<?php
$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");
?>

上面的代码用的很简单。谢谢。

如果你有以秒为单位的时间(即unix时间戳)，那么你可以简单地减去时间并除以86400(秒/天)

我阅读了所有以前的解决方案，没有一个使用PHP 5.3工具:DateTime::Diff和DateInterval::Days

DateInterval::Days精确地包含日期之间的天数。没有必要创造一些特别和奇异的东西。

/**
 * We suppose that PHP is configured in UTC
 * php.ini configuration:
 * [Date]
 * ; Defines the default timezone used by the date functions
 * ; http://php.net/date.timezone
 * date.timezone = UTC
 * @link http://php.net/date.timezone
 */

/**
 * getDaysBetween2Dates
 *
 * Return the difference of days between $date1 and $date2 ($date1 - $date2)
 * if $absolute parameter is false, the return value is negative if $date2 is after than $date1
 *
 * @param DateTime $date1
 * @param DateTime $date2
 * @param Boolean $absolute
 *            = true
 * @return integer
 */
function getDaysBetween2Dates(DateTime $date1, DateTime $date2, $absolute = true)
{
    $interval = $date2->diff($date1);
    // if we have to take in account the relative position (!$absolute) and the relative position is negative,
    // we return negatif value otherwise, we return the absolute value
    return (!$absolute and $interval->invert) ? - $interval->days : $interval->days;
}

echo '<h3>2020-03-01 - 2020-02-01: 29 days as it\'s a standard leap year</h3>';
echo getDaysBetween2Dates(new DateTime("2020-03-01"), new DateTime("2020-02-01"), false);

echo '<h3>1900-03-01 - 1900-02-01: 28 days as it\'s a "standard" century</h3>';
echo getDaysBetween2Dates(new DateTime("1900-03-01"), new DateTime("1900-02-01"), false);

echo '<h3>2000-03-01 - 2000-02-01: 29 days as it\'s a century multiple of 400: 2000=400x5</h3>';
echo getDaysBetween2Dates(new DateTime("2000-03-01"), new DateTime("2000-02-01"), false);

echo '<h3>2020-03-01 - 2020-04-01: -28 days as 2020-03-01 is before 2020-04-01</h3>';
echo getDaysBetween2Dates(new DateTime("2020-02-01"), new DateTime("2020-03-01"), false);

将日期转换为unix时间戳，然后从另一个时间戳中减去一个日期。这将得到以秒为单位的差值，然后除以86400(一天中的秒数)，得到该范围内的大约天数。

如果你的日期格式为25.1.2010,01/25/2010或2010-01-25，你可以使用strtotime函数:

$start = strtotime('2010-01-25');
$end = strtotime('2010-02-20');

$days_between = ceil(abs($end - $start) / 86400);

使用ceil将天数四舍五入到下一个全天。如果您希望获得这两个日期之间的完整天数，则使用floor。

如果日期已经是unix时间戳格式，则可以跳过转换，只执行$days_between部分。对于更奇特的日期格式，您可能必须进行一些自定义解析以使其正确。

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);
}

$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));