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

我发现结果如下:

1年2个月0天0小时53分1秒

正因为如此,我实现了一个尊重复数的函数,删除空值,并有可能缩短输出:

function since($timestamp, $level=6) {
    global $lang;
    $date = new DateTime();
    $date->setTimestamp($timestamp);
    $date = $date->diff(new DateTime());
    // build array
    $since = array_combine(array('year', 'month', 'day', 'hour', 'minute', 'second'), explode(',', $date->format('%y,%m,%d,%h,%i,%s')));
    // remove empty date values
    $since = array_filter($since);
    // output only the first x date values
    $since = array_slice($since, 0, $level);
    // build string
    $last_key = key(array_slice($since, -1, 1, true));
    $string = '';
    foreach ($since as $key => $val) {
        // separator
        if ($string) {
            $string .= $key != $last_key ? ', ' : ' ' . $lang['and'] . ' ';
        }
        // set plural
        $key .= $val > 1 ? 's' : '';
        // add date value
        $string .= $val . ' ' . $lang[ $key ];
    }
    return $string;
}

看起来好多了:

1年2个月53分1秒

可以选择使用$level = 2来缩短它,如下所示:

1年2个月

如果你只需要它的英文版本,就删除$lang部分,或者编辑这个翻译来满足你的需要:

$lang = array(
    'second' => 'Sekunde',
    'seconds' => 'Sekunden',
    'minute' => 'Minute',
    'minutes' => 'Minuten',
    'hour' => 'Stunde',
    'hours' => 'Stunden',
    'day' => 'Tag',
    'days' => 'Tage',
    'month' => 'Monat',
    'months' => 'Monate',
    'year' => 'Jahr',
    'years' => 'Jahre',
    'and' => 'und',
);

其他回答

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); 
$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;

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;

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;
}

这是我的解决方案,请检查并根据您的要求修改

function getHowLongAgo($date, $display = array('Year', 'Month', 'Day', 'Hour', 'Minute', 'Second'), $ago = '') {
        date_default_timezone_set('Australia/Sydney');
        $timestamp = strtotime($date);
        $timestamp = (int) $timestamp;
        $current_time = time();
        $diff = $current_time - $timestamp;

        //intervals in seconds
        $intervals = array(
            'year' => 31556926, 'month' => 2629744, 'week' => 604800, 'day' => 86400, 'hour' => 3600, 'minute' => 60
        );

        //now we just find the difference
        if ($diff == 0) {
            return ' Just now ';
        }

        if ($diff < 60) {
            return $diff == 1 ? $diff . ' second ago ' : $diff . ' seconds ago ';
        }

        if ($diff >= 60 && $diff < $intervals['hour']) {
            $diff = floor($diff / $intervals['minute']);
            return $diff == 1 ? $diff . ' minute ago ' : $diff . ' minutes ago ';
        }

        if ($diff >= $intervals['hour'] && $diff < $intervals['day']) {
            $diff = floor($diff / $intervals['hour']);
            return $diff == 1 ? $diff . ' hour ago ' : $diff . ' hours ago ';
        }

        if ($diff >= $intervals['day'] && $diff < $intervals['week']) {
            $diff = floor($diff / $intervals['day']);
            return $diff == 1 ? $diff . ' day ago ' : $diff . ' days ago ';
        }

        if ($diff >= $intervals['week'] && $diff < $intervals['month']) {
            $diff = floor($diff / $intervals['week']);
            return $diff == 1 ? $diff . ' week ago ' : $diff . ' weeks ago ';
        }

        if ($diff >= $intervals['month'] && $diff < $intervals['year']) {
            $diff = floor($diff / $intervals['month']);
            return $diff == 1 ? $diff . ' month ago ' : $diff . ' months ago ';
        }

        if ($diff >= $intervals['year']) {
            $diff = floor($diff / $intervals['year']);
            return $diff == 1 ? $diff . ' year ago ' : $diff . ' years ago ';
        }
    }

谢谢