How to select only the fields without Display: none CSS property using JQuery?

Question

Hi I am trying to select only the fields who's CSS property is not Display: "None". Please have a look at the code and please let me know what an I doing wrong?

function (controlId) { var returnValidFlag = true;

        //var requiredClass = $this.find('.required');
        //console.log(requiredClass);

        $('.required:not([display=none])').each(function () {
            // Get 'this' HTML object's ID
            var thisInputID = $(this).attr('id');

            // Get 'this' error message 
            var thisErrorMessageID = thisInputID + 'ErrorMessage';
            // console.log('Error ID: ' + thisErrorMessageID);

            // Get the user input value
            var thisInputValue = $(this).val();
            console.log('Input ID: ' + thisInputValue + " " + thisErrorMessageID);

            if (thisInputValue =='') {
                returnValidFlag = false;
                $('#' + thisErrorMessageID).show();
            } else {
                returnValidFlag = true;
                $('#' + thisErrorMessageID).hide();
            }
        });

        return returnValidFlag;
    }

Possible duplicate of Using jQuery, how do you find only visible elements and leave hidden elements alone? — You Old Fool
– You Old Fool, Commented Nov 26, 2018 at 0:28

charlietfl · Accepted Answer · 2018-11-26 00:01:22Z

1

You can use :visible selector.

display is not part of an element's attributes , it is a style property

$('.required:visible)').each(...

answered Nov 26, 2018 at 0:01

charlietfl

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user44 · Accepted Answer · 2018-11-26 00:13:07Z

0

In jQuery, you can use filter.

$('div').filter((index, node) => $(node).css('display') !== 'none')

answered Nov 26, 2018 at 0:13

user44

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