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

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;
    }

我想有一个荷兰版本,支持单复数。仅仅在结尾加一个“s”是不够的,我们用的是完全不同的词,所以我重写了这篇文章的顶部答案。

这将导致:

2年1个月2周1天1分2秒

or

1年2个月1周2天1分1秒

    public function getTimeAgo($full = false){

    $now = new \DateTime;
    $ago = new \DateTime($this->datetime());
    $diff = $now->diff($ago);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'jaren',
        'm' => 'maanden',
        'w' => 'weken',
        'd' => 'dagen',
        'h' => 'uren',
        'i' => 'minuten',
        's' => 'seconden',
    );
    $singleString = array(
        'y' => 'jaar',
        'm' => 'maand',
        'w' => 'week',
        'd' => 'dag',
        'h' => 'uur',
        'i' => 'minuut',
        's' => 'seconde',
    );
    // M.O. 2022-02-11 I rewrote this function to support dutch singles and plurals. Added some docs for next programmer to break his brain :)
    // For each possible notation, if corresponding value of current key is true (>1) otherwise remove its key/value from array
    // If the value from current key is 1, use value from $singleString array. Otherwise use value from $string array
    foreach ($string as $k => &$v) {
        if ($diff->$k) {
            if($diff->$k == 1){
                $v = $diff->$k . ' ' . $singleString[$k];
            } else {
                $v = $diff->$k . ' ' . $v;
            }
        } else {
            if($diff->$k == 1){
                unset($singleString[$k]);
            } else {
                unset($string[$k]);
            }
        }
    }

    // If $full = true, print all values.
    // Values have already been filtered with foreach removing keys that contain a 0 as value
    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . '' : 'zojuist';
}

你应该先测试一下,因为我不是一个优秀的程序员:)

$time_elapsed = timeAgo($time_ago); //The argument $time_ago is in timestamp (Y-m-d H:i:s)format.

//Function definition

function timeAgo($time_ago)
{
    $time_ago = strtotime($time_ago);
    $cur_time   = time();
    $time_elapsed   = $cur_time - $time_ago;
    $seconds    = $time_elapsed ;
    $minutes    = round($time_elapsed / 60 );
    $hours      = round($time_elapsed / 3600);
    $days       = round($time_elapsed / 86400 );
    $weeks      = round($time_elapsed / 604800);
    $months     = round($time_elapsed / 2600640 );
    $years      = round($time_elapsed / 31207680 );
    // Seconds
    if($seconds <= 60){
        return "just now";
    }
    //Minutes
    else if($minutes <=60){
        if($minutes==1){
            return "one minute ago";
        }
        else{
            return "$minutes minutes ago";
        }
    }
    //Hours
    else if($hours <=24){
        if($hours==1){
            return "an hour ago";
        }else{
            return "$hours hrs ago";
        }
    }
    //Days
    else if($days <= 7){
        if($days==1){
            return "yesterday";
        }else{
            return "$days days ago";
        }
    }
    //Weeks
    else if($weeks <= 4.3){
        if($weeks==1){
            return "a week ago";
        }else{
            return "$weeks weeks ago";
        }
    }
    //Months
    else if($months <=12){
        if($months==1){
            return "a month ago";
        }else{
            return "$months months ago";
        }
    }
    //Years
    else{
        if($years==1){
            return "one year ago";
        }else{
            return "$years years ago";
        }
    }
}

我试过了,效果很好

$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-10');
$difference = $datetime1->diff($datetime2);
echo formatOutput($difference);

function formatOutput($diff){
    /* function to return the highrst defference fount */
    if(!is_object($diff)){
        return;
    }

    if($diff->y > 0){
        return $diff->y .(" year".($diff->y > 1?"s":"")." ago");
    }

    if($diff->m > 0){
        return $diff->m .(" month".($diff->m > 1?"s":"")." ago");
    }

    if($diff->d > 0){
        return $diff->d .(" day".($diff->d > 1?"s":"")." ago");
    }

    if($diff->h > 0){
        return $diff->h .(" hour".($diff->h > 1?"s":"")." ago");
    }

    if($diff->i > 0){
        return $diff->i .(" minute".($diff->i > 1?"s":"")." ago");
    }

    if($diff->s > 0){
        return $diff->s .(" second".($diff->s > 1?"s":"")." ago");
    }
}

在这里查看这个链接作为参考

谢谢!玩得开心。

您必须将时间戳的每一部分都转换为Unix时间。例如时间戳:2009-09-12 20:57:19。

((2008-1970)*365)+(8*30)+12)*24+20可以大致估算出自1970年1月1日以来的工作时间。

用这个数字乘以60,再加57,就得到了分钟数。

用这个,乘以60,再加19。

这将非常粗略和不准确地转换它。

有什么原因让你不能开始使用正常的Unix时间吗?