我正在尝试转换格式为2009-09-12 20:57:19的时间戳,并将其转换为3分钟前用PHP。

我找到了一个有用的脚本来做这件事,但我认为它正在寻找一种不同的格式来用作时间变量。我想修改的脚本与此格式的工作是:

function _ago($tm,$rcs = 0) {
    $cur_tm = time(); 
    $dif = $cur_tm-$tm;
    $pds = array('second','minute','hour','day','week','month','year','decade');
    $lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);

    for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]);
        $no = floor($no);
        if($no <> 1)
            $pds[$v] .='s';
        $x = sprintf("%d %s ",$no,$pds[$v]);
        if(($rcs == 1)&&($v >= 1)&&(($cur_tm-$_tm) > 0))
            $x .= time_ago($_tm);
        return $x;
    }

我认为在前几行脚本试图做的事情看起来像这样(不同的日期格式数学):

$dif = 1252809479 - 2009-09-12 20:57:19;

如何将我的时间戳转换成那种(unix?)格式?


当前回答

此函数不是为英语语言而设计的。我把这些单词翻译成英语。在用于英语之前,这需要更多的修正。

function ago($d) {
$ts = time() - strtotime(str_replace("-","/",$d));

        if($ts>315360000) $val = round($ts/31536000,0).' year';
        else if($ts>94608000) $val = round($ts/31536000,0).' years';
        else if($ts>63072000) $val = ' two years';
        else if($ts>31536000) $val = ' a year';

        else if($ts>24192000) $val = round($ts/2419200,0).' month';
        else if($ts>7257600) $val = round($ts/2419200,0).' months';
        else if($ts>4838400) $val = ' two months';
        else if($ts>2419200) $val = ' a month';


        else if($ts>6048000) $val = round($ts/604800,0).' week';
        else if($ts>1814400) $val = round($ts/604800,0).' weeks';
        else if($ts>1209600) $val = ' two weeks';
        else if($ts>604800) $val = ' a week';

        else if($ts>864000) $val = round($ts/86400,0).' day';
        else if($ts>259200) $val = round($ts/86400,0).' days';
        else if($ts>172800) $val = ' two days';
        else if($ts>86400) $val = ' a day';

        else if($ts>36000) $val = round($ts/3600,0).' year';
        else if($ts>10800) $val = round($ts/3600,0).' years';
        else if($ts>7200) $val = ' two years';
        else if($ts>3600) $val = ' a year';

        else if($ts>600) $val = round($ts/60,0).' minute';
        else if($ts>180) $val = round($ts/60,0).' minutes';
        else if($ts>120) $val = ' two minutes';
        else if($ts>60) $val = ' a minute';

        else if($ts>10) $val = round($ts,0).' second';
        else if($ts>2) $val = round($ts,0).' seconds';
        else if($ts>1) $val = ' two seconds';
        else $val = $ts.' a second';


        return $val;
    }

其他回答

试试这个,我从我的旧代码中找到的,它显示了正确的结果

function ago($datefrom, $dateto = -1) {
    // Defaults and assume if 0 is passed in that
    // its an error rather than the epoch

    if ($datefrom == 0) {
        return "A long time ago";
    }
    if ($dateto == -1) {
        $dateto = time();
    }

    // Make the entered date into Unix timestamp from MySQL datetime field

    $datefrom = strtotime($datefrom);

    // Calculate the difference in seconds betweeen
    // the two timestamps

    $difference = $dateto - $datefrom;

    // Based on the interval, determine the
    // number of units between the two dates
    // From this point on, you would be hard
    // pushed telling the difference between
    // this function and DateDiff. If the $datediff
    // returned is 1, be sure to return the singular
    // of the unit, e.g. 'day' rather 'days'

    switch (true) {
        // If difference is less than 60 seconds,
        // seconds is a good interval of choice
        case(strtotime('-1 min', $dateto) < $datefrom):
            $datediff = $difference;
            $res = ($datediff == 1) ? $datediff . ' second' : $datediff . ' seconds';
            break;
        // If difference is between 60 seconds and
        // 60 minutes, minutes is a good interval
        case(strtotime('-1 hour', $dateto) < $datefrom):
            $datediff = floor($difference / 60);
            $res = ($datediff == 1) ? $datediff . ' minute' : $datediff . ' minutes';
            break;
        // If difference is between 1 hour and 24 hours
        // hours is a good interval
        case(strtotime('-1 day', $dateto) < $datefrom):
            $datediff = floor($difference / 60 / 60);
            $res = ($datediff == 1) ? $datediff . ' hour' : $datediff . ' hours';
            break;
        // If difference is between 1 day and 7 days
        // days is a good interval                
        case(strtotime('-1 week', $dateto) < $datefrom):
            $day_difference = 1;
            while (strtotime('-' . $day_difference . ' day', $dateto) >= $datefrom) {
                $day_difference++;
            }

            $datediff = $day_difference;
            $res = ($datediff == 1) ? 'yesterday' : $datediff . ' days';
            break;
        // If difference is between 1 week and 30 days
        // weeks is a good interval            
        case(strtotime('-1 month', $dateto) < $datefrom):
            $week_difference = 1;
            while (strtotime('-' . $week_difference . ' week', $dateto) >= $datefrom) {
                $week_difference++;
            }

            $datediff = $week_difference;
            $res = ($datediff == 1) ? 'last week' : $datediff . ' weeks';
            break;
        // If difference is between 30 days and 365 days
        // months is a good interval, again, the same thing
        // applies, if the 29th February happens to exist
        // between your 2 dates, the function will return
        // the 'incorrect' value for a day
        case(strtotime('-1 year', $dateto) < $datefrom):
            $months_difference = 1;
            while (strtotime('-' . $months_difference . ' month', $dateto) >= $datefrom) {
                $months_difference++;
            }

            $datediff = $months_difference;
            $res = ($datediff == 1) ? $datediff . ' month' : $datediff . ' months';

            break;
        // If difference is greater than or equal to 365
        // days, return year. This will be incorrect if
        // for example, you call the function on the 28th April
        // 2008 passing in 29th April 2007. It will return
        // 1 year ago when in actual fact (yawn!) not quite
        // a year has gone by
        case(strtotime('-1 year', $dateto) >= $datefrom):
            $year_difference = 1;
            while (strtotime('-' . $year_difference . ' year', $dateto) >= $datefrom) {
                $year_difference++;
            }

            $datediff = $year_difference;
            $res = ($datediff == 1) ? $datediff . ' year' : $datediff . ' years';
            break;
    }
    return $res;
}

示例:echo ago('2020-06-03 00:14:21 AM');

产出:6天

我不知道为什么还没有人提到碳。

https://github.com/briannesbitt/Carbon

这实际上是php dateTime的扩展(这里已经使用了),它有:diffForHumans方法。所以你所需要做的就是:

$dt = Carbon::parse('2012-9-5 23:26:11.123789');
echo $dt->diffForHumans();

更多例子:http://carbon.nesbot.com/docs/#api-humandiff

这个解决方案的优点:

它适用于未来的日期,并将在2个月等时间内返回。 你可以使用本地化来获取其他语言,多元化也很好 如果你开始用碳来做其他事情,处理日期就会变得非常容易。

直接回答这个问题…你可以用…

strtotime()

https://www.php.net/manual/en/function.strtotime.php

$dif = time() - strtotime("2009-09-12 20:57:19");

E.G:

echo round(((( time() - strtotime("2021-08-01 21:57:50") )/60)/60)/24).' day(s) ago';

结果:1天前

如果你正在使用PostgreSQL,那么它将为你做的工作:

const DT_SQL = <<<SQL
WITH lapse AS (SELECT (?::timestamp(0) - now()::timestamp(0))::text t)
SELECT CASE
  WHEN (select t from lapse) ~ '^\s*-' THEN replace((select t from lapse), '-', '') ||' ago' 
  ELSE (select t from lapse) END;
SQL;

function timeSpanText($ts, $conn)
// $ts: date-time string, $conn: PostgreSQL PDO connection
{
 return $conn -> prepare(DT_SQL) -> execute([ts]) -> fetchColumn();
}

下面是我不久前构建的通知模块的解决方案。它返回的输出类似于Facebook的通知下拉列表(例如。1天前,刚才,等等)。

public function getTimeDifference($time) {
    //Let's set the current time
    $currentTime = date('Y-m-d H:i:s');
    $toTime = strtotime($currentTime);

    //And the time the notification was set
    $fromTime = strtotime($time);

    //Now calc the difference between the two
    $timeDiff = floor(abs($toTime - $fromTime) / 60);

    //Now we need find out whether or not the time difference needs to be in
    //minutes, hours, or days
    if ($timeDiff < 2) {
        $timeDiff = "Just now";
    } elseif ($timeDiff > 2 && $timeDiff < 60) {
        $timeDiff = floor(abs($timeDiff)) . " minutes ago";
    } elseif ($timeDiff > 60 && $timeDiff < 120) {
        $timeDiff = floor(abs($timeDiff / 60)) . " hour ago";
    } elseif ($timeDiff < 1440) {
        $timeDiff = floor(abs($timeDiff / 60)) . " hours ago";
    } elseif ($timeDiff > 1440 && $timeDiff < 2880) {
        $timeDiff = floor(abs($timeDiff / 1440)) . " day ago";
    } elseif ($timeDiff > 2880) {
        $timeDiff = floor(abs($timeDiff / 1440)) . " days ago";
    }

    return $timeDiff;
}