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

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); 

其他回答

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';
}

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?

function time_elapsed_string($ptime)
{
    $etime = time() - $ptime;

    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)
    {
        $d = $etime / $secs;
        if ($d >= 1)
        {
            $r = round($d);
            return $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ago';
        }
    }
}

只需将日期时间传递给这个func。它会按照之前的格式打印出来

date_default_timezone_set('your-time-zone');
function convert($datetime){
  $time=strtotime($datetime);
  $diff=time()-$time;
  $diff/=60;
  $var1=floor($diff);
  $var=$var1<=1 ? 'min' : 'mins';
  if($diff>=60){
    $diff/=60;
    $var1=floor($diff);
    $var=$var1<=1 ? 'hr' : 'hrs';
    if($diff>=24){$diff/=24;$var1=floor($diff);$var=$var1<=1 ? 'day' : 'days';
    if($diff>=30.4375){$diff/=30.4375;$var1=floor($diff);$var=$var1<=1 ? 'month' : 'months';
    if($diff>=12){$diff/=12;$var1=floor($diff);$var=$var1<=1 ? 'year' : 'years';}}}}
    echo $var1,' ',$var,' ago';
  }

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();
}