没有周末的日差

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【中文标题】没有周末的日差【英文标题】:Day difference without weekends 【发布时间】:2012-09-04 03:10:59 【问题描述】:

我想计算用户输入的总天差

例如当用户输入时

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

现在我正在使用此代码来查找差异

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

total 的结果是 6。但问题是我不想包括周末(周六和周日)的日子

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

所以结果将是 4

----更新---

我有一个包含日期的表格,表格名称是假期日期

例如表格包含2012-09-07

所以,总天数将是 3,因为它没有计算假期日期

如何将输入的日期等同于表格中的日期?

【问题讨论】:

所以现在任何正在寻找这个 Carbon API 的人都有这个carbon.nesbot.com/docs/#api-difference 【参考方案1】:

很容易使用我的最爱:DateTimeDateIntervalDatePeriod

$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

【讨论】:

我不能使用 diff 和 DatePeriod....因为 php 版本 5.2....我有办法更改 diff...但我不明白如何更改日期期... ..你能帮我吗? @Belajar 您可以通过每次迭代增加一天来模仿这种行为。请参阅此论坛帖子以供参考:DateInterval & DatePeriod alternative for PHP 5.2x 我相信在这段代码中,如果假期是周六或周日,那么会被扣除两次。应该有一个“elseif”。 我从最终计数中又删除了一天。我通过说直到今天(2016 年 9 月 1 日)还有多少个工作日来测试它,结果是 1。好吧,没有!因此,我可以多休息 1 天。 非常感谢@dan-lee。它像魔术一样工作。我所做的只是将它复制并粘贴到我的代码中,它工作了!!!【参考方案2】:

在我的情况下,我需要与 OP 相同的答案,但想要更小的答案。 @Bojan 的答案有效,但我不喜欢它不适用于 DateTime 对象,需要使用时间戳,并且正在与 strings 进行比较,而不是与实际对象本身进行比较(感觉很hacky)......这是一个他的答案的修订版。

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;

如果您希望这包括开始结束日期,请根据@xzdead 的评论:

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;

【讨论】:

【参考方案3】:

在没有周末的情况下获得差异的最简单和最快的方法是使用Carbon 库。

这是一个如何使用它的示例:

<?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);

【讨论】:

【参考方案4】:

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

【讨论】:

应该是$interval-&gt;days 而不是$interval-&gt;d【参考方案5】:

date('N') 获取星期几(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;

【讨论】:

不能完全工作。试试这些日期: $start = strtotime('2015-12-01'); $end = strtotime('2015-12-15');结果为 10 天,但必须为 11。为什么,范围内的第一个星期一 (2015-12-07) 没有以正确的方式计算。也许是一个错误?【参考方案6】:

这是@dan-lee函数的改进版:

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;

使用它:

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

【讨论】:

【参考方案7】:

看看这篇文章: Calculate business days

(在您的情况下,您可以省略“假期”部分,因为您只是在工作/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");
?>

【讨论】:

【参考方案8】:

如果您不需要全天但需要准确的秒数,请尝试使用此代码。这接受 unix 时间戳作为输入。

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;

【讨论】:

【参考方案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;

【讨论】:

【参考方案10】:

请查看这个精确的 php 函数返回天数,不包括周末。

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;

试试看,真的很管用..

【讨论】:

【参考方案11】:

使用 Carbon\Caborn 的一个非常简单的解决方案

这是从控制器存储函数调用的存储库文件

<?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;
    

控制器函数接收用户输入请求并在调用存储库函数之前进行验证

<?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);
        
    

  

【讨论】:

【参考方案12】:

这是计算两个日期之间的工作日的替代方法,并且还使用来自http://pear.php.net/package/Date_Holidays 的 Pear 的 Date_Holidays 排除美国假期。

$start_date 和 $end_date 应该是 DateTime 对象(您可以使用new DateTime('@'.$timestamp) 将时间戳转换为 DateTime 对象)。

<?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;

?>

【讨论】:

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