Eliminate flash of unstyled content

2019-01-05 10:02发布

站内问答 / HTML/CSS
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How do I stop the flash of unstyled content (FOUC) on a web page?

标签： css fouc
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7条回答
▲ chillily
2楼-- · 2019-01-05 10:56

The problem with using a css style to initially hide some page elements, and then using javascript to change the style back to visible after page load, is that people who don't have javascript enabled will never get to see those elements. So it's a solution which does not degrade gracefully.

A better way therefore, is to use javascript to both initially hide as well as redisplay those elements after page load. Using jQuery, we might be tempted to do something like this: 

$(document).ready(function() {
    $('body').hide();
    $(window).on('load', function() {
        $('body').show();
    });
});

However, if your page is very big with a lot of elements, then this code won't be applied soon enough (the document body won't be ready soon enough) and you might still see a FOUC. However, there is one element that we CAN hide as soon as script is encountered in the head, even before the document is ready: the HTML tag. So we could do something like this:

<html>
  <head>
  <!-- Other stuff like title and meta tags go here -->
  <style type="text/css">
    .hidden {display:none;}
  </style>
  <script type="text/javascript" src="/scripts/jquery.js"></script>
  <script type="text/javascript">
    $('html').addClass('hidden');
    $(document).ready(function() {    // EDIT: From Adam Zerner's comment below: Rather use load: $(window).on('load', function () {...});
      $('html').show();  // EDIT: Can also use $('html').removeClass('hidden'); 
     });  
   </script>
   </head>
   <body>
   <!-- Body Content -->
   </body>
</html>

Note that the jQuery addClass() method is called *outside* of the .ready() (or better, .on('load')) method.

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