I have start date and end date.
I need to find out the day that is Sunday or Monday etc dependent upon user click on check box.
How can I find/calculate that in PHP?
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问题:
回答1:
You could create a function that uses strtotime() recursively to count the number of days. Since strtotime("next monday"); works just fine.
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); Hopefully this is something you're looking for :)
回答2:
no loops and no recursivity
<?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 ?> 回答3:
The answer by w35I3y was almost correct, but I was getting errors using that function. This function correctly calculates the number of Mondays or any specific day between two given dates:
/** * 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; } Example Usage:
$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"; Outputs something like: 09/28/2010 - 10/19/2010 has 3 Sundays.
回答4:
This question is just crying out for an updated answer that uses PHP's DateTime classes, so here it is:-
/** * @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; } 回答5:
<?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); ?> 回答6:
To count The Total Friday Between Two Dates##
$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 $counter; 回答7:
function daycount($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); } } This is a different version of the first answer which takes a start and end point and works for me. All of the examples given on this page seemed to return the answer plus an additional day for some reason.
回答8:
w35l3y's answer seems to work well, so I up-voted it. Still, I preferred something a bit easier to follow, and with less math and looping (not sure it matters from a performance standpoint, though). I think I covered all the possible scenarios... Here's what I came up with:
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 } Notes: $sday and $eday are the day numbers corresponding to the start and end of the range to be checked, and $i is the day number being counted (I have it in a 0-6 loop). I moved $wks outside the function, as there's no point recalculating it each time.
$wks = floor(($endstamp - $startstamp)/7*24*60*60); $numDays = numDays($sday, $eday, $i, $wks); Make sure the start/end timestamps you're comparing have the same timezone adjustment, if any, otherwise you'll always be off a bit with $cnt and $wks. (I ran into that when counting from an unadjusted first of the year to an adjusted day/time X.)
回答9:
Correct day count. This will count all the Days from Monday to Sunday between 2 given date
<?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; ?> Hope this will help someone
回答10:
I got the answer.Its working for sunday only.But I dont know how to make it for another days
"; $days_until_sunday = date('w', $start) > 0 ? 7 - date('w', $start) : 0; $date = $start + (ONE_DAY * $days_until_sunday); $sundays = array(); while ($date "; $count=0; // Loop and output Y-m-d foreach (sundays_in_range($start, $end) as $sunday) { print "".date("Y-m-d", $sunday).""; $count++; } echo $count; ?>