我想使用php将一个日期减去另一个日期,并以天 - 小时 - 分 - 秒格式显示结果.我是怎么用php做的.我尝试使用时间戳,但它没有给出正确的值.请给出建议
例如:从2012-04-27 19:30:56到2012-04-27 19:37:56
我用这个代码
if(strtotime($history['datetimestamp']) > strtotime($lasttime)) {
$totalelapsed1 = (strtotime($history['datetimestamp'])-strtotime($lasttime));
if($totalelapsed1 > 60 ) {
$sec = $totalelapsed1%60;
$min = round($totalelapsed1/60 , 0);
$minute = $min + $minute;
$second = $sec + $second;
// echo $min. " min " .$sec." sec";
} else {
//echo "0 min " . $totalelapsed1." sec";
$minute = 0 + $minute;
$second = $totalelapsed1 + $second;
}
} else {
$minute = 0 + $minute;
$second = 0 + $second;
// echo "0 min 0 sec";
}
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Jav*_*der 10
你必须这样使用: -
Run Code Online (Sandbox Code Playgroud)<?php //$now = new DateTime(); // current date/time $now = new DateTime("2010-07-28 01:11:50"); $ref = new DateTime("2010-07-30 05:56:40"); $diff = $now->diff($ref); printf('%d days, %d hours, %d minutes', $diff->d, $diff->h, $diff->i); prints 2 days, 4 hours, 44 minutes见http://docs.php.net/datetime.diff
编辑:但您也可以将问题更多地转移到数据库端,例如通过在表中存储过期日期/时间然后执行类似的查询...
WHERE key='7gedufgweufg' AND expires<Now()许多rdbms对日期/时间算法有合理/良好的支持.
链接网址: - 如何减去两个日期和时间以获得差异
http://www.webpronews.com/calculating-the-difference-between-two-dates-using-php-2005-11
我建议使用DateTime和DateInterval对象.
$date1 = new DateTime("2007-03-24");
$date2 = new DateTime("2009-06-26");
$interval = $date1->diff($date2);
echo "days difference ".$interval->d." days ";
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阅读更多php DateTime :: diff手册