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Home/ Questions/Q 6631659
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T22:34:39+00:00 2026-05-25T22:34:39+00:00

I have two functions: $(function() { $(.deactivated).click(function() { var Container = $(this).parent(); var id

  • 0

I have two functions:

$(function() {
    $(".deactivated").click(function() {           
          var Container = $(this).parent();             
       var id = $(this).attr("id");             
        var string = 'id='+ id ;    
        $.ajax({   
            url: "<?php echo site_url('social/activate') ?>",
            type: "POST",
            data: string,
            cache: false,
                 success: function(){
                     Container.fadeOut(1000, function(){
                     $(this).load("<?php echo site_url('social/social_icon') ?>", {id: id}, function(){
                         $(this).hide().fadeIn(700);
                         $(this).click();
                     })
                 });
            }   
        });
        return false;
    });
});

 $(function() {
    $(".activated").click(function() {           
          var Container = $(this).parent();             
       var id = $(this).attr("id");             
        var string = 'id='+ id ;    
        $.ajax({   
            url: "<?php echo site_url('social/deactivate') ?>",
            type: "POST",
            data: string,
            cache: false,
                 success: function(){
                     Container.fadeOut(1000, function(){
                     $(this).load("<?php echo site_url('social/social_icon') ?>", {id: id}, function(){
                         $(this).hide().fadeIn(700);
                     })
                 });
            }   
        });
        return false;
    });
});

One is to activate link and other is to deactivate it. Functions are working fine, but when link is activated or deactivated, it can’t be clicked again to change it (page needs to be refreshed in order for functions to work again). What I need to do to make it work?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T22:34:40+00:00Added an answer on May 25, 2026 at 10:34 pm

    Change:

    $(".activated").click(function() {
    $(".deactivated").click(function() { 
    

    To:

    $(".activated").live('click',function() {  
    $(".deactivated").live('click',function() { 
    
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