jQuery：一年中的一周剧本表演 [英] jQuery: week of year script acting up

问题描述

前段时间我需要一个脚本来每周更新一些内容，我的问题在这个论坛。

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

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现在，我不是像你这样的jQuery专业人员:)，但我所遇到的问题是今天（这篇文章的那一刻）它的第52周，但脚本没有工作，它没有显示第52周的内容，所以我将我的HTML更改为第53周，并且它有效。

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.

问题是没有第53周，所以我'我担心我将不得不改变我的HTML以继续计算第54周，第56周，＃57等等，当那些星期不存在时。

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.

以下是HTML的摘录（结构在几周内重复，当然内容更改）：

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>

脚本：

<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>

知道发生了什么事吗？

非常感谢您对此的任何帮助。

Thanks a lot for any help on this.

推荐答案

编辑：您使用的脚本认为星期日 - 星期六的周。由于一年并不总是在星期日开始，因此它将第一个部分周视为第1周。

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.

如果你只想要从第一个星期开始的7个星期年，请使用下面的脚本。如果你想要它基于星期日，那么你正在使用的脚本是正确的。

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.

有一个星期＃53 。这不是整整一周。

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

但我们在一周 52 。 我猜测当前的脚本没有说明闰年。

相反，你可以计算今年的日期，并将其除以7。

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

示例： 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>

你仍会得到 53 在一年的最后一天（或闰年的最后两天）。

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

编辑：修正了<$ c的偏移$ c> 1 dayOfYear 。

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