CODEWITHSUNDEEP
javascriptjquerytypeerror
StopKran
StopKran

Reputation: 395

TypeError in JavaScript function

I have some JavaScript code:

var update_money = function(money_am){
    var d = {'ammount': money_am};
    $.post("/adr/", d, load_money)
}

z = parseFloat($('#money').val());
$('#money').change(z, update_money);

When it executes, it gives me a runtime error:

TypeError: 'stopPropagation' called on an object that does not implement interface Event.

In debug, I found that money_am is not a float. It is an object. But if I change my code like this:

var update_money = function(money_am){
    var d = {'ammount': parseFloat($('#money').val())};
    $.post("/adr/", d, load_money)
}

It works great. What should I do to fix this problem?

Upvotes: 3

Views: 883

Answers (2)

Regent
Regent

Reputation: 5178

Data, that was passed to event handler, can be accessed by event.data:

Fiddle.

var update_money = function(event)
{
    var money_am = event.data;
    alert(money_am);
    var d = {'ammount': money_am};
    $.post("/adr/", d, load_money);
}

function load_money() { }

z = parseFloat($('#money').val());
$('#money').change(z, update_money);

Upvotes: 2

Adrian Forsius
Adrian Forsius

Reputation: 1438

Make sure that the money-value is parseable by parseFloat:

You also might want to change your code to this:

$('#money').change(function() {
    var z = parseFloat($('#money').val());
    load_money(z);
});

This will allow you to get the new money value everytime #money-field is changed. Instead of always using the same value. Along with properly defining z.

Upvotes: 0

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