Question

A while ago I needed a script to update some content every week, and my question was answered in this forum.

—

Now, and I’m not a jQuery pro like you :), but the “problem” I have is that today (the moment of this post) it’s week #52 but the script wasn’t working, it wasn’t showing the content for week #52, so I changed my HTML to week #53, and it worked.

The “problem” is that there’s no week #53, so I’m afraid I’m going to have to change my HTML to continue counting for week #54, #56, #57 and so on, when those weeks don’t exist.

Here’s an extract of the HTML (the structure repeats over the weeks, content changes of course):

<div class="quotes-container quote-53">
 <div class="quote">Quote here...</div>
 <div class="author">&mdash; Author </div>
</div>

Script:

<script type="text/javascript">
Date.prototype.getWeek = function() {
var onejan = new Date(this.getFullYear(),0,1);
return Math.ceil((((this - onejan) / 86400000) + onejan.getDay()+1)/7);
}

jQuery(function(){  
 var today = new Date();
 var weekno = today.getWeek();
 jQuery('#quotes-wrapper').load('/common/testimonials.html div.quote-'+weekno);
});
</script>

Any idea what’s going on?

Thanks a lot for any help on this.

Answer 1

EDIT: The script you’re using considers a “week” to be Sunday – Saturday. Since a year doesn’t always start on a Sunday, it considers the first partial week to be week 1.

If you simply want 7 say periods since the first of the year, use the script below. If you want it based on Sunday, then the script you’re using would be correct.

---

There is a week #53. It’s just not a full week.

But we are in week 52. I’m guessing the current script doesn’t account for leap years.

Instead you can calculate the day of this year, and divide that by 7.

Example: http://jsfiddle.net/etTS2/2/

<script type="text/javascript">

    Date.prototype.getWeek = function() {
      var onejan = new Date(this.getFullYear(),0,1);
      var today = new Date(this.getFullYear(),this.getMonth(),this.getDate());
      var dayOfYear = ((today - onejan +1)/86400000);
      return Math.ceil(dayOfYear/7)
    };


    jQuery(function(){  
     var today = new Date();
     var weekno = today.getWeek();
     jQuery('#quotes-wrapper').load('/common/testimonials.html div.quote-'+weekno);
    });

</script>

You’ll still get a result of 53 on the last day of the year (or last two days in a leap year).

EDIT: Fixed a offset of 1 on the dayOfYear.