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PHP 怎么计算年龄,域名年龄?已知 YYYY-mm-dd。求域名年龄

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Release: 2016-06-13 11:09:57
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PHP 如何计算年龄,域名年龄?已知 YYYY-mm-dd。求域名年龄。
如何计算域名年龄困扰了很久,自己写了一个函数,但是是错误的,因为有的年份是闰年的。

<br /><?php <br /><br /><br />//时间输入必须单位<br />function mathAge($ymd2){<br />    $ymd1 = "2012-2-27";<br />    @list($y1,$m1,$d1) = explode("-",date("Y-m-d", strtotime($ymd1)));<br />    @list($y2,$m2,$d2) = explode("-",date("Y-m-d", strtotime($ymd2)));<br />    <br />    $arr = array(<br />    "年" =>  round( $y1-$y2 ),<br />    "月" =>  round( $m1-$m2 ),<br />    "天" =>  round( $d1-$d2 ),<br />    );<br /><br />    $arr["天"] = round($d1-$d2);<br />    while ($arr["天"] < 0) {<br />        $arr["天"] += 30;<br />        $arr["月"] -= 1;<br />    }<br />    <br />    while ($arr["月"] < 0) {<br />    	$arr["月"] += 12;<br />    	$arr["年"] -= 1;<br />    }<br />    $txt = '';<br />    foreach ($arr as $k => $v)<br />    {<br />        if(!$v)  continue;<br />        $txt .= $v.$k;<br />    }<br />    echo "$ymd1 - $ymd2 = $txt \n";<br />    return $txt;<br />  }<br /><br />  $dateArr = array(<br />  "2011-12-28",<br />  "2011-12-29",<br />  "2011-12-27",<br />  <br />  "2010-12-28",<br />  "2010-12-29",<br />  "2010-12-27",<br />  <br />  "2011-10-1",<br />  "2010-5-1",<br />  "2010-2-28",<br />  <br />  "1995-1-1",<br />  "1995-12-31",<br />  );<br />  foreach ($dateArr as $date)<br />  {<br />      mathAge($date);<br />      <br />  }<br />?><br /><br />
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以上输出测试内容为:


2012-2-27 - 2011-12-28 = 1月29天 <br />2012-2-27 - 2011-12-29 = 1月28天 <br />2012-2-27 - 2011-12-27 = 2月 <br />2012-2-27 - 2010-12-28 = 1年1月29天 <br />2012-2-27 - 2010-12-29 = 1年1月28天 <br />2012-2-27 - 2010-12-27 = 1年2月 <br />2012-2-27 - 2011-10-1 = 4月26天 <br />2012-2-27 - 2010-5-1 = 1年9月26天 <br />2012-2-27 - 2010-2-28 = 1年11月29天 <br />2012-2-27 - 1995-1-1 = 17年1月26天 <br />2012-2-27 - 1995-12-31 = 16年1月26天 
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如何计算好两个日期之间相差的年月?考虑闰年 2月份。
------解决方案--------------------
好吧,我也不是很清楚,不过特意过来顶一下楼主

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