我想检查php中两个日期之间是否有任何坐姿或周日。
$fromDate = '03/21/2016';
$toDate = '03/28/2016';
我尝试了这里提出的其他解决方案，但它们效果不佳。我想要它在 laravel 5.2 中，所以如果有任何简单的方法来解决这个问题，请指导我。谢谢！
更新我还想了解两个日期之间是否有工作日（周一至周五）。我需要它，因为用户只能选择周六和周日，所以在这种情况下，我必须对他隐藏工作日选项。所以我需要的是一种告诉我开始和结束日期是否有周末或工作日或两者兼而有之的方法。
您可以使用
date("N")来获取一周中的当前日期，并添加日期之间的天数差...如果此值大于或等于 6，则为周末之间或周末之间的一个日期。
//Get current day of week. For example Friday = 5
$day_of_week = date("N", strtotime($fromDate));
$days = $day_of_week + (strtotime($toDate) - strtotime($fromDate)) / (60*60*24);
//strtotime(...) - convert a string date to unixtimestamp in seconds.
//The difference between strtotime($toDate) - strtotime($fromDate) is the number of seconds between this 2 dates.
//We divide by (60*60*24) to know the number of days between these 2 dates (60 - seconds, 60 - minutes, 24 - hours)
//After that add the number of days between these 2 dates to a day of the week. So if the first date is Friday and days between these 2 dates are: 3 the sum will be 5+3 = 8 and it's bigger than 6 so we know it's a weekend between.
if($days >= 6){
//we have a weekend. Because day of week of first date + how many days between this 2 dates are greater or equal to 6 (6=Saturday)
} else {
//we don't have a weekend
}
试试
<?php
$is_weekend = 0;
$fromDate = strtotime('2016-03-21');
$toDate = strtotime('2016-03-28');
while (date("Y-m-d", $fromDate) != date("Y-m-d", $toDate)) {
$day = date("w", $fromDate);
if ($day == 0 || $day == 6) {
$is_weekend = 1;
}
$fromDate = strtotime(date("Y-m-d", $fromDate) . "+1 day");
}
echo $is_weekend;
锄头它有助于:)