如何使用php获取假期日期

Question

例如：

1. 一月的第三个星期一 马丁路德金纪念日
2. 2 月的第三个星期一 总统日（华盛顿诞辰）
3. 三月的最后一个星期日 复活节
4. 5 月的最后一个星期一 阵亡将士纪念日

我正在尝试获取这些日期，这样我就可以在我的日历上标记它，而无需手动将未来几年的所有内容都放入。

---

修改答案！！

$curYir = date("Y");//current year

$MLK = date('Y-m-d', strtotime("january $curYir third monday")); //marthin luthor king day
$PD = date('Y-m-d', strtotime("february $curYir third monday")); //presidents day
$Est =  date('Y-m-d', easter_date($curYir))); // easter 
$MDay = date('Y-m-d', strtotime("may $curYir first monday")); // memorial day
//("may $curYir last monday") will give you the last monday of may 1967
//much better to determine it by a loop
      $eMDay = explode("-",$MDay);
      $year = $eMDay[0];
      $month = $eMDay[1];
      $day = $eMDay[2];

      while($day <= 31){
          $day = $day + 7;
      }
      if($day > 31)
      $day = $day - 7;

      $MDay = $year.'-'.$month.'-'.$day;
$LD = date('Y-m-d', strtotime("september $curYir first monday"));  //labor day
$CD = date('Y-m-d', strtotime("october $curYir third monday")); //columbus day
$TH = date('Y-m-d', strtotime("november $curYir first thursday")); // thanks giving 
//("november $curYir last thursday") will give you the last thursday of november 1967
//much better to determine it by a loop
      $eTH = explode("-",$TH);
      $year = $eTH[0];
      $month = $eTH[1];
      $day = $eTH[2];

      while($day <= 30){
          $day = $day + 7;
      }
      if($day > 30)
      //watch out for the days in the month November only have 30
      $day = $day - 7;

      $TH = $year.'-'.$month.'-'.$day;

Answer 1

您可以利用 php.ini 的这个功能。 strtotime

$currentYear = date("Y");

MLK 日 -

echo date('Y-m-d', strtotime("third monday of january $currentYear"));

总统日 -

echo date('Y-m-d', strtotime("third monday/OD February $currentYear"));

复活节 -

echo date('Y-m-d', strtotime("last sunday of march $currentYear"));

阵亡将士纪念日 -

echo date('Y-m-d', strtotime("last monday of may $currentYear"));

Answer 2

strtotime 函数在这里很有用，但它似乎在解释英语词汇方面有一些特殊性。请务必仔细检查您的措辞并检查您的结果。

例如，这似乎应该可以工作，但它将返回 4 月的最后一个星期一 ($currentYear = 2014)！

echo "Memorial Day " . date('F d, Y', strtotime("may $currentYear last monday")) . "<br />";

但是，此措辞将返回正确的 2014 年假期。

echo "Memorial Day " . date('F d, Y', strtotime("last monday of May $currentYear")) . "<br />"; 

这将为您提供明年假期的正确日期。

echo "Memorial Day " . date('F d, Y', strtotime("last monday of May $currentYear+1")) . "<br />";

PHP 文档页面在其使用上有额外的 cmets。 保重

Answer 3

这种方式对我有用。

function observed_date($holiday){
    $day = date("w", strtotime($holiday));
    if($day == 6) {
        $observed_date = $holiday -1;
    } elseif ($day == 0) {
        $observed_date = $holiday +1;
    } else {
        $observed_date = $holiday;
    }
    return $observed_date;
}

function get_holiday($holiday_name) {

    $currentYear = date('Y');

    switch ($holiday_name) {
        // New Years Day
        case "new_year":
            $holiday = observed_date(date('Ymd', strtotime("first day of january $currentYear")));
            break;
        // Martin Luther King, Jr. Day
        case "mlk_day":
            $holiday = date('Ymd', strtotime("january $currentYear third monday"));
            break;
        // President's Day
        case "presidents_day":
            $holiday = date('Ymd', strtotime("february $currentYear third monday"));
            break;
        // Memorial Day
        case "memorial_day":
            $holiday = (new DateTime("Last monday of May"))->format("Ymd");
            break;
        // Independence Day
        case "independence_day":
            $holiday = observed_date(date('Ymd', strtotime("july 4 $currentYear")));
            break;
        // Labor Day
        case "labor_day":
            $holiday = date('Ymd', strtotime("september $currentYear first monday"));
            break;
        // Columbus Day
        case "columbus_day":
            $holiday = date('Ymd', strtotime("october $currentYear second monday"));
            break;
        // Veteran's Day
        case "veterans_day":
            $holiday = observed_date(date('Ymd', strtotime("november 11 $currentYear")));
            break;
        // Thanksgiving Day
        case "thanksgiving_day":
            $holiday = date('Ymd', strtotime("november $currentYear fourth thursday"));
            break;
        // Christmas Day
        case "christmas_day":
        $holiday = observed_date(date('Ymd', strtotime("december 25 $currentYear")));
            break;

        default:
            $holiday = "";
            break;
    }
    return $holiday;

}

然后就可以调用函数echo get_holiday('new_year');打印出假期了

2015 年的结果应该是20150101

Answer 4

在德国移动圣日的解决方案：返回一个 'Y-m-d' 格式的字符串数组，包括耶稣受难日、复活节星期一、升天和五旬节星期一。 每天为复活节创建一个新的 DateTime 对象以增强可读性。

function beweglicheFeiertage( int $year ):array
{
    //Karfreitag - good friday
    $return = array();
    $easterDate = DateTime::createFromFormat('U', easter_date($year) );
    $easterDate->modify('- 1 days');
    $return[] = $easterDate->format('Y-m-d');
    //Ostermontag easter monday
    $easterDate = DateTime::createFromFormat('U', easter_date($year) );
    $easterDate->modify('+ 2 day');
    $return[] = $easterDate->format('Y-m-d');
    //Himmelfahrt ascencion
    $easterDate = DateTime::createFromFormat('U', easter_date($year) );
    $easterDate->modify('+ 40 days');//go to Ascencionday
    $return[] = $easterDate->format('Y-m-d');
    //Pfingstmontag - pentecost monday
    $easterDate = DateTime::createFromFormat('U', easter_date($year) );
    $easterDate->modify('+ 51 days');//go to Pentecost Monday
    $return[] = $easterDate->format('Y-m-d');

    return $return;
}

Answer 5

我创建了一个函数，用于计算任何给定年份的所有联邦假日（美国），并将它们返回到一个数组中。我已经用几个不同的年份（从 2016 年到 2024 年）对其进行了测试。以下是代码，如果您发现任何计算错误，请告诉我：

    //function that checks if a holiday lands on saturday/sunday and so we can move them to a friday/monday respectively
    private function getObservedDate($holidayDate){

        $dayofweek = date("w", strtotime($holidayDate));

        if ($dayofweek == 6) $holidayDate = date('m/d/Y', strtotime("$holidayDate - 1 days")); //saturday moves to friday
        else if ($dayofweek == 0)  $holidayDate = date('m/d/Y', strtotime("$holidayDate + 1 days"));  //sunday moves monday

        return $holidayDate;
    }


    //function that calculates the holidays for any given year
    private function getFederalHolidaysForYear($year){

        $NY = $this->getObservedDate( date('m/d/Y', strtotime("1/1/$year")) ); //new years day

        $MLK = $this->getObservedDate( date('m/d/Y', strtotime("third monday of january $year")) );  //martin luther king day

        $PD = $this->getObservedDate( date('m/d/Y', strtotime("third monday of february $year")) ); ; //presidents day

        $MDay = $this->getObservedDate( date('m/d/Y', strtotime("last monday of May $year")) ); //memorial day  

        $IDay = $this->getObservedDate( date('m/d/Y', strtotime("7/4/$year")) );  // independence day

        $LD = $this->getObservedDate( date('m/d/Y', strtotime("first monday of september $year")) ); //labor day

        $VD = $this->getObservedDate( date('m/d/Y', strtotime("11/11/$year")) ); //veterans day

        $ColD =$this->getObservedDate( date('m/d/Y', strtotime("second monday of october $year")) ); //columbus day   

        $TG = $this->getObservedDate( date('m/d/Y', strtotime("last thursday of november $year")) ); // thanksgiving       

        $CD = $this->getObservedDate( date('m/d/Y', strtotime("12/25/$year")) );  //christmas day 

        $nonWorkingDays = array();

        array_push($nonWorkingDays, $NY, $MLK, $PD, $MDay, $IDay, $LD, $ColD, $VD, $TG, $CD);

        return $nonWorkingDays;
    }

Answer 6

如果您愿意一劳永逸地在这里手动输入日期，您可以在这里使用我的自定义功能。它简短，简单，可以满足每个人的需要。如果需要，您可以自己修改它。此外，如果您可以找到一种方法以给定格式动态传递数组中的日期，那么您可以选择以这种方式对其进行修改。但是，我下面的函数只适用于那些有 STATIC 日期的假期。您可能必须手动管理每年有动态或变化天的假期，例如阵亡将士纪念日、劳动节等。

我的自定义 PHP 函数

如果您要检查国际假期，这里是我为它编写的一个函数。

function checkHoliday($date){
  if(date('l', strtotime($date)) == 'Saturday'){
    return "Saturday";
  }else if(date('l', strtotime($date)) == 'Sunday'){
    return "Sunday";
  }else{
    $receivedDate = date('d M', strtotime($date));

    $holiday = array(
      '01 Jan' => 'New Year Day',
      '18 Jan' => 'Martin Luther King Day',
      '22 Feb' => 'Washington\'s Birthday',
      '05 Jul' => 'Independence Day',
      '11 Nov' => 'Veterans Day',
      '24 Dec' => 'Christmas Eve',
      '25 Dec' => 'Christmas Day',
      '31 Dec' => 'New Year Eve'
    );

    foreach($holiday as $key => $value){
      if($receivedDate == $key){
        return $value;
      }
    }
  }
}

这样称呼它

$date = '01/01/2021';

echo checkHoliday($date);