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。

它能帮你检查

   function calculate_time_span($seconds)
{  
 $year = floor($seconds /31556926);
$months = floor($seconds /2629743);
$week=floor($seconds /604800);
$day = floor($seconds /86400); 
$hours = floor($seconds / 3600);
 $mins = floor(($seconds - ($hours*3600)) / 60); 
$secs = floor($seconds % 60);
 if($seconds < 60) $time = $secs." seconds ago";
 else if($seconds < 3600 ) $time =($mins==1)?$mins."now":$mins." mins ago";
 else if($seconds < 86400) $time = ($hours==1)?$hours." hour ago":$hours." hours ago";
 else if($seconds < 604800) $time = ($day==1)?$day." day ago":$day." days ago";
 else if($seconds < 2629743) $time = ($week==1)?$week." week ago":$week." weeks ago";
 else if($seconds < 31556926) $time =($months==1)? $months." month ago":$months." months ago";
 else $time = ($year==1)? $year." year ago":$year." years ago";
return $time; 
}  
  $seconds = time() - strtotime($post->post_date); 
echo calculate_time_span($seconds); 

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';
}

这就是我要的。这是阿巴斯汗帖子的修改版本:

<?php

  function calculate_time_span($post_time)
  {  
  $seconds = time() - strtotime($post);
  $year = floor($seconds /31556926);
  $months = floor($seconds /2629743);
  $week=floor($seconds /604800);
  $day = floor($seconds /86400); 
  $hours = floor($seconds / 3600);
  $mins = floor(($seconds - ($hours*3600)) / 60); 
  $secs = floor($seconds % 60);
  if($seconds < 60) $time = $secs." seconds ago";
  else if($seconds < 3600 ) $time =($mins==1)?$mins."now":$mins." mins ago";
  else if($seconds < 86400) $time = ($hours==1)?$hours." hour ago":$hours." hours ago";
  else if($seconds < 604800) $time = ($day==1)?$day." day ago":$day." days ago";
  else if($seconds < 2629743) $time = ($week==1)?$week." week ago":$week." weeks ago";
  else if($seconds < 31556926) $time =($months==1)? $months." month ago":$months." months ago";
  else $time = ($year==1)? $year." year ago":$year." years ago";
  return $time; 
  }  



 // uses
 // $post_time="2017-12-05 02:05:12";
 // echo calculate_time_span($post_time); 

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。

我修改了原来的函数一点(在我看来更有用，或更符合逻辑)。

// display "X time" ago, $rcs is precision depth
function time_ago ($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 ($v = count($lngh) - 1; ($v >= 0) && (($no = $dif / $lngh[$v]) <= 1); $v--);
    if ($v < 0)
      $v = 0;
  $_tm = $cur_tm - ($dif % $lngh[$v]);

  $no = ($rcs ? floor($no) : round($no)); // if last denomination, round

  if ($no != 1)
    $pds[$v] .= 's';
  $x = $no . ' ' . $pds[$v];

  if (($rcs > 0) && ($v >= 1))
    $x .= ' ' . $this->time_ago($_tm, $rcs - 1);

  return $x;
}