我想计算给定月份和年份的工作日天数。工作日意味着星期一至星期五。我该怎么做？
答案 0 :(得分：22)
您无需计算每月的每一天。你已经知道前28天包含20个工作日，无论如何。您所要做的就是确定最近几天。将起始值更改为29.然后将20个工作日添加到返回值。
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;
}
答案 1 :(得分：4)
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错误警告未显示通知，请删除@。
答案 2 :(得分：2)
尝试这个
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 $workingDays;
}
答案 3 :(得分：1)
这是我能提出的最简单的代码。 你真的需要创建一个数组或数据库表来保存假期以获得真正的“工作日”计数，但这不是所要求的，所以你去，希望这有助于某人。
function get_weekdays($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;
}
答案 4 :(得分：1)
获取两个日期之间没有假期的工作日的数量：
echo number_of_working_days('2013-12-23', '2013-12-29');
3
答案 5 :(得分：1)
DateObject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;
}
答案 6 :(得分：1)
计算任何日期的一个月内的工作日：
{{1}}
答案 7 :(得分：0)
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，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）总数
答案 8 :(得分：0)
我想出了一个非循环函数。在性能方面要好得多。它可能看起来很乱，但它只需要问PHP第一天的工作日和月份的数天：其余的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 + $extraWorkdays;
}
答案 9 :(得分：0)
这些功能可以无循环。
这些函数使用以下方法计算工作日数：
// main functions
// weekdays in month of year
function calculateNumberOfWeekDaysAtDate($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 ? $rest : 5);
}
答案 10 :(得分：0)
function workingDays($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;
}
答案 11 :(得分：0)
我创建了一个简单的函数，它接受 $first_day_of_month（星期日/星期一等工作日）。您可以像这样找到一个月的第一天：
date('N', strtotime(date("01-m-Y")));
并使用 $month_last_date 可以这样获取：
date("t");
功能如下：
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;
}
答案 12 :(得分：-1)
这将起作用
// 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);