Exploring jQuery: Easy method to toggle multiple elements individually (and related inquiries)

As a newcomer to jQuery, I am facing a specific scenario:
I have set up the following example: http://jsfiddle.net/dVf8m/
My concern is regarding streamlining the slideToggle functionality. Currently, I am using two ids on menu elements (#trigger1 and #trigger2) along with two ids on hidden divs (#one and #two). This results in redundant jQuery code. Is there a more efficient way to eliminate all the ids for a simpler implementation?
Furthermore, when clicking on both menu elements (First and Second), both divs are displayed simultaneously. Is there a method to ensure only one hidden div is visible at a time? How can I make the first div disappear when the second one appears?
Lastly, if I opt for fadeIn/fadeOut effects instead of slideToggle, how can I apply this when both involve the .click event?

javascriptjqueryhtmlcss

Answer №1

Revise your code to resemble the following structure. Create a class for the div, attach a click listener to it, and assign any attribute to the div while providing the id of the div to be toggled.

<div id="top">
    <ul>
        <li><span id="trigger1" class="toggler" data-item="item1">First</span></li>
        <li><span id="trigger2" class="toggler" data-item="item2">Second</span></li>
        <li>Third</li>
    </ul>
</div>
<div class="hidden" id="item1">
    <ul>
        <li><a href="#">Something</a></li>
        <li><a href="#">Something Else</a></li>
        <li><a href="#">Another Thing</a></li>
    </ul>
</div>
<div class="hidden" id="item2">
    <ul>
        <li><a href="#">Apple</a></li>
        <li><a href="#">Orange</a></li>
        <li><a href="#">...</a></li>
    </ul>
</div>

$(document).ready(function() {
    $('.toggler').click(function(e) {            
        $("#"+$(this).attr("data-item")).slideToggle(500); 
    });
});

JSFIDDLE

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