JQuery function not found

Question

following code:

// also tried function getDeletedDates()
var getDeletedDates = function()
{
    var s = new Array();

    $(".deleted").each(function(i, e) {
        s.push($(e).attr("data-day"));
    });
};

    $(function()
    {
        $("#delete_send").click(function() {
            alert("drin");
            $.ajax({
                  url: "delete.php",
                  type: "POST",
                  data: ({deleteDates : getDeletedDates()}),
                  dataType: "json",
                  success: function(msg){
                     alert(msg);
                  },
                  beforeSend: function(){
                      alert("Lösche folgende Urlaubstage: "+ getDeletedDates().join(", "));
                  },
                  error: function(x, s, e) {
                      alert("Fehler: " + s);
                  }
               }
            );
        });
    });

But i come into beforeSend() he always says "getDeletedDates() undefined" Why is this, i declared the function in global scope?

thanks in advance.

Matthew Abbott · Accepted Answer · 2010-07-21 08:01:45Z

5

Your function doesn't return anything, so the result will be undefined. Change the method to return the array.

UPDATE:

When you do getDeletedDates() it is evaluated to undefined, because of the lack of return result. This is why getDeletedDates() is undefined is the error message.

edited Jul 21, 2010 at 8:01

answered Jul 21, 2010 at 6:57

Matthew Abbott

61.7k9 gold badges108 silver badges131 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Weidling C Over a year ago

ohh thank you this helped, but I'm confused why it says "getDeletedDates() is undefined" but thank you very much!

Prutswonder · Accepted Answer · 2010-07-21 07:01:27Z

2

You are calling your function, but the function is defined as a variable/pointer and doesn't return anything. The following modification should work (not tested):

function getDeletedDates()
{
    var s = new Array();

    $(".deleted").each(function(i, e) {
        s.push($(e).attr("data-day"));
    });

    return s;
};

answered Jul 21, 2010 at 7:01

Prutswonder

10.1k3 gold badges29 silver badges41 bronze badges

Collectives™ on Stack Overflow

JQuery function not found

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related