获取 3 个下一个工作日期，跳过周末和节假日

Question

我正在尝试使用 php 获取接下来的 3 天，但如果日期是周末或定义的假期，我想跳过该日期。

例如，如果日期是 2013 年 12 月 23 日星期一，我的假期日期是array('2013-12-24', '2013-12-25'); 脚本将返回

Monday, December 23, 2013
Thursday, December 26, 2013
Friday, December 27, 2013
Monday, December 30, 2013

她是我当前的代码：

$arr_date = explode('-', '2013-12-23');
$counter = 0;
for ($iLoop = 0; $iLoop < 4; $iLoop++)
{
    $dayOfTheWeek = date("N", mktime(0, 0, 0, $arr_date[1], $arr_date[2]+$counter, $arr_date[0]));
    if ($dayOfTheWeek == 6) { $counter += 2; }
    $date = date("Y-m-d", mktime(0, 0, 0, $arr_date[1], $arr_date[2]+$counter, $arr_date[0]));
    echo $date;
    $counter++;
}

我遇到的问题是我不知道如何排除假期日期。

Answer 1

使用 aDateTime递归接下来的几天，并使用字符串比较检查下一个生成的日期是否属于假期或周末数组in_array

$holidays = array('12-24', '12-25');
$weekend = array('Sun','Sat');
$date = new DateTime('2013-12-23');
$nextDay = clone $date;
$i = 0; // We have 0 future dates to start with
$nextDates = array(); // Empty array to hold the next 3 dates
while ($i < 3)
{
    $nextDay->add(new DateInterval('P1D')); // Add 1 day
    if (in_array($nextDay->format('m-d'), $holidays)) continue; // Don't include year to ensure the check is year independent
    // Note that you may need to do more complicated things for special holidays that don't use specific dates like "the last Friday of this month"
    if (in_array($nextDay->format('D'), $weekend)) continue;
    // These next lines will only execute if continue isn't called for this iteration
    $nextDates[] = $nextDay->format('Y-m-d');
    $i++;
}

isset()使用for的建议O(1)代替O(n)：

$holidays = array('12-24' => '', '12-25' => '');
$weekend = array('Sun' => '','Sat' => '');
$date = new DateTime('2013-12-23');
$dayInterval = new DateInterval('P1D');
$nextDay = clone $date;
$i = 0;
$nextDates = array();
while ($i < 3)
{
    $nextDay->add($dayInterval);
    if (isset($holidays[$nextDay->format('m-d')])) continue;
    if (isset($weekend[$nextDay->format('D')])) continue;
    $nextDates[] = $nextDay->format('Y-m-d');
    $i++;
}

Answer 2

这是一个简单的示例，说明如何获取两个日期之间的工作日数；您可以简单地修改它以满足您的需求：

function number_of_working_dates($from, $days) {
    $workingDays = [1, 2, 3, 4, 5]; # date format = N (1 = Monday, ...)
    $holidayDays = ['*-12-25', '*-01-01', '2013-12-24', '2013-12-25']; # variable and fixed holidays

    $from = new DateTime($from);
    $dates = [];
    $dates[] = $from->format('Y-m-d');
    while ($days) {
        $from->modify('+1 day');

        if (!in_array($from->format('N'), $workingDays)) continue;
        if (in_array($from->format('Y-m-d'), $holidayDays)) continue;
        if (in_array($from->format('*-m-d'), $holidayDays)) continue;

        $dates[] = $from->format('Y-m-d');
        $days--;
    }
    return $dates;
}

print_r( number_of_working_dates('2013-12-23', 3) );

demo