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。

其他回答

一些语言显示时间之前有一些问题,例如阿拉伯语,有3种格式需要显示日期。 我在我的项目中使用这个函数,希望他们能帮助到别人(任何建议或改进我都会很感激:))

/**
 *
 * @param   string $date1 
 * @param   string $date2 the date that you want to compare with $date1
 * @param   int $level  
 * @param   bool $absolute  
 */

function app_date_diff( $date1, $date2, $level = 3, $absolute = false ) {

    $date1 = date_create($date1);   
    $date2 = date_create($date2);
    $diff = date_diff( $date1, $date2, $absolute );

    $d = [
        'invert' => $diff->invert
    ];  

    $diffs = [
        'y' => $diff->y, 
        'm' => $diff->m, 
        'd' => $diff->d
    ];

    $level_reached = 0;

    foreach($diffs as $k=>$v) {

        if($level_reached >= $level) {
            break;
        }

        if($v > 0) {
            $d[$k] = $v;
            $level_reached++;
        }

    }

    return  $d;

}

/**
 * 
 */

function date_timestring( $periods, $format = 'latin', $separator = ',' ) {

    $formats = [
        'latin' => [
            'y' => ['year','years'],
            'm' => ['month','months'],
            'd' => ['day','days']
        ],
        'arabic' => [
            'y' => ['سنة','سنتين','سنوات'],
            'm' => ['شهر','شهرين','شهور'],
            'd' => ['يوم','يومين','أيام']
        ]
    ];

    $formats = $formats[$format];

    $string = [];

    foreach($periods as $period=>$value) {

        if(!isset($formats[$period])) {
            continue;
        }

        $string[$period] = $value.' ';
        if($format == 'arabic') {
            if($value == 2) {
                $string[$period] = $formats[$period][1];
            }elseif($value > 2 && $value <= 10) {
                $string[$period] .= $formats[$period][2];
            }else{
                $string[$period] .= $formats[$period][0];
            }

        }elseif($format == 'latin') {
            $string[$period] .= ($value > 1) ? $formats[$period][1] : $formats[$period][0];
        }

    }

    return implode($separator, $string);


}

function timeago( $date ) {

    $today = date('Y-m-d h:i:s');

    $diff = app_date_diff($date,$today,2);

    if($diff['invert'] == 1) {
        return '';
    }

    unset($diff[0]);

    $date_timestring = date_timestring($diff,'latin');

    return 'About '.$date_timestring;

}

$date1 = date('Y-m-d');
$date2 = '2018-05-14';

$diff = timeago($date2);
echo $diff;

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); 

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';
  }

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?

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';
}