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Home/ Questions/Q 6790629
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Editorial Team
  • 0
Editorial Team
Asked: May 26, 20262026-05-26T17:44:51+00:00 2026-05-26T17:44:51+00:00

i have this code, that works without any problem: <script> $(document).ready(function () { $(.block1).click(function

  • 0

i have this code, that works without any problem:

<script>
    $(document).ready(function () {
            $(".block1").click(function () {
                $("#fade1").fadeIn("slow").fadeOut("slow");
            });
    });

    $(document).ready(function () {
            $(".block2").click(function () {
                $("#fade2").fadeIn("slow").fadeOut("slow");
            });
    });
</script>

However, as i have six blocks i am trying change it to a loop:

$(document).ready(function () {
     for (i=1; i<7; i++) {
         alert(i);
         $(".block"+i).click(function () {
             $("#fade"+i).fadeIn("slow").fadeOut("slow");
             alert (i);
         });
     }
});

This for loop isn’t working as expected. It’s giving all 6 blocks an alert, as expected, but instead of each alert reading “block1”, “block2”, “block3”, etc., they all say “block7”.

Does anyone know why this is happening?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T17:44:52+00:00Added an answer on May 26, 2026 at 5:44 pm

    If you add a shared class to your divs, you could do it without a loop.

    Update:

    Its not 100% clear what you are trying to do, but here’s a method that could work using jquery data.

    <div class="animate-block" data-fadeid="fade1"></div>
    <div class="animate-block" data-fadeid="fade2"></div>
    <div class="animate-block" data-fadeid="fade3"></div>
    <div class="animate-block" data-fadeid="fade4"></div>
    
    <div id="fade1"></div>
    <div id="fade2"></div>
    <div id="fade3"></div>
    <div id="fade4"></div>
    
    <script>
    $(document).ready(function () {
            $(".animate-block").click(function () {
                var fadeId = $(this).data("fadeid");
                $("#"+fadeId).fadeIn("slow").fadeOut("slow");
            });
    });
    </script>
    

    If your block divs and fade divs are next to each other, it becomes even simpler since you can use the next() method.

    <div class="animate-block"></div>
    <div id="fade1"></div>
    <div class="animate-block"></div>
    <div id="fade2"></div>
    <div class="animate-block"></div>
    <div id="fade3"></div>
    <div class="animate-block"></div>
    <div id="fade4"></div>
    
    <script>
    $(document).ready(function () {
            $(".animate-block").click(function () {
                $(this).next().fadeIn("slow").fadeOut("slow");
            });
    });
    </script>
    
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