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

$time_elapsed = timeAgo($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";
        }
    }
}

其他回答

下面是一个非常简单和非常有效的解决方案。

function timeElapsed($originalTime){

        $timeElapsed=time()-$originalTime;

        /*
          You can change the values of the following 2 variables 
          based on your opinion. For 100% accuracy, you can call
          php's cal_days_in_month() and do some additional coding
          using the values you get for each month. After all the
          coding, your final answer will be approximately equal to
          mine. That is why it is okay to simply use the average
          values below.
        */
        $averageNumbDaysPerMonth=(365.242/12);
        $averageNumbWeeksPerMonth=($averageNumbDaysPerMonth/7);

        $time1=(((($timeElapsed/60)/60)/24)/365.242);
        $time2=floor($time1);//Years
        $time3=($time1-$time2)*(365.242);
        $time4=($time3/$averageNumbDaysPerMonth);
        $time5=floor($time4);//Months
        $time6=($time4-$time5)*$averageNumbWeeksPerMonth;
        $time7=floor($time6);//Weeks
        $time8=($time6-$time7)*7;
        $time9=floor($time8);//Days
        $time10=($time8-$time9)*24;
        $time11=floor($time10);//Hours
        $time12=($time10-$time11)*60;
        $time13=floor($time12);//Minutes
        $time14=($time12-$time13)*60;
        $time15=round($time14);//Seconds

        $timeElapsed=$time2 . 'yrs ' . $time5 . 'months ' . $time7 . 
                     'weeks ' . $time9 .  'days ' . $time11 . 'hrs '
                     . $time13 . 'mins and ' . $time15 . 'secs.';

        return $timeElapsed;

}

回显时间已用(1201570814);

样例输出:

6年4个月3周4天12小时40分36秒。

使用:

echo elapsed_time('2016-05-09 17:00:00'); // 18 saat 8 dakika önce yazıldı.

功能:

function elapsed_time($time){// Nekadar zaman geçmiş

        $diff = time() - strtotime($time); 

        $sec = $diff;
        $min = floor($diff/60);
        $hour = floor($diff/(60*60));
        $hour_min = floor($min - ($hour*60));
        $day = floor($diff/(60*60*24));
        $day_hour = floor($hour - ($day*24));
        $week = floor($diff/(60*60*24*7));
        $mon = floor($diff/(60*60*24*7*4));
        $year = floor($diff/(60*60*24*7*4*12));

        //difference calculate to string
        if($sec < (60*5)){
            return 'şimdi yazıldı.';
        }elseif($min < 60){
            return 'biraz önce yazıldı.';
        }elseif($hour < 24){
            return $hour.' saat '.$hour_min.' dakika önce yazıldı.';
        }elseif($day < 7){
            if($day_hour!=0){$day_hour=$day_hour.' saat ';}else{$day_hour='';}
            return $day.' gün '.$day_hour.'önce yazıldı.';
        }elseif($week < 4){
            return $week.' hafta önce yazıldı.';
        }elseif($mon < 12){
            return $mon.' ay önce yazıldı.';
        }else{
            return $year.' yıl önce yazıldı.';
        }
    }

这其实是我找到的一个更好的解决方案。使用jQuery,但它的工作完美。此外,它会自动刷新类似于SO和Facebook的方式,所以你不需要刷新页面就能看到更新。

这个插件将读取<time>标签中的datetime attr,并为您填写它。

e.g. "4 minutes ago" or "about 1 day ago

http://timeago.yarp.com/

我知道这里有几个答案,但这是我想到的。这只处理MySQL DATETIME值,根据我回应的原始问题。数组$a需要做一些工作。我欢迎就如何改进提出意见。电话为:

Echo time_elapsed_string('2014-11-14 09:42:28');

function time_elapsed_string($ptime)
{
    // Past time as MySQL DATETIME value
    $ptime = strtotime($ptime);

    // Current time as MySQL DATETIME value
    $csqltime = date('Y-m-d H:i:s');

    // Current time as Unix timestamp
    $ctime = strtotime($csqltime); 

    // Elapsed time
    $etime = $ctime - $ptime;

    // If no elapsed time, return 0
    if ($etime < 1){
        return '0 seconds';
    }

    $a = array( 365 * 24 * 60 * 60  =>  'year',
                 30 * 24 * 60 * 60  =>  'month',
                      24 * 60 * 60  =>  'day',
                           60 * 60  =>  'hour',
                                60  =>  'minute',
                                 1  =>  'second'
    );

    $a_plural = array( 'year'   => 'years',
                       'month'  => 'months',
                       'day'    => 'days',
                       'hour'   => 'hours',
                       'minute' => 'minutes',
                       'second' => 'seconds'
    );

    foreach ($a as $secs => $str){
        // Divide elapsed time by seconds
        $d = $etime / $secs;
        if ($d >= 1){
            // Round to the next lowest integer 
            $r = floor($d);
            // Calculate time to remove from elapsed time
            $rtime = $r * $secs;
            // Recalculate and store elapsed time for next loop
            if(($etime - $rtime)  < 0){
                $etime -= ($r - 1) * $secs;
            }
            else{
                $etime -= $rtime;
            }
            // Create string to return
            $estring = $estring . $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ';
        }
    }
    return $estring . ' ago';
}
$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;