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';
}

查看小时、分钟和秒。。

$date1 = "2008-11-01 22:45:00"; 

$date2 = "2009-12-04 13:44:01"; 

$diff = abs(strtotime($date2) - strtotime($date1)); 

$years   = floor($diff / (365*60*60*24)); 
$months  = floor(($diff - $years * 365*60*60*24) / (30*60*60*24)); 
$days    = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

$hours   = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24)/ (60*60)); 

$minuts  = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60)/ 60); 

$seconds = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60 - $minuts*60)); 

printf("%d years, %d months, %d days, %d hours, %d minuts\n, %d seconds\n", $years, $months, $days, $hours, $minuts, $seconds); 

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);

简单的功能

function time_difference($time_1, $time_2, $limit = null)
{

    $val_1 = new DateTime($time_1);
    $val_2 = new DateTime($time_2);

    $interval = $val_1->diff($val_2);

    $output = array(
        "year" => $interval->y,
        "month" => $interval->m,
        "day" => $interval->d,
        "hour" => $interval->h,
        "minute" => $interval->i,
        "second" => $interval->s
    );

    $return = "";
    foreach ($output AS $key => $value) {

        if ($value == 1)
            $return .= $value . " " . $key . " ";
        elseif ($value >= 1)
            $return .= $value . " " . $key . "s ";

        if ($key == $limit)
            return trim($return);
    }
    return trim($return);
}

像这样使用

回波时间差（$time_1，$time_2，“天”）；

将返回2年8个月2天

对于遗留代码（PHP＜5.3）使用此选项。有关最新的解决方案，请参阅下面jurka的答案

您可以使用strtotime（）将两个日期转换为unix时间，然后计算它们之间的秒数。由此很容易计算出不同的时间段。

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

$diff = abs(strtotime($date2) - strtotime($date1));

$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

printf("%d years, %d months, %d days\n", $years, $months, $days);

编辑：显然，尤尔卡在下面描述了这样做的首选方式。如果您没有PHP5.3或更高版本，通常只建议使用我的代码。

评论中的一些人指出，上面的代码只是一个近似值。我仍然认为，对于大多数目的来说，这是很好的，因为范围的使用更多地是为了提供时间流逝或剩余时间的感觉，而不是为了提供精度-如果你想这样做，只需输出日期。

尽管如此，我还是决定解决这些投诉。如果您确实需要一个精确的范围，但还没有访问PHP 5.3，请使用下面的代码（它也可以在PHP 4中使用）。这是PHP内部用于计算范围的代码的直接端口，但不考虑夏时制。这意味着它最多休息一个小时，但除此之外，它应该是正确的。

<?php

/**
 * Calculate differences between two dates with precise semantics. Based on PHPs DateTime::diff()
 * implementation by Derick Rethans. Ported to PHP by Emil H, 2011-05-02. No rights reserved.
 * 
 * See here for original code:
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/tm2unixtime.c?revision=302890&view=markup
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/interval.c?revision=298973&view=markup
 */

function _date_range_limit($start, $end, $adj, $a, $b, $result)
{
    if ($result[$a] < $start) {
        $result[$b] -= intval(($start - $result[$a] - 1) / $adj) + 1;
        $result[$a] += $adj * intval(($start - $result[$a] - 1) / $adj + 1);
    }

    if ($result[$a] >= $end) {
        $result[$b] += intval($result[$a] / $adj);
        $result[$a] -= $adj * intval($result[$a] / $adj);
    }

    return $result;
}

function _date_range_limit_days($base, $result)
{
    $days_in_month_leap = array(31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
    $days_in_month = array(31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);

    _date_range_limit(1, 13, 12, "m", "y", &$base);

    $year = $base["y"];
    $month = $base["m"];

    if (!$result["invert"]) {
        while ($result["d"] < 0) {
            $month--;
            if ($month < 1) {
                $month += 12;
                $year--;
            }

            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;
        }
    } else {
        while ($result["d"] < 0) {
            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;

            $month++;
            if ($month > 12) {
                $month -= 12;
                $year++;
            }
        }
    }

    return $result;
}

function _date_normalize($base, $result)
{
    $result = _date_range_limit(0, 60, 60, "s", "i", $result);
    $result = _date_range_limit(0, 60, 60, "i", "h", $result);
    $result = _date_range_limit(0, 24, 24, "h", "d", $result);
    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    $result = _date_range_limit_days(&$base, &$result);

    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    return $result;
}

/**
 * Accepts two unix timestamps.
 */
function _date_diff($one, $two)
{
    $invert = false;
    if ($one > $two) {
        list($one, $two) = array($two, $one);
        $invert = true;
    }

    $key = array("y", "m", "d", "h", "i", "s");
    $a = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $one))));
    $b = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $two))));

    $result = array();
    $result["y"] = $b["y"] - $a["y"];
    $result["m"] = $b["m"] - $a["m"];
    $result["d"] = $b["d"] - $a["d"];
    $result["h"] = $b["h"] - $a["h"];
    $result["i"] = $b["i"] - $a["i"];
    $result["s"] = $b["s"] - $a["s"];
    $result["invert"] = $invert ? 1 : 0;
    $result["days"] = intval(abs(($one - $two)/86400));

    if ($invert) {
        _date_normalize(&$a, &$result);
    } else {
        _date_normalize(&$b, &$result);
    }

    return $result;
}

$date = "1986-11-10 19:37:22";

print_r(_date_diff(strtotime($date), time()));
print_r(_date_diff(time(), strtotime($date)));

我有一些简单的逻辑：

<?php
    per_days_diff('2011-12-12','2011-12-29')
    function per_days_diff($start_date, $end_date) {
        $per_days = 0;
        $noOfWeek = 0;
        $noOfWeekEnd = 0;
        $highSeason=array("7", "8");

        $current_date = strtotime($start_date);
        $current_date += (24 * 3600);
        $end_date = strtotime($end_date);

        $seassion = (in_array(date('m', $current_date), $highSeason))?"2":"1";

        $noOfdays = array('');

        while ($current_date <= $end_date) {
            if ($current_date <= $end_date) {
                $date = date('N', $current_date);
                array_push($noOfdays,$date);
                $current_date = strtotime('+1 day', $current_date);
            }
        }

        $finalDays = array_shift($noOfdays);
        //print_r($noOfdays);
        $weekFirst = array("week"=>array(),"weekEnd"=>array());
        for($i = 0; $i < count($noOfdays); $i++)
        {
            if ($noOfdays[$i] == 1)
            {
                //echo "This is week";
                //echo "<br/>";
                if($noOfdays[$i+6]==7)
                {
                    $noOfWeek++;
                    $i=$i+6;
                }
                else
                {
                    $per_days++;
                }
                //array_push($weekFirst["week"],$day);
            }
            else if($noOfdays[$i]==5)
            {
                //echo "This is weekend";
                //echo "<br/>";
                if($noOfdays[$i+2] ==7)
                {
                    $noOfWeekEnd++;
                    $i = $i+2;
                }
                else
                {
                    $per_days++;
                }
                //echo "After weekend value:- ".$i;
                //echo "<br/>";
            }
            else
            {
                $per_days++;
            }
        }

        /*echo $noOfWeek;
          echo "<br/>";
          echo $noOfWeekEnd;
          echo "<br/>";
          print_r($per_days);
          echo "<br/>";
          print_r($weekFirst);
        */

        $duration = array("weeks"=>$noOfWeek, "weekends"=>$noOfWeekEnd, "perDay"=>$per_days, "seassion"=>$seassion);
        return $duration;
      ?>