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!