我有两个表格日期:

Start Date: 2007-03-24 
End Date: 2009-06-26

现在我需要通过以下形式找到这两者之间的区别:

2 years, 3 months and 2 days

如何在PHP中执行此操作?


当前回答

这是我的职责。所需PHP>=5.3.4。它使用DateTime类。非常快,很快,可以区分两个日期,甚至所谓的“开始时间”。

if(function_exists('grk_Datetime_Since') === FALSE){
    function grk_Datetime_Since($From, $To='', $Prefix='', $Suffix=' ago', $Words=array()){
        #   Est-ce qu'on calcul jusqu'à un moment précis ? Probablement pas, on utilise maintenant
        if(empty($To) === TRUE){
            $To = time();
        }

        #   On va s'assurer que $From est numérique
        if(is_int($From) === FALSE){
            $From = strtotime($From);
        };

        #   On va s'assurer que $To est numérique
        if(is_int($To) === FALSE){
            $To = strtotime($To);
        }

        #   On a une erreur ?
        if($From === FALSE OR $From === -1 OR $To === FALSE OR $To === -1){
            return FALSE;
        }

        #   On va créer deux objets de date
        $From = new DateTime(@date('Y-m-d H:i:s', $From), new DateTimeZone('GMT'));
        $To   = new DateTime(@date('Y-m-d H:i:s', $To), new DateTimeZone('GMT'));

        #   On va calculer la différence entre $From et $To
        if(($Diff = $From->diff($To)) === FALSE){
            return FALSE;
        }

        #   On va merger le tableau des noms (par défaut, anglais)
        $Words = array_merge(array(
            'year'      => 'year',
            'years'     => 'years',
            'month'     => 'month',
            'months'    => 'months',
            'week'      => 'week',
            'weeks'     => 'weeks',
            'day'       => 'day',
            'days'      => 'days',
            'hour'      => 'hour',
            'hours'     => 'hours',
            'minute'    => 'minute',
            'minutes'   => 'minutes',
            'second'    => 'second',
            'seconds'   => 'seconds'
        ), $Words);

        #   On va créer la chaîne maintenant
        if($Diff->y > 1){
            $Text = $Diff->y.' '.$Words['years'];
        } elseif($Diff->y == 1){
            $Text = '1 '.$Words['year'];
        } elseif($Diff->m > 1){
            $Text = $Diff->m.' '.$Words['months'];
        } elseif($Diff->m == 1){
            $Text = '1 '.$Words['month'];
        } elseif($Diff->d > 7){
            $Text = ceil($Diff->d/7).' '.$Words['weeks'];
        } elseif($Diff->d == 7){
            $Text = '1 '.$Words['week'];
        } elseif($Diff->d > 1){
            $Text = $Diff->d.' '.$Words['days'];
        } elseif($Diff->d == 1){
            $Text = '1 '.$Words['day'];
        } elseif($Diff->h > 1){
            $Text = $Diff->h.' '.$Words['hours'];
        } elseif($Diff->h == 1){
            $Text = '1 '.$Words['hour'];
        } elseif($Diff->i > 1){
            $Text = $Diff->i.' '.$Words['minutes'];
        } elseif($Diff->i == 1){
            $Text = '1 '.$Words['minute'];
        } elseif($Diff->s > 1){
            $Text = $Diff->s.' '.$Words['seconds'];
        } else {
            $Text = '1 '.$Words['second'];
        }

        return $Prefix.$Text.$Suffix;
    }
}

其他回答

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 "difference " . $interval->y . " years, " . $interval->m." months, ".$interval->d." days "; 

// shows the total amount of days (not divided into years, months and days like above)
echo "difference " . $interval->days . " days ";

阅读更多php DateTime::diff手册

根据手册:

从PHP 5.2.2开始,DateTime对象可以使用比较运算符进行比较。

$date1 = new DateTime("now");
$date2 = new DateTime("tomorrow");

var_dump($date1 == $date2); // bool(false)
var_dump($date1 < $date2);  // bool(true)
var_dump($date1 > $date2);  // bool(false)

您还可以使用以下代码通过向上舍入分数来返回日期差异$date1=$duedate;//指定到期日echo$date2=日期(“Y-m-d”);//当前日期$ts1=字符串时间($date1);$ts2=字符串时间($date2);$seconds_diff=$ts1-$ts2;echo$datediff=ceil(($seconds_diff/3600)/24);//天内返回

如果您使用php的floor方法而不是ceil,它将返回舍入分数。请检查此处的差异,有时,如果您的临时服务器时区与现场站点时区不同,在这种情况下,您可能会得到不同的结果,因此请相应地更改条件。

“如果”日期存储在MySQL中,我发现在数据库级别进行差异计算更容易。。。然后根据“天”、“小时”、“分钟”、“秒”输出,分析并显示相应的结果。。。

mysql> select firstName, convert_tz(loginDate, '+00:00', '-04:00') as loginDate, TIMESTAMPDIFF(DAY, loginDate, now()) as 'Day', TIMESTAMPDIFF(HOUR, loginDate, now())+4 as 'Hour', TIMESTAMPDIFF(MINUTE, loginDate, now())+(60*4) as 'Min', TIMESTAMPDIFF(SECOND, loginDate, now())+(60*60*4) as 'Sec' from User_ where userId != '10158' AND userId != '10198' group by emailAddress order by loginDate desc;
 +-----------+---------------------+------+------+------+--------+
 | firstName | loginDate           | Day  | Hour | Min  | Sec    |
 +-----------+---------------------+------+------+------+--------+
 | Peter     | 2014-03-30 18:54:40 |    0 |    4 |  244 |  14644 |
 | Keith     | 2014-03-30 18:54:11 |    0 |    4 |  244 |  14673 |
 | Andres    | 2014-03-28 09:20:10 |    2 |   61 | 3698 | 221914 |
 | Nadeem    | 2014-03-26 09:33:43 |    4 |  109 | 6565 | 393901 |
 +-----------+---------------------+------+------+------+--------+
 4 rows in set (0.00 sec)
function showTime($time){

    $start      = strtotime($time);
    $end        = strtotime(date("Y-m-d H:i:s"));
    $minutes    = ($end - $start)/60;


    // years 
    if(($minutes / (60*24*365)) > 1){
        $years = floor($minutes/(60*24*365));
        return "From $years year( s ) ago";
    }


    // monthes 
    if(($minutes / (60*24*30)) > 1){
        $monthes = floor($minutes/(60*24*30));
        return "From $monthes monthe( s ) ago";
    }


    // days 
    if(($minutes / (60*24)) > 1){
        $days = floor($minutes/(60*24));
        return "From $days day( s ) ago";
    }

    // hours 
    if(($minutes / 60) > 1){
        $hours = floor($minutes/60);
        return "From $hours hour( s ) ago";
    }

    // minutes 
    if($minutes > 1){
        $minutes = floor($minutes);
        return "From $minutes minute( s ) ago";
    }
}

echo showTime('2022-05-05 21:33:00');
$date = '2012.11.13';
$dateOfReturn = '2017.10.31';

$substract = str_replace('.', '-', $date);

$substract2 = str_replace('.', '-', $dateOfReturn);



$date1 = $substract;
$date2 = $substract2;

$ts1 = strtotime($date1);
$ts2 = strtotime($date2);

$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);

$month1 = date('m', $ts1);
$month2 = date('m', $ts2);

echo $diff = (($year2 - $year1) * 12) + ($month2 - $month1);