TypeError in JavaScript function

Question

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?

Did you declare z as a variable earlier in the code, forgetting to declare var can give wierd behaviour? — Adrian Forsius
– Adrian Forsius, Commented Sep 16, 2014 at 12:20
@Paradoxis load_money is another function. I think it doesn't matter, becouse actually I have error before post. When update_money executed money_am has wrong type, and thats actually problem. — StopKran
– StopKran, Commented Sep 16, 2014 at 12:26
@Adrian-Forsius I didn't declare z as a variable earlier. Actually I try it now, but it didn't work too, and gives me same error. — StopKran
– StopKran, Commented Sep 16, 2014 at 12:30

Regent · Accepted Answer · 2014-09-16 12:25:11Z

2

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);

answered Sep 16, 2014 at 12:25

Regent

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Comments

Adrian Forsius · Accepted Answer · 2014-09-16 12:25:03Z

0

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.

answered Sep 16, 2014 at 12:25

Adrian Forsius

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