
        return(false);
    })
});
</script>

I've another problem. I get an error in FireFox and I don't know what my fault is. I always did it like this way and I never got an error. I already check lower/uppercase mistakes but I can't find anything.

Thanks

$.ajax({type: "POST", url: "ajax/check_username.php", data: {username: username}}).done is not a function

<script type="text/javascript">
$(document).ready(function(){
    $("#username").keyup(function(){
        var username = $("#username").val();
        $(".usernameFeedback").fadeIn("fast");

        $.ajax({
            type: "POST",
            url: "ajax/check_username.php",
            data: { username: username }
        }).done(function( msg ) {
            $("#loadingImage").hide();
            if(msg.status != "error")
                {
                    if(msg.available == "yes")
                    {
                        $(".usernameFeedback span").text(msg.message);
                        $(".usernameFeedback span").removeClass("notok");
                        $(".usernameFeedback span").addClass("ok");
                    }
                    else
                    {
                        $(".usernameFeedback span").text(msg.message);
                        $(".usernameFeedback span").addClass("notok");
                    }
                }
        });
        return(false);
    })
});
</script>

• javascript
• jquery

• What does the ajax function return if not a Deferred? Try to log that to the console. – Bergi Commented May 2, 2012 at 15:32

1 Answer 1

Probably your jQuery version is too old. You need at least jQuery 1.5 for jqXHR objects to implement the Promise interface you are using.

If you cannot upgrade for some reason, simply use the success option:

$.ajax({
    type: "POST",
    url: "ajax/check_username.php",
    data: { username: username },
    success: function(msg) {

    }
});