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JavaScript

Angelic Sanoy
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Angelic Sanoy
Full Stack JavaScript Techdegree Student 6,583 Points
Posted by Angelic Sanoy
.a{fill-rule:evenodd;}techdegree
Angelic Sanoy
Full Stack JavaScript Techdegree Student 6,583 Points

Jquery

Any idea why it's not working. I want to a result of 1.5 if they type 2.

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $('.face-button').click(function() {

    var PostCount = parseFloat($('.post_count:input').val());
    var PostValue;

    if (PostCount == 1) {
                PostValue = 1;
                } else if ( PostCount == 2) {
                 PostValue = 1.5;

                } else if ( PostCount == 3) {
                 PostValue = 2;

                }else if ( PostCount == 4) {
                 PostValue = 2.5;

                }else if ( PostCount == 5) {
                 PostValue = 3;

    } else {    

    $('.result').text('Total Cost: $' + PostValue );


      }

}); });

</script>

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

The selector '.post_count:input' includes a pseudo-class which is not one of the standard ones according to MDN. With no HTML shown I can only guess at a fix. If it's an input element that is inside a container with the "post_count" class, perhaps you want ".post_count input". Or if it's an input with that class, "input.post_count". Otherwise, I'd need to see the HTML.

Then, because of the final "else", the ".result" will only be updated when the input is not recognized and the "PostValue" is still undefined. If you want the update to occur on every click, that code should not be in an "else".