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Home/ Questions/Q 7585983
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T19:18:39+00:00 2026-05-30T19:18:39+00:00

I have made a script which slides thought some customers quotes. See code below

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I have made a script which slides thought some customers quotes. See code below

<script>

jQuery(document).ready(function () {
  jQuery(".da-feedback .da-quote").hide();
  jQuery(".da-feedback .da-quote:first-child").show();
  setTimeout('NextItem(1)',6000);
});

function NextItem($n) {

 jQuery('.da-feedback .da-quote:nth-child('+$n+')').slideUp('slow');
 $n = $n + 1;
 jQuery('.da-feedback .da-quote:nth-child('+$n+')').slideDown('slow');
 setTimeout('NextItem('+$n+')',6000);
}

</script>

<div class="da-feedback">
   <div class="da-quote" style="display: none; ">....</div>
   <div class="da-quote" style="display: none; ">....</div>
   <div class="da-quote" style="display: none; ">....</div>
</div>

I would like to know if jquerys gives you a function to make this easier.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T19:18:40+00:00Added an answer on May 30, 2026 at 7:18 pm

    Have you looked into .delay()?

    function NextItem(n) {
        $('.da-feedback .da-quote:nth-child('+$n+')').slideUp('slow')
            .delay(6000).slideDown('slow', NextItem(n++)); 
        // callback fires upon completion
    };
    NextItem(1);
    

    Incidentally, it doesn’t actually affect anything, but with JavaScript you don’t need to start your variables with a $, and with jQuery it’s considered a convention to do so if and only if the variable is a jQuery object.

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