php - 如何比较自定义格式的两个时间戳？

Question

我有两个格式如下的时间戳:substr( date( 'YmdHisu' ), 0, 17 )

20120921105240000
20120921115626000

现在，我该如何比较它们？

我试着做一个简单的方法，比如:

$diff = abs( 20120921105240000 - 20120921115626000 );

但这并没有给我所需的结果，因为时间把它搞砸了。我想做的就是找出它们之间经过了多少分钟(或秒)。

提前致谢。

Answer 1

$date1 = DateTime::createFromFormat('Ymdhisu', '20120921105240000');
$date2 = DateTime::createFromFormat('Ymdhisu', '20120921115626000');
$interval = $date1->diff($date2);
print_r($interval);

/*DateInterval Object
(
    [y] => 0
    [m] => 0
    [d] => 0
    [h] => 1
    [i] => 3
    [s] => 46
    [invert] => 0
    [days] => 0
)*/

$seconds = $date2->getTimestamp()-$date1->getTimestamp();
echo $seconds;

Example

查看更多 info on the object and method .

对于 PHP < 5.3:

方法sscanf + mktime :

$date1 = '20120921105240000';
$date2 = '20120921115626000';

function parse_mydate_string($string) {
    if ($a = sscanf($string, '%4s%2s%2s%2s%2s%2s') {
        if (FALSE !== $r = mktime($a[3], $a[4], $a[5], $a[1], $a[2], $a[0])) {
            return $r;
        }
    }
    throw new InvalidArgumentException('Not the expect date string.');
}

$diff    = parse_mydate_string($date1) - parse_mydate_string($date2);    
$absdiff = abs($diff);

或sscanf + vsprintf + strtotime :

$date1 = '20120921105240000';
$date2 = '20120921115626000';

$date1 = vsprintf('%s-%s-%s %s:%s:%s', sscanf($date1, '%4s%2s%2s%2s%2s%2s'));
$date2 = vsprintf('%s-%s-%s %s:%s:%s', sscanf($date2, '%4s%2s%2s%2s%2s%2s'));

$diff = abs(strtotime($date2) - strtotime($date1));

或多个substr字符串操作 + strtotime:

$date1 = '20120921105240000';
$date2 = '20120921115626000';

$date1 = substr($date1, 0, 4).'-'.substr($date1, 4, 2).'-'.substr($date1, 6, 2).' '.substr($date1, 8, 2).':'.substr($date1, 10, 2).':'.substr($date1, 12, 2);
$date2 = substr($date2, 0, 4).'-'.substr($date2, 4, 2).'-'.substr($date2, 6, 2).' '.substr($date2, 8, 2).':'.substr($date2, 10, 2).':'.substr($date2, 12, 2);

$diff = abs(strtotime($date2) - strtotime($date1));