如何使用PHP找到两个日期之间的天数?
当前回答
这段代码为我工作,并用PHP 8版本测试:
function numberOfDays($startDate, $endDate)
{
//1) converting dates to timestamps
$startSeconds = strtotime($startDate);
$endSeconds = strtotime($endDate);
//2) Calculating the difference in timestamps
$diffSeconds = $startSeconds - $endSeconds;
//3) converting timestamps to days
$days=round($diffSeconds / 86400);
/* note :
1 day = 24 hours
24 * 60 * 60 = 86400 seconds
*/
//4) printing the number of days
printf("Difference between two dates: ". abs($days) . " Days ");
return abs($days);
}
其他回答
// Change this to the day in the future
$day = 15;
// Change this to the month in the future
$month = 11;
// Change this to the year in the future
$year = 2012;
// $days is the number of days between now and the date in the future
$days = (int)((mktime (0,0,0,$month,$day,$year) - time(void))/86400);
echo "There are $days days until $day/$month/$year";
使用这个:)
$days = (strtotime($endDate) - strtotime($startDate)) / (60 * 60 * 24);
print $days;
现在起作用了
面向对象的风格:
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
程序上的风格:
$datetime1 = date_create('2009-10-11');
$datetime2 = date_create('2009-10-13');
$interval = date_diff($datetime1, $datetime2);
echo $interval->format('%R%a days');
$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;
echo round($datediff / (60 * 60 * 24));
从PHP 5.3及以上版本开始,添加了新的日期/时间函数来获得不同:
$datetime1 = new DateTime("2010-06-20");
$datetime2 = new DateTime("2011-06-22");
$difference = $datetime1->diff($datetime2);
echo 'Difference: '.$difference->y.' years, '
.$difference->m.' months, '
.$difference->d.' days';
print_r($difference);
结果如下:
Difference: 1 years, 0 months, 2 days
DateInterval Object
(
[y] => 1
[m] => 0
[d] => 2
[h] => 0
[i] => 0
[s] => 0
[invert] => 0
[days] => 367
)
希望能有所帮助!
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