Php 获取给定月份的工作日数
我想计算给定月份和年份中的工作日数。工作日是指周一至周五。我该怎么做?一些基本代码:Php 获取给定月份的工作日数,php,calendar,days,weekday,weekend,Php,Calendar,Days,Weekday,Weekend,我想计算给定月份和年份中的工作日数。工作日是指周一至周五。我该怎么做?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
$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错误警告没有显示通知,请删除@
。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 $workingDays;
}
函数getWeekdays($m,$y=NULL){
$arrDtext=数组('Mon','Tue','Wed','Thu','Fri');
if(is_null($y)|(!is_null($y)&&$y='')
$y=日期('y');
$d=1;
$timestamp=mktime(0,0,0,$m,$d,$y);
$lastDate=日期('t',$timestamp);
$workingDays=0;
对于($i=$d;$i查找给定月份的最后一天和工作日
然后执行一个简单的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--;
}
$dates=explode(’,’,date('t,N',strottime('2013-11-01'));
$day=$dates[1];
$tot=$dates[0];
$cnt=0;
而($tot>1)
{
如果($day<6)
{
$cnt++;
}
如果($day==1)
{
$day=7;
}
其他的
{
美元日--;
}
$tot--;
}
$cnt=给定月份的工作日(周一到周五)总数这是我能想到的最简单的代码。 您确实需要创建一个数组或数据库表来保存假期,以获得真正的“工作日”计数,但这并不是要求的,所以现在开始吧,希望这对其他人有所帮助
function get_weekdays($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 $weekdays;
}
函数get_平日($m,$y){
$lastday=日期(“t”,mktime(0,0,0,$m,1,$y));
$weekdays=0;
对于($d=1;$d0&$wd<6)$weekdays++;
}
返回$weekdays;
}
您不需要计算月内的每一天。您已经知道前28天包含20个工作日,无论发生什么情况。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 $weekdays+20;
}
函数get_平日($m,$y){
$lastday=日期(“t”,mktime(0,0,0,$m,1,$y));
$weekdays=0;
对于($d=29;$d0&$wd<6)$weekdays++;
}
返回$weekdays+20;
}
获取两个日期之间无节假日的工作日数:
使用示例:
工作日的回音数(“2013-12-23”、“2013-12-29”);
输出:
3
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);
//2013年10月
$month=10;
//循环月日
对于($i=1;$iDateObject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 + $workingDays;
}
函数getWorkingDays(DateTime$date){
$month=clone$date;
$month->modify(‘本月最后一天’);
$workingDays=0;
对于($i=$month->format('t');$i>28;--$i){
如果($month->format('N')<6){
++$工作日;
}
$month->modify('-1天');
}
返回20+美元工作日;
}
从任何日期开始计算一个月内的工作日:
public function getworkd($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 $wdays;
}
公共函数getworkd($mday)
{
$dn=新日期时间($mday);
$dfrom=$dn->格式('Y-m-01');
$dtill=$dn->格式('Y-m-t');
$df=新日期时间($dfrom);
$dt=新日期时间($dtill);
$wdays=0;
而($dfformat('D');
if($dof=='Sun'|$dof=='Sat');else$wdays++;
$df->add(新的日期间隔('P1D'));
}
返回$wdays;
}
我提出了一个非循环函数。性能要好得多。它可能看起来很混乱,但它只需要询问PHP第一天的工作日和月份的天数: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 + $extraWorkdays;
}
函数countWorkDays($year,$month)
{
$workingWeekdays=5;
$firstDayTimestamp=mktime(0,0,0,$month,1,$year);
$firstDayWeekDay=(int)日期(“N”,$firstDayTimestamp);//1:monday,7:saturday
$upToDay=(int)日期(“t”,$firstDayTimestamp);
$firstMonday=1==$firstDayWeekDay?1:9-$firstDayWeekDay;
$wholeWeeks=$firstMonday<$uptday?(int)楼层(($uptday-$firstMonday+1)/7):0;
$extraDays=($upToDay-$firstMonday+1)%7;
$initialWorkdays=$firstMonday>1&&$firstDayWeekDay
这些函数在没有循环的情况下工作
这些函数使用以下公式计算工作日数:
- 当月第一个星期一的天数
- 月份天数
//主要功能
//一年中月份的工作日
函数CalculateEnumberOfWeekdaySatDate($month,$year)
{
//对不起,我不知道$month和$year的正确格式,我希望这是正确的。
//如有错误,请改正
$firstMondayInjurentMonth=(int)日期(“j”,strotime(“01的第一个星期一-$month-$year”);//获取当月的第一个星期一进行计算
$numberofdaysofcurrentmount=(int)日期(“t”,strotime(“01-$month-$year”);//月份天数
返回CalculateEnumberOfWeekdaysFromFirstMonday和NumberOfMonthDays($FirstMondayInjurentMonth,$numberOfDaysOfCurrentMonth);
}
//当月的周天数
函数CalculateEnumberOfWeekDaysInCurrentMonth()
{
$firstMondayInjurentMonth=(int)日期(“j”,strotime(“本月的第一个星期一”);//获取本月的第一个星期一进行计算
$numberOfDaysOfCurrentMonth=(int)日期(“t”);//本月的天数
返回CalculateEnumberOfWeekdaysFromFirstMonday和NumberOfMonthDays($FirstMondayInjurentMonth,$numberOfDaysOfCurrentMonth);
}
//辅助函数
函数CalculateEnumberOfWeekDays fromFirstMonday和NumberOfMonthDays($firstMondayInCurrentMonth,$numberOfDaysOfCurrentMonth)
{
返回$numberOfWeekDays=($start=($firstmondayirrentmonth-3))<0?0:$start)+floor($numberofdaysofcurrentmount-($firstmondayirrentmonth-1))/7)*5+($rest=($numberofdaysofcurrentmount-($firstmondayirrentmonth-1))函数工作日($m,$y){
$days=cal_days,以月为单位(cal_GREGORIAN,$m,$y);
$working_days=0;
$days_rest=array(5,6);//星期五、星期六
对于($d=1;$d<$days+1;$d++){
if(!in_数组(日期(“w”,strotime({$d}-{$m}-{$y}”),$days_rest)){
$working_days++;
}
}
返回$working_天;
}
我创建了一个简单的函数,它取每月的$first\u day\u(星期日/星期一等)
// main functions
// weekdays in month of year
function calculateNumberOfWeekDaysAtDate($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 ? $rest : 5);
}
function workingDays($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;
}
date('N', strtotime(date("01-m-Y")));
date("t");
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;
}