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, $days);
解决方案作者:Emil H
PHP5.3以下版本,可以尝试以下代码。但是作者特别说明,这种算法算出的可能是一个近似值,而不是精确值。不过作为常规应用应该是没有问题的。
<?php /** * Calculate differences between two dates with precise semantics. Based on PHPs DateTime::diff() * implementation by Derick Rethans. Ported to PHP by Emil H, 2011-05-02. No rights reserved. * * See here for original code: * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/tm2unixtime.c?revision=302890&view=markup * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/interval.c?revision=298973&view=markup */ function _date_range_limit($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)));
解决方案作者:Toby Allen
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本文核心内容节选自Stack Overflow网站的讨论与解答。原文地址:How to calculate the difference between two dates using PHP?。Stack Overflow文章版权遵循CC BY-SA 3.0及署名要求许可协议。转载请保留来源网站、来源页面、文章作者及文章作者的个人资料页面。