本文主要解决PHP如何判断两个日期之间相距的天数，并可换算为月、年。

在PHP5.3以上版本，可以使用strtotime()后的数值直接相减，然后换算为年月日。举例：

$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);

解决方案作者：Emil H

PHP5.3以下版本，可以尝试以下代码。但是作者特别说明，这种算法算出的可能是一个近似值，而不是精确值。不过作为常规应用应该是没有问题的。

/**

* 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)));