What is causing the nth-child selector to fail when attempting to style every other visible element?

Currently, I am experimenting with enhancing the style of Alternate Visible Elements. My initial idea was to utilize the nth-child(2n+1) concept, but unfortunately it did not yield the desired results. To simplify my issue, here is a sample scenario:

HTML:

<div class='hide find'>TEST</div>
<div class='hide find'>TEST</div> 
<div class='find'>TEST</div> 
<div class='hide find'>TEST</div> 
<div class='find'>TEST</div> 
<div class='hide find'>TEST</div> 
<div class='hide find'>TEST</div> 
<div class='find'>TEST</div> 
<div class='hide find'>TEST</div> 
<div class='find'>TEST</div>

CSS:

.hide{
    display:none;
}
.alternate{
    background-color:grey;
}

Jquery:

$('.find:visible:nth-child(2n+1)').addClass('alternate');    //This method is not functioning as expected. Why?

I am unsure about the reason for this failure. However, I managed to devise a workaround solution using a function that achieves the desired outcome. Despite this workaround, I believe it would be more elegant and efficient if the issue can be resolved using the original approach outlined above. Below is the workaround I implemented:

function addAlternateStyle(){
var alt= true;
    $('.find:visible').each(function(){
        if(alt){
            alt=!alt;
            $(this).addClass('alternate');
        }
        else{
            alt=!alt;
        }
    });
}

Any assistance on this matter would be greatly appreciated.

Answer №2

To target alternate elements in your selection with jQuery, you can utilize the following formulas:

$('div[class!="hide find"]:even')  

$('div[class!="hide find"]:odd')

The odd/even pseudo-class was applied here along with a "not" selector to verify visibility (class != "hide find")

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