如何找到两个日期之间的星期一或星期二的数量?

w35I3y的答案几乎是正确的,但我在使用该函数时遇到了错误。此函数正确计算两个给定日期之间的星期一或任何特定日期的数量:

/** 
* Counts the number occurrences of a certain day of the week between a start and end date
* The $start and $end variables must be in UTC format or you will get the wrong number 
* of days  when crossing daylight savings time
* @param - $day - the day of the week such as "Monday", "Tuesday"...
* @param - $start - a UTC timestamp representing the start date
* @param - $end - a UTC timestamp representing the end date
* @return Number of occurences of $day between $start and $end
*/
function countDays($day, $start, $end)
{        
    //get the day of the week for start and end dates (0-6)
    $w = array(date('w', $start), date('w', $end));

    //get partial week day count
    if ($w[0] < $w[1])
    {            
        $partialWeekCount = ($day >= $w[0] && $day <= $w[1]);
    }else if ($w[0] == $w[1])
    {
        $partialWeekCount = $w[0] == $day;
    }else
    {
        $partialWeekCount = ($day >= $w[0] || $day <= $w[1]);
    }

    //first count the number of complete weeks, then add 1 if $day falls in a partial week.
    return floor( ( $end-$start )/60/60/24/7) + $partialWeekCount;
}

示例用法:

$start = strtotime("tuesday UTC");    
$end = strtotime("3 tuesday UTC");       
echo date("m/d/Y", $start). " - ".date("m/d/Y", $end). " has ". countDays(0, $start, $end). " Sundays";

输出类似于: 2010年9月28日至2010年10月19日有3个星期日。

您可以创建一个递归使用strtotime()来计算天数的函数。自从 strtotime("next monday"); 工作得很好。

function daycount($day, $startdate, $counter)
{
    if($startdate >= time())
    {
        return $counter;
    }
    else
    {
        return daycount($day, strtotime("next ".$day, $startdate), ++$counter);
    }
}

echo daycount("monday", strtotime("01.01.2009"), 0);

希望这是你正在寻找的东西:)

这个问题迫切需要一个使用PHP的更新答案 DateTime 类,所以这里是:-

/**
 * @param String $dayName eg 'Mon', 'Tue' etc
 * @param DateTimeInterface $start
 * @param DateTimeInterface $end
 * @return int
 */
function countDaysByName($dayName, \DateTimeInterface $start, \DateTimeInterface $end)
{
    $count = 0;
    $interval = new \DateInterval('P1D');
    $period = new \DatePeriod($start, $interval, $end);

    foreach($period as $day){
        if($day->format('D') === ucfirst(substr($dayName, 0, 3))){
            $count ++;
        }
    }
    return $count;
}

无循环,无递归

<?php
define('ONE_WEEK', 604800); // 7 * 24 * 60 * 60

function number_of_days($days, $start, $end) {
    $w = array(date('w', $start), date('w', $end));
    $x = floor(($end-$start)/ONE_WEEK);
    $sum = 0;

    for ($day = 0;$day < 7;++$day) {
        if ($days & pow(2, $day)) {
            $sum += $x + ($w[0] > $w[1]?$w[0] <= $day || $day <= $w[1] : $w[0] <= $day && $day <= $w[1]);
        }
    }

    return $sum;
}

//$start = $end = time();

// 0x10 == pow(2, 4) == 1 << 4 // THURSDAY
// 0x20 == pow(2, 5) == 1 << 5 // FRIDAY
echo number_of_days(0x01, $start, $end); // SUNDAY
echo number_of_days(pow(2, 0), $start, $end); // SUNDAY
echo number_of_days(0x02, $start, $end); // MONDAY
echo number_of_days(pow(2, 1), $start, $end); // MONDAY
echo number_of_days(0x04, $start, $end); // TUESDAY
echo number_of_days(1 << 2, $start, $end); // TUESDAY
echo number_of_days(0x08, $start, $end); // WEDNESDAY
echo number_of_days(1 << 3, $start, $end); // WEDNESDAY
echo number_of_days(0x10, $start, $end); // THURSDAY
echo number_of_days(0x20, $start, $end); // FRIDAY
echo number_of_days(0x40, $start, $end); // SATURDAY
echo number_of_days(0x01 | 0x40, $start, $end); // WEEKENDS : SUNDAY | SATURDAY
echo number_of_days(0x3E, $start, $end); // WORKDAYS : MONDAY | TUESDAY | WEDNESDAY | THURSDAY | FRIDAY
?>

计算两个日期之间的星期五总数##

$from_date=(2015-01-01);

$to_date=(2015-01-20);

while(strtotime($from_date) <= strtotime($to_date)){

    //5 for count Friday, 6 for Saturday , 7 for Sunday

    if(date("N",strtotime($from_date))==5){
        $counter++;
    }
    $from_date = date ("Y-m-d", strtotime("+1 day", strtotime($from_date)));

}

echo $counter;

<?php
$date = strtotime('2009-01-01');
$dateMax = strtotime('2009-02-23');

$nbr = 0;
while ($date < $dateMax) {
  var_dump(date('Y-m-d', $date));
  $nbr++;
  $date += 7 * 24 * 3600;
}
echo "<pre>";
 var_dump($nbr);
?>

function daycount($day, $startdate, $enddate, $counter) {
    if($startdate >= $enddate) {
        return $counter-1;  // A hack to make this function return the correct number of days.
    } else {
        return $this->daycount($day, strtotime("next ".$day, $startdate), $enddate, ++$counter);
    }
}

这是第一个答案的不同版本,它有起点和终点,对我来说都很有效。由于某种原因,本页给出的所有示例似乎都返回了答案加上额外的一天。

w35l3y的答案似乎很有效,所以我投了赞成票。不过,我更喜欢一些更容易理解、数学和循环更少的东西(不过从性能的角度来看,这并不重要)。我想我涵盖了所有可能的情况。..以下是我的想法:

function numDays($sday, $eday, $i, $cnt) {
    if (($sday < $eday && $i >= $sday && $i <= $eday) || ($sday > $eday && ($i >= $sday || $i <= $eday))) {
        // partial week (implied by $sday != $eday), so $i (day iteration) may have occurred one more time
        // a) end day is ahead of start day; $i is within start/end of week range
        // b) start day is ahead of end day (i.e., Tue start, Sun end); $i is either in back half of first week or front half of second week
        $cnt++;
    } elseif ($sday == $eday && $i == $sday) {
        // start day and end day are the same, and $i is that day, i.e., Tue occurs twice from Tue-Tue (1 wk, so $wks = $cnt)
        $cnt++;
    }

    return $cnt;    // # of complete weeks + partial week, if applicable
}

注意:$sday和$eday是与要检查的范围的开始和结束相对应的天数,$i是正在计数的天数(我在0-6的循环中)。我将$wks移出了函数,因为每次重新计算都没有意义。

$wks = floor(($endstamp - $startstamp)/7*24*60*60);
$numDays = numDays($sday, $eday, $i, $wks);

确保你比较的开始/结束时间戳具有相同的时区调整(如果有的话),否则你总是会有点偏离$cnt和$wks。(我在从未调整的第一年到调整后的日期/时间X计数时遇到了这个问题。)

正确的日计数。这将计算2个给定日期之间从周一到周日的所有天数

 <?php

    $startdate='2018-04-01';
    $enddate='2018-04-20';

    $mondayCount = 0;
    $tuesday = 0;
    $wednesday = 0;
    $thursday = 0;

    $friday = 0;
    $saturday = 0;
    $sunday = 0;

    $begin = new DateTime($startdate );
    $end = new DateTime( $enddate );
    $end = $end->modify( '+1 day' );

    $interval = new DateInterval('P1D');
    $daterange = new DatePeriod($begin, $interval ,$end);
    // looping each days from FROM and TO dates

    foreach($daterange as $date) {
        //$eachDate = $date->format("d-m-Y");
        echo $eachDateName = $date->format("l");
        switch ($eachDateName)
        {
            case 'Monday' :
                $mondayCount++;break;
            case 'Tuesday' :
                $tuesday++;break;
            case 'Wednesday' :
                $wednesday++;break;
            case 'Thursday' :
                $thursday++;break;
            case 'Friday' :
                $friday++;break;
            case 'Saturday' :
                $saturday++;break;
            case 'Sunday' :
                $sunday++;break;
        }

    }
    echo $mondayCount;echo "---";
    echo $tuesday;echo "---";
    echo $wednesday;echo "---";
    echo $thursday;echo "---";
    echo $friday;echo "----";
    echo $saturday;echo "---";
    echo $sunday;
?>

希望这能帮助到别人

我得到了答案。它只在星期天工作。但我不知道如何再坚持几天

<?php

// Define a constant of 1 day in seconds
define(ONE_DAY, 86400); 
date_default_timezone_set('America/New_York'); 

// Accepts two timestamps, start and end 
// Returns an array of timestamps that fall on a sunday 
function sundays_in_range($start, $end) {    
    echo date('N', $start);
    echo "<br/>";                     
    $days_until_sunday = date('w', $start) > 0 ? 7 - date('w', $start) : 0; 

    $date = $start + (ONE_DAY * $days_until_sunday); 
    $sundays = array(); 
    while ($date <= $end) { 
        array_push($sundays, $date); 
        $date += (7 * ONE_DAY); 
    } 
    return $sundays; 
} 

// Calculate some example dates. Today, and 30 days from now 
$start = time($start); 
$end = time($end) + (30 * ONE_DAY); 
echo ONE_DAY;
echo "<br/>";
 $count=0;
// Loop and output Y-m-d 
foreach (sundays_in_range($start, $end) as $sunday)
{
print "<option>".date("Y-m-d", $sunday)."</option><br/>";
 $count++;
 }

echo  $count;
?>