计算两个日期之间的星期天 [英] calculate sundays between two dates

问题描述

我想计算给定两个日期之间的所有星期日.我尝试了以下代码.如果天数较少，但如果我输入的天数较多，则效果很好.它可以保持处理，并且最大执行时间超过我更改的时间，但是即使执行时间为200秒，它也可以保持处理.

I want to calculate all Sunday's between given two dates. I tried following code. It works fine if days are less but if i enter more days. It keeps processing and Maximum execution time exceeds i changed the time but it even keeps processing even execution time is 200sec.

代码是

<?php
$one="2013-01-01";
$two="2013-02-30";

$no=0;
for($i=$one;$i<=$two;$i++)
{

    $day=date("N",strtotime($i));
    if($day==7)
    {
    $no++;
    }
}
echo $no;

?>

请帮助.

推荐答案

约翰·康德(John Conde)的答案是正确的，但这是一种更有效，更数学的解决方案:

John Conde's answer is correct, but here is a more efficient and mathy solution:

$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');
$days = $start->diff($end, true)->days;

$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);

echo $sundays;

让我为您分解一下.

$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');

首先，创建一些 DateTime 对象，它们是功能强大的内置对象，在PHP对象中就恰好解决了这类问题.

First, create some DateTime objects, which are powerful built-in PHP objects meant for exactly this kind of problem.

$days = $start->diff($end, true)->days;

接下来，使用 DateTime :: diff 来查找与$start到$end(在此将true作为第二个参数传递，以确保该值始终为正)，并获取它们之间的天数.

Next, use DateTime::diff to find the difference from $start to $end (passing true here as the second parameter ensures that this value is always positive), and get the number of days between them.

$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);

这是最重要的-确实不是那么复杂.首先，我们知道每个星期都有一个星期日，因此我们至少要有$days / 7个星期日开始，然后用intval向下舍入到最接近的int.

Here comes the big one - but it's not so complicated, really. First, we know there is one Sunday for every week, so we have at least $days / 7 Sundays to begin with, rounded down to the nearest int with intval.

最重要的是，在不到一周的时间内，可能会有一个星期日；例如，下周的星期五到星期一包含4天；其中之一是星期天.因此，取决于我们开始和结束的时间，可能还会有另一个.这很容易解释:

On top of that, there could be a Sunday in a span of time less than a week; for example, Friday to Monday of the next week contains 4 days; one of them is a Sunday. So, depending on when we start and end, there could be another. This is easy to account for:

• $start->format('N')(请参见 DateTime :: format )为我们提供了 ISO-8601 作为开始日期的星期几，即从1到7的数字(1是星期一，7是星期日).
• $days % 7为我们提供了剩余天数，这些剩余天数不能平均分为几周.

• $start->format('N') (see DateTime::format) gives us the ISO-8601 day of the week for the start date, which is a number from 1 to 7 (1 is Monday, 7 is Sunday).
• $days % 7 gives us the number of leftover days that don't divide evenly into weeks.

如果我们的开始日期和剩余天数加起来等于或大于7，则我们到了星期日.知道这一点，我们只需要添加该表达式即可，如果我们将1设置为true，则为0赋值，因为我们将其添加到int值.

If our starting day and the number of leftover days add up to 7 or more, then we reached a Sunday. Knowing that, we just have to add that expression, which will give us 1 if it's true or 0 if it's false, since we're adding it to an int value.

就在那里！这种方法的优点是，它不需要在给定时间之间每天进行迭代，也不需要检查是否是星期天，这样可以节省大量计算量，也可以使您看起来很聪明.希望有帮助！

And there you have it! The advantage of this method is that it doesn't require iterating over every day between the given times and checking to see if it's a Sunday, which will save you a lot computation, and also it will make you look really clever. Hope that helps!

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