Issue on Adding Values to a Variable in jQuery

Question

Can you please take a look at This Demo and let me know why I am not able to add values to the var result variable in my code? I think I declared the result in a global scope so all methods will have access to it but it is not functioning here for me!

Here is the code I have:

<ul>
    <li><input type="checkbox" id="25" />25</li>
    <li><input type="checkbox" id="50" />50</li>
    <li><input type="checkbox" id="75" />75</li>
</ul>
<div id="result"></div>

and jQuery code:

$(document).ready(function () {
    var result = 0;
    $("#25").on("click", function () {
        var newResult = result + 25;
        $("#result").html(newResult);
    });
    $("#50").on("click", function () {
        var newResult = result + 50;
        $("#result").html(newResult);
    });
    $("#75").on("click", function () {
        var newResult = result + 75;
        $("#result").html(newResult);
    });
});

Thanks.

Arun P Johny · Accepted Answer · 2014-03-05 00:07:19Z

5

First jQuery is not added in the fiddle
You are not updating the value of result when an item is selected

Try

$(document).ready(function () {
    var result = 0;
    $("#25, #50, #75").on("change", function () {
        if (this.checked) {
            result += +this.id;
        } else {
            result -= +this.id;
        }
        $("#result").html(result);
    });
});

Demo: Fiddle

answered Mar 5, 2014 at 0:07

Arun P Johny

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4 Comments

Behseini Over a year ago

Thanks Arun , but can you please let me know what does "+this" mean in your code? I mean adding the plus sign before "this"?

Arun P Johny Over a year ago

@Behseini it is the unary plus operator, I used it to convert the string id to an int value(used it instead of parseInt(this.id))

Behseini Over a year ago

Thanks but the solution is not working on this Example jsfiddle.net/Behseini/q9V35/4 ! can you please take alook at that and let me know what I am doing wrong there?

Arun P Johny Over a year ago

@Behseini look at your console there are no variables named newval - jsfiddle.net/arunpjohny/q9V35/5

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