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获取给定月份的工作日数

发布于 2024-12-19 20:19:41 字数 43 浏览 4 评论 0原文

我想计算给定月份和年份中的工作日天数。工作日是指周一至周五。我该怎么做？

原文

I want to calculate the number of weekdays days in a give month and year. Weekdays means monday to friday. How do i do it ?

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评论（13）

终陌 2024-12-26 20:19:41

你不需要计算这个月的每一天。您已经知道无论如何，前 28 天包含 20 个工作日。您所要做的就是确定最后几天。将起始值更改为 29。然后将 20 个工作日添加到返回值中。

function get_weekdays($m,$y) {
$lastday = date("t",mktime(0,0,0,$m,1,$y));
$weekdays=0;
for($d=29;$d<=$lastday;$d++) {
    $wd = date("w",mktime(0,0,0,$m,$d,$y));
    if($wd > 0 && $wd < 6) $weekdays++;
    }
return $weekdays+20;
}

You don't need to count every day in the month. You already know the first 28 days contain 20 weekdays no matter what. All you have to do is determine the last few days. Change the start value to 29. Then add 20 weekdays to your return value.

function get_weekdays($m,$y) {
$lastday = date("t",mktime(0,0,0,$m,1,$y));
$weekdays=0;
for($d=29;$d<=$lastday;$d++) {
    $wd = date("w",mktime(0,0,0,$m,$d,$y));
    if($wd > 0 && $wd < 6) $weekdays++;
    }
return $weekdays+20;
}
回复 原文
み青杉依旧 2024-12-26 20:19:41

一些基本代码：

$month = 12;
$weekdays = array();
$d = 1;

do {
    $mk = mktime(0, 0, 0, $month, $d, date("Y"));
    @$weekdays[date("w", $mk)]++;
    $d++;
} while (date("m", $mk) == $month);

print_r($weekdays);

如果您的 PHP 错误警告不显示通知，请删除 @。

Some basic code:

$month = 12;
$weekdays = array();
$d = 1;

do {
    $mk = mktime(0, 0, 0, $month, $d, date("Y"));
    @$weekdays[date("w", $mk)]++;
    $d++;
} while (date("m", $mk) == $month);

print_r($weekdays);

Remove the @ if your PHP error warning doesn't show notices.

回复 原文
烟酉 2024-12-26 20:19:41

试试这个

function getWeekdays($m, $y = NULL){
    $arrDtext = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri');

    if(is_null($y) || (!is_null($y) && $y == ''))
        $y = date('Y');

    $d = 1;
    $timestamp = mktime(0,0,0,$m,$d,$y);
    $lastDate = date('t', $timestamp);
    $workingDays = 0;
    for($i=$d; $i<=$lastDate; $i++){
        if(in_array(date('D', mktime(0,0,0,$m,$i,$y)), $arrDtext)){
            $workingDays++;
        }
    }
    return $workingDays;
}

try this one

function getWeekdays($m, $y = NULL){
    $arrDtext = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri');

    if(is_null($y) || (!is_null($y) && $y == ''))
        $y = date('Y');

    $d = 1;
    $timestamp = mktime(0,0,0,$m,$d,$y);
    $lastDate = date('t', $timestamp);
    $workingDays = 0;
    for($i=$d; $i<=$lastDate; $i++){
        if(in_array(date('D', mktime(0,0,0,$m,$i,$y)), $arrDtext)){
            $workingDays++;
        }
    }
    return $workingDays;
}
回复 原文
°如果伤别离去 2024-12-26 20:19:41

这是我能想到的最简单的代码。
您确实需要创建一个数组或数据库表来保存假期以获得真正的“工作日”计数，但这不是所要求的，所以就在这里，希望这对某人有帮助。

function get_weekdays($m,$y) {
$lastday = date("t",mktime(0,0,0,$m,1,$y));
$weekdays=0;
for($d=1;$d<=$lastday;$d++) {
    $wd = date("w",mktime(0,0,0,$m,$d,$y));
    if($wd > 0 && $wd < 6) $weekdays++;
    }
return $weekdays;
}

This is the simplest code I could come up with.
You really would need to create an array or a database table to hold the holidays to get a true, "Working Days" count, but that wasn't what was asked, so here you go, hope this helps someone.

function get_weekdays($m,$y) {
$lastday = date("t",mktime(0,0,0,$m,1,$y));
$weekdays=0;
for($d=1;$d<=$lastday;$d++) {
    $wd = date("w",mktime(0,0,0,$m,$d,$y));
    if($wd > 0 && $wd < 6) $weekdays++;
    }
return $weekdays;
}
回复 原文
笑咖 2024-12-26 20:19:41

获取两个日期之间无节假日的工作日数：

使用示例：

echo number_of_working_days('2013-12-23', '2013-12-29');

输出：

3

链接到功能

Get the number of working days without holidays between two dates :

Use example:

echo number_of_working_days('2013-12-23', '2013-12-29');

Output:

3

Link to the function

回复 原文
坏尐絯 2024-12-26 20:19:41

日期对象方法：

function getWorkingDays(DateTime $date) {
    $month = clone $date;
    $month->modify('last day of this month');
    $workingDays = 0;
    for ($i = $month->format('t'); $i > 28; --$i) {
        if ($month->format('N') < 6) {
            ++$workingDays;
        }
        $month->modify('-1 day');
    }

    return 20 + $workingDays;
}

DateObject method:

function getWorkingDays(DateTime $date) {
    $month = clone $date;
    $month->modify('last day of this month');
    $workingDays = 0;
    for ($i = $month->format('t'); $i > 28; --$i) {
        if ($month->format('N') < 6) {
            ++$workingDays;
        }
        $month->modify('-1 day');
    }

    return 20 + $workingDays;
}
回复 原文
秉烛思 2024-12-26 20:19:41

从任意日期计算一个月内的工作日：

public function getworkd($mday)
{
    $dn = new DateTime($mday);
    $dfrom = $dn->format('Y-m-01');
    $dtill = $dn->format('Y-m-t');
    $df = new DateTime($dfrom);
    $dt = new DateTime($dtill);
    $wdays = 0;
    while($df<=$dt)
    {
        $dof= $df->format('D') ;
        if( $dof == 'Sun' || $dof == 'Sat' ) ; else $wdays++;
        $df->add(new DateInterval('P1D'));
    }
    return $wdays;
}

Calculate working days in a month from any date:

public function getworkd($mday)
{
    $dn = new DateTime($mday);
    $dfrom = $dn->format('Y-m-01');
    $dtill = $dn->format('Y-m-t');
    $df = new DateTime($dfrom);
    $dt = new DateTime($dtill);
    $wdays = 0;
    while($df<=$dt)
    {
        $dof= $df->format('D') ;
        if( $dof == 'Sun' || $dof == 'Sat' ) ; else $wdays++;
        $df->add(new DateInterval('P1D'));
    }
    return $wdays;
}
回复 原文
傲世九天 2024-12-26 20:19:41

查找给定月份的最后一天和工作日
然后做一个简单的 while 循环，例如：-

$dates = explode(',', date('t,N', strtotime('2013-11-01')));
$day = $dates[1]; 
$tot = $dates[0]; 
$cnt = 0;
while ($tot>1)
{   
    if ($day < 6)
    {   
        $cnt++;
    }   
    if ($day == 1)
    {   
        $day = 7;
    }   
    else
    {   
        $day--;
    }   
    $tot--;
}

$cnt = 给定月份的工作日（周一至周五）总计

Find the last day and the weekday for the given month
then do a simple while loop like :-

$dates = explode(',', date('t,N', strtotime('2013-11-01')));
$day = $dates[1]; 
$tot = $dates[0]; 
$cnt = 0;
while ($tot>1)
{   
    if ($day < 6)
    {   
        $cnt++;
    }   
    if ($day == 1)
    {   
        $day = 7;
    }   
    else
    {   
        $day--;
    }   
    $tot--;
}

$cnt = total of weekday (Monday to Friday) for a given month

回复 原文
她比我温柔 2024-12-26 20:19:41

我想出了一个非循环函数。在性能方面要好得多。看起来可能很乱，但它只需要询问 PHP 第一天是工作日和该月的天数即可：其余的都是基于逻辑的算术运算。

function countWorkDays($year, $month)
{
    $workingWeekdays   = 5;
    $firstDayTimestamp = mktime(0, 0, 0, $month, 1, $year);
    $firstDayWeekDay   = (int)date("N", $firstDayTimestamp); //1: monday, 7: saturday
    $upToDay           = (int)date("t", $firstDayTimestamp);

    $firstMonday = 1 === $firstDayWeekDay ? 1 : 9 - $firstDayWeekDay;
    $wholeWeeks  = $firstMonday < $upToDay ? (int)floor(($upToDay - $firstMonday + 1) / 7) : 0;
    $extraDays   = ($upToDay - $firstMonday + 1) % 7;

    $initialWorkdays      = $firstMonday > 1 && $firstDayWeekDay <= $workingWeekdays ? $workingWeekdays - $firstDayWeekDay + 1 : 0;
    $workdaysInWholeWeeks = $wholeWeeks * $workingWeekdays;
    $extraWorkdays        = $extraDays <= $workingWeekdays ? $extraDays : $workingWeekdays;

    return $initialWorkdays + $workdaysInWholeWeeks + $extraWorkdays;
}

I've come up with a non-loop function. Much better in terms of performance. It might seem messy but it just needs to ask PHP the first day's weekday and the month's number days: the rest are arithmetical operations based on logic.

function countWorkDays($year, $month)
{
    $workingWeekdays   = 5;
    $firstDayTimestamp = mktime(0, 0, 0, $month, 1, $year);
    $firstDayWeekDay   = (int)date("N", $firstDayTimestamp); //1: monday, 7: saturday
    $upToDay           = (int)date("t", $firstDayTimestamp);

    $firstMonday = 1 === $firstDayWeekDay ? 1 : 9 - $firstDayWeekDay;
    $wholeWeeks  = $firstMonday < $upToDay ? (int)floor(($upToDay - $firstMonday + 1) / 7) : 0;
    $extraDays   = ($upToDay - $firstMonday + 1) % 7;

    $initialWorkdays      = $firstMonday > 1 && $firstDayWeekDay <= $workingWeekdays ? $workingWeekdays - $firstDayWeekDay + 1 : 0;
    $workdaysInWholeWeeks = $wholeWeeks * $workingWeekdays;
    $extraWorkdays        = $extraDays <= $workingWeekdays ? $extraDays : $workingWeekdays;

    return $initialWorkdays + $workdaysInWholeWeeks + $extraWorkdays;
}
回复 原文
圈圈圆圆圈圈 2024-12-26 20:19:41

这些函数可以无需循环工作。

这些函数使用以下方式计算工作日数：

  • 当月第一个星期一的天数
  • 当月的天数
// main functions 
// weekdays in month of year
function calculateNumberOfWeekDaysAtDate($month, $year)
{
    // I'm sorry, I don't know the right format for the $month and $year, I hope this is right.
    // PLEASE CORRECT IF WRONG
    $firstMondayInCurrentMonth = (int) date("j", strtotime("first monday of 01-$month-$year")); //get first monday in month for calculations
    $numberOfDaysOfCurrentMonth = (int) date("t", strtotime("01-$month-$year")); // number of days in month

    return calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth);
}

// week days in current month
function calculateNumberOfWeekDaysInCurrentMonth()
{
    $firstMondayInCurrentMonth = (int) date("j", strtotime("first monday of this month")); //get first monday in month for calculations
    $numberOfDaysOfCurrentMonth = (int) date("t"); // number of days in this month

    return calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth);
}

// helper functions
function calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth)
{
    return $numberOfWeekDays = (($start = ($firstMondayInCurrentMonth - 3)) < 0 ? 0 : $start) + floor(($numberOfDaysOfCurrentMonth - ($firstMondayInCurrentMonth - 1)) / 7) * 5 + (($rest = (($numberOfDaysOfCurrentMonth - ($firstMondayInCurrentMonth - 1)) % 7)) <= 5 ? $rest : 5);
}

These functions work Without Loops.

The functions calculate the number of weekdays using:

  • day-number of first monday in month
  • number of days in month
// main functions 
// weekdays in month of year
function calculateNumberOfWeekDaysAtDate($month, $year)
{
    // I'm sorry, I don't know the right format for the $month and $year, I hope this is right.
    // PLEASE CORRECT IF WRONG
    $firstMondayInCurrentMonth = (int) date("j", strtotime("first monday of 01-$month-$year")); //get first monday in month for calculations
    $numberOfDaysOfCurrentMonth = (int) date("t", strtotime("01-$month-$year")); // number of days in month

    return calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth);
}

// week days in current month
function calculateNumberOfWeekDaysInCurrentMonth()
{
    $firstMondayInCurrentMonth = (int) date("j", strtotime("first monday of this month")); //get first monday in month for calculations
    $numberOfDaysOfCurrentMonth = (int) date("t"); // number of days in this month

    return calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth);
}

// helper functions
function calculateNumberOfWeekDaysFromFirstMondayAndNumberOfMonthDays($firstMondayInCurrentMonth, $numberOfDaysOfCurrentMonth)
{
    return $numberOfWeekDays = (($start = ($firstMondayInCurrentMonth - 3)) < 0 ? 0 : $start) + floor(($numberOfDaysOfCurrentMonth - ($firstMondayInCurrentMonth - 1)) / 7) * 5 + (($rest = (($numberOfDaysOfCurrentMonth - ($firstMondayInCurrentMonth - 1)) % 7)) <= 5 ? $rest : 5);
}
回复 原文
胡大本事 2024-12-26 20:19:41
function workingDays($m,$y) {
    $days = cal_days_in_month(CAL_GREGORIAN, $m, $y);
    $workig_days = 0;
    $days_rest = array(5,6); //friday,saturday
    for ( $d=1 ; $d < $days+1 ; $d++ ) {
        if ( !in_array(date("w",strtotime("{$d}-{$m}-{$y}")),$days_rest)  ) {
            $workig_days++;
        }
    }
    return $workig_days;
}

function workingDays($m,$y) {
    $days = cal_days_in_month(CAL_GREGORIAN, $m, $y);
    $workig_days = 0;
    $days_rest = array(5,6); //friday,saturday
    for ( $d=1 ; $d < $days+1 ; $d++ ) {
        if ( !in_array(date("w",strtotime("{$d}-{$m}-{$y}")),$days_rest)  ) {
            $workig_days++;
        }
    }
    return $workig_days;
}
回复 原文
没︽人懂的悲伤 2024-12-26 20:19:41

我创建了一个简单的函数，它接受 $first_day_of_month （星期天/星期一等）。您可以像这样找到每月的第一天：

date('N', strtotime(date("01-m-Y")));

并使用可以像这样获取的 $month_last_date ：

date("t");

这是函数：

function workingDaysInMonth(int $first_day_of_month, int $month_last_date) : array
{
    $working_days = [];
    $day = $first_day_of_month;
    $working_day_count = 0;
    for ($i = 1; $i <= $month_last_date; $i++) {
        if ($day == 8) {
            $day = 1;
        }
        if (!($day == 6 || $day == 7)) {
            $working_day_count++;
            $working_days[$i] = $working_day_count;
        }
        $day++;
    }
    return $working_days;
}

I created a simple function that takes the $first_day_of_month (week day like Sunday/Monday etc). You can find out the first day of month like this:

date('N', strtotime(date("01-m-Y")));

And using the $month_last_date which can be procured like this:

date("t");

Here is the function:

function workingDaysInMonth(int $first_day_of_month, int $month_last_date) : array
{
    $working_days = [];
    $day = $first_day_of_month;
    $working_day_count = 0;
    for ($i = 1; $i <= $month_last_date; $i++) {
        if ($day == 8) {
            $day = 1;
        }
        if (!($day == 6 || $day == 7)) {
            $working_day_count++;
            $working_days[$i] = $working_day_count;
        }
        $day++;
    }
    return $working_days;
}
回复 原文
烟─花易冷 2024-12-26 20:19:41

这会起作用

// oct. 2013
$month = 10;

// loop through month days
for ($i = 1; $i <= 31; $i++) {

    // given month timestamp
    $timestamp = mktime(0, 0, 0, $month, $i, 2012);

    // to be sure we have not gone to the next month
    if (date("n", $timestamp) == $month) {

        // current day in the loop
        $day = date("N", $timestamp);

        // if this is between 1 to 5, weekdays, 1 = Monday, 5 = Friday
        if ($day == 1 OR $day <= 5) {

            // write it down now
            $days[$day][] = date("j", $timestamp);
        }
    }
}

// to see if it works :)
print_r($days);

this will work

// oct. 2013
$month = 10;

// loop through month days
for ($i = 1; $i <= 31; $i++) {

    // given month timestamp
    $timestamp = mktime(0, 0, 0, $month, $i, 2012);

    // to be sure we have not gone to the next month
    if (date("n", $timestamp) == $month) {

        // current day in the loop
        $day = date("N", $timestamp);

        // if this is between 1 to 5, weekdays, 1 = Monday, 5 = Friday
        if ($day == 1 OR $day <= 5) {

            // write it down now
            $days[$day][] = date("j", $timestamp);
        }
    }
}

// to see if it works :)
print_r($days);
回复 原文
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