由于每个人都在发布代码示例，这里有另一个版本。

我想要一个函数来显示从秒到年的差异（仅一个单位）。对于超过1天的时段，我希望它在午夜滚动（周一上午10点到周三上午9点是2天前，而不是1天前）。对于超过一个月的时间段，我希望滚动在当月的同一天（包括30/31天的月份和闰年）。

这就是我想到的：

/**
 * Returns how long ago something happened in the past, showing it
 * as n seconds / minutes / hours / days / weeks / months / years ago.
 *
 * For periods over a day, it rolls over at midnight (so doesn't depend
 * on current time of day), and it correctly accounts for month-lengths
 * and leap-years (months and years rollover on current day of month).
 *
 * $param string $timestamp in DateTime format
 * $return string description of interval
 */
function ago($timestamp)
{
    $then = date_create($timestamp);

    // for anything over 1 day, make it rollover on midnight
    $today = date_create('tomorrow'); // ie end of today
    $diff = date_diff($then, $today);

    if ($diff->y > 0) return $diff->y.' year'.($diff->y>1?'s':'').' ago';
    if ($diff->m > 0) return $diff->m.' month'.($diff->m>1?'s':'').' ago';
    $diffW = floor($diff->d / 7);
    if ($diffW > 0) return $diffW.' week'.($diffW>1?'s':'').' ago';
    if ($diff->d > 1) return $diff->d.' day'.($diff->d>1?'s':'').' ago';

    // for anything less than 1 day, base it off 'now'
    $now = date_create();
    $diff = date_diff($then, $now);

    if ($diff->d > 0) return 'yesterday';
    if ($diff->h > 0) return $diff->h.' hour'.($diff->h>1?'s':'').' ago';
    if ($diff->i > 0) return $diff->i.' minute'.($diff->i>1?'s':'').' ago';
    return $diff->s.' second'.($diff->s==1?'':'s').' ago';
}

我更喜欢使用date_create和date_diff对象。

代码：

$date1 = date_create("2007-03-24");
$date2 = date_create("2009-06-26");

$dateDifference = date_diff($date1, $date2)->format('%y years, %m months and %d days');

echo $dateDifference;

输出：

2 years, 3 months and 2 days

有关更多信息，请阅读PHP date_diff手册

根据手册date_diff是的别名日期时间：：diff（）

我投票支持jurka的答案，因为这是我最喜欢的，但我有一个pre-php.5.3版本。。。

我发现自己在解决一个类似的问题——这就是我最初如何回答这个问题——但只是需要时间上的差异。但我的函数也很好地解决了这个问题，而且我自己的库中没有任何地方可以将它保存在不会丢失和遗忘的地方，所以……希望这对某人有用。

/**
 *
 * @param DateTime $oDate1
 * @param DateTime $oDate2
 * @return array 
 */
function date_diff_array(DateTime $oDate1, DateTime $oDate2) {
    $aIntervals = array(
        'year'   => 0,
        'month'  => 0,
        'week'   => 0,
        'day'    => 0,
        'hour'   => 0,
        'minute' => 0,
        'second' => 0,
    );

    foreach($aIntervals as $sInterval => &$iInterval) {
        while($oDate1 <= $oDate2){ 
            $oDate1->modify('+1 ' . $sInterval);
            if ($oDate1 > $oDate2) {
                $oDate1->modify('-1 ' . $sInterval);
                break;
            } else {
                $iInterval++;
            }
        }
    }

    return $aIntervals;
}

测试：

$oDate = new DateTime();
$oDate->modify('+111402189 seconds');
var_dump($oDate);
var_dump(date_diff_array(new DateTime(), $oDate));

结果是：

object(DateTime)[2]
  public 'date' => string '2014-04-29 18:52:51' (length=19)
  public 'timezone_type' => int 3
  public 'timezone' => string 'America/New_York' (length=16)

array
  'year'   => int 3
  'month'  => int 6
  'week'   => int 1
  'day'    => int 4
  'hour'   => int 9
  'minute' => int 3
  'second' => int 8

我从这里得到了最初的想法，我对其进行了修改以供使用（我希望我的修改也会显示在该页面上）。

通过从$aIntervals数组中删除不需要的间隔（例如“周”），或者添加$aExclude参数，或者在输出字符串时过滤掉它们，可以非常容易地删除它们。

// If you just want to see the year difference then use this function.
// Using the logic I've created you may also create month and day difference
// which I did not provide here so you may have the efforts to use your brain.
// :)
$date1='2009-01-01';
$date2='2010-01-01';
echo getYearDifference ($date1,$date2);
function getYearDifference($date1=strtotime($date1),$date2=strtotime($date2)){
    $year = 0;
    while($date2 > $date1 = strtotime('+1 year', $date1)){
        ++$year;
    }
    return $year;
}

查看以下链接。这是迄今为止我找到的最好的答案

function dateDiff ($d1, $d2) {

    // Return the number of days between the two dates:    
    return round(abs(strtotime($d1) - strtotime($d2))/86400);

} // end function dateDiff

当你通过日期参数。函数使用PHP ABS（）绝对值始终返回正数作为两者之间的天数日期。请记住，两个日期之间的天数不是包括两个日期。因此，如果您正在寻找天数由输入日期之间的所有日期表示，您需要向该函数的结果添加一（1）。例如，差异（由上述函数返回）2013-02-09和2013-02-14之间的值为5。但天数或日期范围2013-02-09-2013-02-14表示的日期为6。

http://www.bizinfosys.com/php/date-difference.html

这将尝试检测是否给定了时间戳，并将返回未来的日期/时间作为负值：

<?php

function time_diff($start, $end = NULL, $convert_to_timestamp = FALSE) {
  // If $convert_to_timestamp is not explicitly set to TRUE,
  // check to see if it was accidental:
  if ($convert_to_timestamp || !is_numeric($start)) {
    // If $convert_to_timestamp is TRUE, convert to timestamp:
    $timestamp_start = strtotime($start);
  }
  else {
    // Otherwise, leave it as a timestamp:
    $timestamp_start = $start;
  }
  // Same as above, but make sure $end has actually been overridden with a non-null,
  // non-empty, non-numeric value:
  if (!is_null($end) && (!empty($end) && !is_numeric($end))) {
    $timestamp_end = strtotime($end);
  }
  else {
    // If $end is NULL or empty and non-numeric value, assume the end time desired
    // is the current time (useful for age, etc):
    $timestamp_end = time();
  }
  // Regardless, set the start and end times to an integer:
  $start_time = (int) $timestamp_start;
  $end_time = (int) $timestamp_end;

  // Assign these values as the params for $then and $now:
  $start_time_var = 'start_time';
  $end_time_var = 'end_time';
  // Use this to determine if the output is positive (time passed) or negative (future):
  $pos_neg = 1;

  // If the end time is at a later time than the start time, do the opposite:
  if ($end_time <= $start_time) {
    $start_time_var = 'end_time';
    $end_time_var = 'start_time';
    $pos_neg = -1;
  }

  // Convert everything to the proper format, and do some math:
  $then = new DateTime(date('Y-m-d H:i:s', $$start_time_var));
  $now = new DateTime(date('Y-m-d H:i:s', $$end_time_var));

  $years_then = $then->format('Y');
  $years_now = $now->format('Y');
  $years = $years_now - $years_then;

  $months_then = $then->format('m');
  $months_now = $now->format('m');
  $months = $months_now - $months_then;

  $days_then = $then->format('d');
  $days_now = $now->format('d');
  $days = $days_now - $days_then;

  $hours_then = $then->format('H');
  $hours_now = $now->format('H');
  $hours = $hours_now - $hours_then;

  $minutes_then = $then->format('i');
  $minutes_now = $now->format('i');
  $minutes = $minutes_now - $minutes_then;

  $seconds_then = $then->format('s');
  $seconds_now = $now->format('s');
  $seconds = $seconds_now - $seconds_then;

  if ($seconds < 0) {
    $minutes -= 1;
    $seconds += 60;
  }
  if ($minutes < 0) {
    $hours -= 1;
    $minutes += 60;
  }
  if ($hours < 0) {
    $days -= 1;
    $hours += 24;
  }
  $months_last = $months_now - 1;
  if ($months_now == 1) {
    $years_now -= 1;
    $months_last = 12;
  }

  // "Thirty days hath September, April, June, and November" ;)
  if ($months_last == 9 || $months_last == 4 || $months_last == 6 || $months_last == 11) {
    $days_last_month = 30;
  }
  else if ($months_last == 2) {
    // Factor in leap years:
    if (($years_now % 4) == 0) {
      $days_last_month = 29;
    }
    else {
      $days_last_month = 28;
    }
  }
  else {
    $days_last_month = 31;
  }
  if ($days < 0) {
    $months -= 1;
    $days += $days_last_month;
  }
  if ($months < 0) {
    $years -= 1;
    $months += 12;
  }

  // Finally, multiply each value by either 1 (in which case it will stay the same),
  // or by -1 (in which case it will become negative, for future dates).
  // Note: 0 * 1 == 0 * -1 == 0
  $out = new stdClass;
  $out->years = (int) $years * $pos_neg;
  $out->months = (int) $months * $pos_neg;
  $out->days = (int) $days * $pos_neg;
  $out->hours = (int) $hours * $pos_neg;
  $out->minutes = (int) $minutes * $pos_neg;
  $out->seconds = (int) $seconds * $pos_neg;
  return $out;
}

示例用法：

<?php
  $birthday = 'June 2, 1971';
  $check_age_for_this_date = 'June 3, 1999 8:53pm';
  $age = time_diff($birthday, $check_age_for_this_date)->years;
  print $age;// 28

Or:

<?php
  $christmas_2020 = 'December 25, 2020';
  $countdown = time_diff($christmas_2020);
  print_r($countdown);