沒有周末的日差

Question

我想計算用戶輸入的總天差

例如當用戶輸入

start_date = 2012-09-06和end-date = 2012-09-11

現在我正在使用此代碼來查找差異

$count = abs(strtotime($start_date) - strtotime($end_date));
$day   = $count+86400;
$total = floor($day/(60*60*24));

total 的結果將是 6。但問題是我不想包括周末（周六和周日）的天數

2012-09-06
2012-09-07
2012-09-08 Saturday
2012-09-09 Sunday
2012-09-10
2012-09-11

所以結果將是 4

- - 更新 - -

我有一個包含日期的表，表名是假期日期

例如該表包含2012-09-07

所以，總天數將是 3，因為它沒有計算假期日期

我如何做到這一點以將日期從輸入到表中的日期等同起來？

Answer 1

我最喜歡的很容易： DateTime ， DateInterval和DatePeriod

$start = new DateTime('2012-09-06');
$end = new DateTime('2012-09-11');
// otherwise the  end date is excluded (bug?)
$end->modify('+1 day');

$interval = $end->diff($start);

// total days
$days = $interval->days;

// create an iterateable period of date (P1D equates to 1 day)
$period = new DatePeriod($start, new DateInterval('P1D'), $end);

// best stored as array, so you can add more than one
$holidays = array('2012-09-07');

foreach($period as $dt) {
    $curr = $dt->format('D');

    // substract if Saturday or Sunday
    if ($curr == 'Sat' || $curr == 'Sun') {
        $days--;
    }

    // (optional) for the updated question
    elseif (in_array($dt->format('Y-m-d'), $holidays)) {
        $days--;
    }
}


echo $days; // 4

Answer 2

在我的情況下，我需要與 OP 相同的答案，但想要更小的東西。 @Bojan 的答案有效，但我不喜歡它不適用於DateTime對象，需要使用時間戳，並且正在與strings而不是實際對象本身進行比較（這感覺很糟糕）......這是他的修訂版回答。

function getWeekdayDifference(\DateTime $startDate, \DateTime $endDate)
{
    $days = 0;

    while($startDate->diff($endDate)->days > 0) {
        $days += $startDate->format('N') < 6 ? 1 : 0;
        $startDate = $startDate->add(new \DateInterval("P1D"));
    }

    return $days;
}

如果您希望包含開始和結束日期，請根據 @xzdead 的評論：

function getWeekdayDifference(\DateTime $startDate, \DateTime $endDate)
{
    $isWeekday = function (\DateTime $date) {
        return $date->format('N') < 6;
    };

    $days = $isWeekday($endDate) ? 1 : 0;

    while($startDate->diff($endDate)->days > 0) {
        $days += $isWeekday($startDate) ? 1 : 0;
        $startDate = $startDate->add(new \DateInterval("P1D"));
    }

    return $days;
}

Answer 3

使用DateTime ：

$datetime1 = new DateTime('2012-09-06');
$datetime2 = new DateTime('2012-09-11');
$interval = $datetime1->diff($datetime2);
$woweekends = 0;
for($i=0; $i<=$interval->d; $i++){
    $datetime1->modify('+1 day');
    $weekday = $datetime1->format('w');

    if($weekday !== "0" && $weekday !== "6"){ // 0 for Sunday and 6 for Saturday
        $woweekends++;  
    }

}

echo $woweekends." days without weekend";

// 4 days without weekends

Answer 4

在沒有周末的情況下獲得差異的最簡單和最快的方法是使用Carbon庫。

這是一個如何使用它的示例：

<?php

$from = Carbon\Carbon::parse('2016-05-21 22:00:00');
$to = Carbon\Carbon::parse('2016-05-21 22:00:00');
echo $to->diffInWeekdays($from);

Answer 5

date('N') 獲取星期幾（1 - 星期一，7 - 星期日）

$start = strtotime('2012-08-06');
$end = strtotime('2012-09-06');

$count = 0;

while(date('Y-m-d', $start) < date('Y-m-d', $end)){
  $count += date('N', $start) < 6 ? 1 : 0;
  $start = strtotime("+1 day", $start);
}

echo $count;

Answer 6

這是@dan-lee 函數的改進版本：

function get_total_days($start, $end, $holidays = [], $weekends = ['Sat', 'Sun']){

    $start = new \DateTime($start);
    $end   = new \DateTime($end);
    $end->modify('+1 day');

    $total_days = $end->diff($start)->days;
    $period = new \DatePeriod($start, new \DateInterval('P1D'), $end);

    foreach($period as $dt) {
        if (in_array($dt->format('D'),  $weekends) || in_array($dt->format('Y-m-d'), $holidays)){
            $total_days--;
        }
    }
    return $total_days;
}

要使用它：

$start    = '2021-06-12';
$end      = '2021-06-17';
$holidays = ['2021-06-15'];
echo get_total_days($start, $end, $holidays); // Result: 3

Answer 7

看看這篇文章： 計算工作日

（在您的情況下，您可以省略“假期”部分，因為您只在工作日/工作日之后）

<?php
//The function returns the no. of business days between two dates
function getWorkingDays($startDate,$endDate){
    // do strtotime calculations just once
    $endDate = strtotime($endDate);
    $startDate = strtotime($startDate);


    //The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24
    //We add one to inlude both dates in the interval.
    $days = ($endDate - $startDate) / 86400 + 1;

    $no_full_weeks = floor($days / 7);
    $no_remaining_days = fmod($days, 7);

    //It will return 1 if it's Monday,.. ,7 for Sunday
    $the_first_day_of_week = date("N", $startDate);
    $the_last_day_of_week = date("N", $endDate);

    //---->The two can be equal in leap years when february has 29 days, the equal sign is added here
    //In the first case the whole interval is within a week, in the second case the interval falls in two weeks.
    if ($the_first_day_of_week <= $the_last_day_of_week) {
        if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week) $no_remaining_days--;
        if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week) $no_remaining_days--;
    }
    else {
        // (edit by Tokes to fix an edge case where the start day was a Sunday
        // and the end day was NOT a Saturday)

        // the day of the week for start is later than the day of the week for end
        if ($the_first_day_of_week == 7) {
            // if the start date is a Sunday, then we definitely subtract 1 day
            $no_remaining_days--;

            if ($the_last_day_of_week == 6) {
                // if the end date is a Saturday, then we subtract another day
                $no_remaining_days--;
            }
        }
        else {
            // the start date was a Saturday (or earlier), and the end date was (Mon..Fri)
            // so we skip an entire weekend and subtract 2 days
            $no_remaining_days -= 2;
        }
    }

    //The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder
//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it
   $workingDays = $no_full_weeks * 5;
    if ($no_remaining_days > 0 )
    {
      $workingDays += $no_remaining_days;
    }    


    return $workingDays;
}

// This will return 4
echo getWorkingDays("2012-09-06","2012-09-11");
?>

Answer 8

如果您不需要全天但需要准確的秒數，請嘗試使用此代碼。 這接受 unix 時間戳作為輸入。

function timeDifferenceWithoutWeekends($from, $to) {
    $start = new DateTime("@".$from);
    $current = clone $start;
    $end = new DateTime("@".$to);
    $sum = 0;
    while ($current<$end) {
        $endSlice = clone $current;
        $endSlice->setTime(0,0,0);
        $endSlice->modify('+1 day');
        if ($endSlice>$end) {
            $endSlice= clone $end;
        }
        $seconds = $endSlice->getTimestamp()-$current->getTimestamp();
        $currentDay = $current->format("D");
        if ($currentDay != 'Sat' && $currentDay != 'Sun') {
            $sum+=$seconds;
        }
        $current = $endSlice;
    }
    return $sum;
}

Answer 9

/**
 * Getting the Weekdays count[ Excludes : Weekends]
 * 
 * @param type $fromDateTimestamp
 * @param type $toDateTimestamp
 * @return int
 */
public static function getWeekDaysCount($fromDateTimestamp = null, $toDateTimestamp=null) {

    $startDateString   = date('Y-m-d', $fromDateTimestamp);
    $timestampTomorrow = strtotime('+1 day', $toDateTimestamp);
    $endDateString     = date("Y-m-d", $timestampTomorrow);
    $objStartDate      = new \DateTime($startDateString);    //intialize start date
    $objEndDate        = new \DateTime($endDateString);    //initialize end date
    $interval          = new \DateInterval('P1D');    // set the interval as 1 day
    $dateRange         = new \DatePeriod($objStartDate, $interval, $objEndDate);

    $count = 0;

    foreach ($dateRange as $eachDate) {
        if (    $eachDate->format("w") != 6 
            &&  $eachDate->format("w") != 0 
        ) {
            ++$count;
        }
    }
    return $count;
}

Answer 10

請看看這個精確的 php 函數返回天數，周末除外。

function Count_Days_Without_Weekends($start, $end){
    $days_diff = floor(((abs(strtotime($end) - strtotime($start))) / (60*60*24)));
    $run_days=0;
    for($i=0; $i<=$days_diff; $i++){
        $newdays = $i-$days_diff;
        $futuredate = strtotime("$newdays days");
        $mydate = date("F d, Y", $futuredate);
        $today = date("D", strtotime($mydate));             
        if(($today != "Sat") && ($today != "Sun")){
            $run_days++;
        }
    }
return $run_days;
}

試試看，確實有效。。

Answer 11

使用 Carbon\\Caborn 的一個非常簡單的解決方案

這是從控制器存儲功能調用的存儲庫文件

<?php

namespace App\Repositories\Leave;

use App\Models\Holiday;
use App\Models\LeaveApplication;
use App\Repositories\BaseRepository;
use Carbon\Carbon;

class LeaveApplicationRepository extends BaseRepository
{
    protected $holiday;

    public function __construct(LeaveApplication $model, Holiday $holiday)
    {
        parent::__construct($model);
        $this->holiday = $holiday;
    }

    /**
     * Get all authenticated user leave
     */
    public function getUserLeave($id)
    {
        return $this->model->where('employee_id',$id)->with(['leave_type','approver'])->get();
    }

    /**
     * @param array $request
     */
    public function create($request)
    {
        $request['total_days'] = $this->getTotalDays($request['start_date'],$request['end_date']);

        return $this->model->create($request->only('send_to','leave_type_id','start_date','end_date','desc','total_days'));
    }

    /**
     * Get total leave days
     */
    private function getTotalDays($startDate, $endDate)
    {
        $holidays = $this->getHolidays(); //Get all public holidays
        $leaveDays = 0; //Declare values which hold leave days
        //Format the dates
        $startDate = Carbon::createFromFormat('Y-m-d',$startDate);
        $endEnd = Carbon::createFromFormat('Y-m-d',$endDate);
        //Check user dates
        for($date = $startDate; $date <= $endEnd; $date->modify('+1 day')) {
            if (!$date->isWeekend() && !in_array($date,$holidays)) {
                $leaveDays++; //Increment days if not weekend and public holidays
            }
        }
        return $leaveDays; //return total days
    }

    /**
     * Get Current Year Public Holidays
     */
    private function getHolidays()
    {
        $holidays = array();
        $dates = $this->holiday->select('date')->where('active',1)->get();
        foreach ($dates as $date) {
            $holidays[]=Carbon::createFromFormat('Y-m-d',$date->date);
        }
        return $holidays;
    }
}

控制器函數接收用戶輸入請求並在調用存儲庫函數之前進行驗證

<?php

namespace App\Http\Controllers\Leave;

use App\Http\Controllers\AuthController;
use App\Http\Requests\Leave\LeaveApplicationRequest;
use App\Repositories\Leave\LeaveApplicationRepository;
use Exception;

class LeaveApplicationController extends AuthController
{
    protected $leaveApplication;

    /**
     * LeaveApplicationsController constructor.
     */
    public function __construct(LeaveApplicationRepository $leaveApplication)
    {
        parent::__construct();
        $this->leaveApplication = $leaveApplication;
    }

    /**
     * Store a newly created resource in storage.
     */
    public function store(LeaveApplicationRequest $request)
    {
        try {
            $this->leaveApplication->create($request);
            return $this->successRoute('leaveApplications.index','Leave Applied');
        }
        catch (Exception $e) {
            return $this->errorWithInput($request);
        }
    }
}
  

Answer 12

這是計算兩個日期之間的工作日的替代方法，並且還使用來自http://pear.php.net/package/Date_Holidays 的Pear 的 Date_Holidays 排除美國假期。

$start_date 和 $end_date 應該是 DateTime 對象（您可以使用new DateTime('@'.$timestamp)從時間戳轉換為 DateTime 對象）。

<?php
function business_days($start_date, $end_date)
{
  require_once 'Date/Holidays.php';
  $dholidays = &Date_Holidays::factory('USA');
  $days = 0;

  $period = new DatePeriod($start_date, new DateInterval('P1D'), $end_date);

  foreach($period as $dt)
  {
    $curr = $dt->format('D');

    if($curr != 'Sat' && $curr != 'Sun' && !$dholidays->isHoliday($dt->format('Y-m-d')))
    {
      $days++;
    }
  }
  return $days;
}
?>