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

再加上另一个选择……

虽然我更喜欢在这里发布的DateTime方法,但我不喜欢它显示0年等事实。

/* 
 * Returns a string stating how long ago this happened
 */

private function timeElapsedString($ptime){
    $diff = time() - $ptime;
    $calc_times = array();
    $timeleft   = array();

    // Prepare array, depending on the output we want to get.
    $calc_times[] = array('Year',   'Years',   31557600);
    $calc_times[] = array('Month',  'Months',  2592000);
    $calc_times[] = array('Day',    'Days',    86400);
    $calc_times[] = array('Hour',   'Hours',   3600);
    $calc_times[] = array('Minute', 'Minutes', 60);
    $calc_times[] = array('Second', 'Seconds', 1);

    foreach ($calc_times AS $timedata){
        list($time_sing, $time_plur, $offset) = $timedata;

        if ($diff >= $offset){
            $left = floor($diff / $offset);
            $diff -= ($left * $offset);
            $timeleft[] = "{$left} " . ($left == 1 ? $time_sing : $time_plur);
        }
    }

    return $timeleft ? (time() > $ptime ? null : '-') . implode(' ', $timeleft) : 0;
}

其他回答

再加上另一个选择……

虽然我更喜欢在这里发布的DateTime方法,但我不喜欢它显示0年等事实。

/* 
 * Returns a string stating how long ago this happened
 */

private function timeElapsedString($ptime){
    $diff = time() - $ptime;
    $calc_times = array();
    $timeleft   = array();

    // Prepare array, depending on the output we want to get.
    $calc_times[] = array('Year',   'Years',   31557600);
    $calc_times[] = array('Month',  'Months',  2592000);
    $calc_times[] = array('Day',    'Days',    86400);
    $calc_times[] = array('Hour',   'Hours',   3600);
    $calc_times[] = array('Minute', 'Minutes', 60);
    $calc_times[] = array('Second', 'Seconds', 1);

    foreach ($calc_times AS $timedata){
        list($time_sing, $time_plur, $offset) = $timedata;

        if ($diff >= $offset){
            $left = floor($diff / $offset);
            $diff -= ($left * $offset);
            $timeleft[] = "{$left} " . ($left == 1 ? $time_sing : $time_plur);
        }
    }

    return $timeleft ? (time() > $ptime ? null : '-') . implode(' ', $timeleft) : 0;
}
$time_ago = ' ';
$time = time() - $time; // to get the time since that moment
$tokens = array (
31536000 => 'year',2592000 => 'month',604800 => 'week',86400 => 'day',3600 => 'hour',
60  => 'minute',1 => 'second');
foreach ($tokens as $unit => $text) {
if ($time < $unit)continue;
$numberOfUnits = floor($time / $unit);
$time_ago = ' '.$time_ago. $numberOfUnits.' '.$text.(($numberOfUnits>1)?'s':'').'  ';
$time = $time % $unit;}echo $time_ago;

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

上面的回答略有修改:

  $commentTime = strtotime($whatever)
  $today       = strtotime('today');
  $yesterday   = strtotime('yesterday');
  $todaysHours = strtotime('now') - strtotime('today');

private function timeElapsedString(
    $commentTime,
    $todaysHours,
    $today,
    $yesterday
) {
    $tokens = array(
        31536000 => 'year',
        2592000 => 'month',
        604800 => 'week',
        86400 => 'day',
        3600 => 'hour',
        60 => 'minute',
        1 => 'second'
    );
    $time = time() - $commentTime;
    $time = ($time < 1) ? 1 : $time;
    if ($commentTime >= $today || $commentTime < $yesterday) {
        foreach ($tokens as $unit => $text) {
            if ($time < $unit) {
                continue;
            }
            if ($text == 'day') {
                $numberOfUnits = floor(($time - $todaysHours) / $unit) + 1;
            } else {
                $numberOfUnits = floor(($time)/ $unit);
            }
            return $numberOfUnits . ' ' . $text . (($numberOfUnits > 1) ? 's' : '') . ' ago';
        }
    } else {
        return 'Yesterday';
    }
}

这里的许多解决方案没有考虑舍入。例如:

事件发生在两天前的下午3点。如果您在下午2点查看,它会显示在一天前。如果你在下午4点查看,它会显示两天前。

如果你使用unix时间,这有助于:

// how long since event has passed in seconds
$secs = time() - $time_ago;

// how many seconds in a day
$sec_per_day = 60*60*24;

// days elapsed
$days_elapsed = floor($secs / $sec_per_day);

// how many seconds passed today
$today_seconds = date('G')*3600 + date('i') * 60 + date('s');

// how many seconds passed in the final day calculation
$remain_seconds = $secs % $sec_per_day;

if($today_seconds < $remain_seconds)
{
    $days_elapsed++;
}

echo 'The event was '.$days_ago.' days ago.';

如果你担心闰秒和夏时制,这并不完美。