计算星期六和星期日的数量

How can I calculate the number of Saturdays and Sundays between two dates in php?

如何在php中计算两个日期之间的星期六和星期日的数量？

Is there any inbuilt function for that purpose?

是否有任何内置功能用于此目的？

6 个解决方案

#1

There is a related question here already, Calculate business days

这里有一个相关的问题，计算工作日

You can use this to subtract from 7 to get the weekend days, or similar.

您可以使用它从7中减去周末天数或类似天数。

#2

I don't think there is a built in for that, but this should do the job :

我不认为有内置的，但这应该做的工作：

$startTime = START_TIMESTAMP;
$endTime = END_TIMESTAMP;
$time = $startTime;
$count = 0;

while(date('w', $time) != 0) { // 0 (for Sunday) through 6 (for Saturday)
    $time += 86400;
}

while($time < $endTime) {
    $count++;
    $time += 7 * 86400;
}

#3

Let us all KISS (Keep It Simple Stupid). Why make it so complicated?

让我们所有的吻（保持简单愚蠢）。为什么要这么复杂？

function countWeekendDays($start, $end)
{
    // $start in timestamp
        // $end in timestamp


    $iter = 24*60*60; // whole day in seconds
    $count = 0; // keep a count of Sats & Suns

    for($i = $start; $i <= $end; $i=$i+$iter)
    {
        if(Date('D',$i) == 'Sat' || Date('D',$i) == 'Sun')
        {
            $count++;
        }
    }
    return $count;
   }

#4

<?php
date_default_timezone_set("Europe/Lisbon");
$d1 = new DateTime("2009-06-01"); /* inclusive */
$d2 = new DateTime("2009-07-01"); /* exclusive */

$interval = $d2->diff($d1);
$number_of_days = $interval->format("%d");

$number_of_weekends = $number_of_days / 7;
$remainder = $number_of_days % 7;

if ($remainder >=2 && $d1->format("D") == "Sat")
    $number_of_weekends++;
elseif ($d1->format("w") + $remainder >= 8)
    $number_of_weekends++;

I may have missed by one in the last condition, be sure to check it with different starting dates. (Feel free to edit this answer if you spot an error).

我可能在最后一个条件中错过了一个，一定要用不同的开始日期检查它。（如果发现错误，请随意编辑此答案）。

#5

there's definitely no built in function for that but you can use strtotime to loop days

肯定没有内置功能，但你可以使用strtotime循环天

$start = strtotime('2010-01-01');
$end = strtotime('2010-01-09');

function numWeekdays( $start_ts, $end_ts, $day, $include_start_end = false ) {

    $day = strtolower( $day );
    $current_ts = $start_ts;
    // loop next $day until timestamp past $end_ts
    while( $current_ts < $end_ts ) {

        if( ( $current_ts = strtotime( 'next '.$day, $current_ts ) ) < $end_ts) {
            $days++;
        }
    }

    // include start/end days
    if ( $include_start_end ) {
        if ( strtolower( date( 'l', $start_ts ) ) == $day ) {
            $days++;
        }
        if ( strtolower( date( 'l', $end_ts ) ) == $day ) {
            $days++;
        }
    }   

    return (int)$days;

}

echo numWeekDays( $start, $end, 'saturday', false );

#6

I searched for awhile for a simple solution that worked and decided to write my own and came up with this

我搜索了一段时间，找到了一个简单的解决方案，并且决定自己编写并提出这个问题

        $start = date('Y-m-d');
        $end = date('Y-m-d', strtotime($start.' +1 year'));
        $current = $start;
        $count = 0;

        while($current != $end){
            if(date('l', strtotime($current)) == 'Saturday'){
                $count++;
            }

            $current = date('Y-m-d', strtotime($current.' +1 day'));
        };

        echo $count;

#1