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)));