试试这个，我从我的旧代码中找到的，它显示了正确的结果

function ago($datefrom, $dateto = -1) {
    // Defaults and assume if 0 is passed in that
    // its an error rather than the epoch

    if ($datefrom == 0) {
        return "A long time ago";
    }
    if ($dateto == -1) {
        $dateto = time();
    }

    // Make the entered date into Unix timestamp from MySQL datetime field

    $datefrom = strtotime($datefrom);

    // Calculate the difference in seconds betweeen
    // the two timestamps

    $difference = $dateto - $datefrom;

    // Based on the interval, determine the
    // number of units between the two dates
    // From this point on, you would be hard
    // pushed telling the difference between
    // this function and DateDiff. If the $datediff
    // returned is 1, be sure to return the singular
    // of the unit, e.g. 'day' rather 'days'

    switch (true) {
        // If difference is less than 60 seconds,
        // seconds is a good interval of choice
        case(strtotime('-1 min', $dateto) < $datefrom):
            $datediff = $difference;
            $res = ($datediff == 1) ? $datediff . ' second' : $datediff . ' seconds';
            break;
        // If difference is between 60 seconds and
        // 60 minutes, minutes is a good interval
        case(strtotime('-1 hour', $dateto) < $datefrom):
            $datediff = floor($difference / 60);
            $res = ($datediff == 1) ? $datediff . ' minute' : $datediff . ' minutes';
            break;
        // If difference is between 1 hour and 24 hours
        // hours is a good interval
        case(strtotime('-1 day', $dateto) < $datefrom):
            $datediff = floor($difference / 60 / 60);
            $res = ($datediff == 1) ? $datediff . ' hour' : $datediff . ' hours';
            break;
        // If difference is between 1 day and 7 days
        // days is a good interval                
        case(strtotime('-1 week', $dateto) < $datefrom):
            $day_difference = 1;
            while (strtotime('-' . $day_difference . ' day', $dateto) >= $datefrom) {
                $day_difference++;
            }

            $datediff = $day_difference;
            $res = ($datediff == 1) ? 'yesterday' : $datediff . ' days';
            break;
        // If difference is between 1 week and 30 days
        // weeks is a good interval            
        case(strtotime('-1 month', $dateto) < $datefrom):
            $week_difference = 1;
            while (strtotime('-' . $week_difference . ' week', $dateto) >= $datefrom) {
                $week_difference++;
            }

            $datediff = $week_difference;
            $res = ($datediff == 1) ? 'last week' : $datediff . ' weeks';
            break;
        // If difference is between 30 days and 365 days
        // months is a good interval, again, the same thing
        // applies, if the 29th February happens to exist
        // between your 2 dates, the function will return
        // the 'incorrect' value for a day
        case(strtotime('-1 year', $dateto) < $datefrom):
            $months_difference = 1;
            while (strtotime('-' . $months_difference . ' month', $dateto) >= $datefrom) {
                $months_difference++;
            }

            $datediff = $months_difference;
            $res = ($datediff == 1) ? $datediff . ' month' : $datediff . ' months';

            break;
        // If difference is greater than or equal to 365
        // days, return year. This will be incorrect if
        // for example, you call the function on the 28th April
        // 2008 passing in 29th April 2007. It will return
        // 1 year ago when in actual fact (yawn!) not quite
        // a year has gone by
        case(strtotime('-1 year', $dateto) >= $datefrom):
            $year_difference = 1;
            while (strtotime('-' . $year_difference . ' year', $dateto) >= $datefrom) {
                $year_difference++;
            }

            $datediff = $year_difference;
            $res = ($datediff == 1) ? $datediff . ' year' : $datediff . ' years';
            break;
    }
    return $res;
}

示例:echo ago('2020-06-03 00:14:21 AM');

产出:6天

function humanTiming ($time)
        {

            $time = time() - $time; // to get the time since that moment
            $time = ($time<1)? 1 : $time;
            $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);
                return $numberOfUnits.' '.$text.(($numberOfUnits>1)?'s':'');
            }

        }

echo humanTiming( strtotime($mytimestring) );

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;

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',
);

我做了这个，它工作得很好，它既适用于Unix时间戳，如1470919932，也适用于格式化时间，如16-08-11 14:53:30

function timeAgo($time_ago) {
    $time_ago =  strtotime($time_ago) ? strtotime($time_ago) : $time_ago;
    $time  = time() - $time_ago;

switch($time):
// seconds
case $time <= 60;
return 'lessthan a minute ago';
// minutes
case $time >= 60 && $time < 3600;
return (round($time/60) == 1) ? 'a minute' : round($time/60).' minutes ago';
// hours
case $time >= 3600 && $time < 86400;
return (round($time/3600) == 1) ? 'a hour ago' : round($time/3600).' hours ago';
// days
case $time >= 86400 && $time < 604800;
return (round($time/86400) == 1) ? 'a day ago' : round($time/86400).' days ago';
// weeks
case $time >= 604800 && $time < 2600640;
return (round($time/604800) == 1) ? 'a week ago' : round($time/604800).' weeks ago';
// months
case $time >= 2600640 && $time < 31207680;
return (round($time/2600640) == 1) ? 'a month ago' : round($time/2600640).' months ago';
// years
case $time >= 31207680;
return (round($time/31207680) == 1) ? 'a year ago' : round($time/31207680).' years ago' ;

endswitch;
}

?>

此函数不是为英语语言而设计的。我把这些单词翻译成英语。在用于英语之前，这需要更多的修正。

function ago($d) {
$ts = time() - strtotime(str_replace("-","/",$d));

        if($ts>315360000) $val = round($ts/31536000,0).' year';
        else if($ts>94608000) $val = round($ts/31536000,0).' years';
        else if($ts>63072000) $val = ' two years';
        else if($ts>31536000) $val = ' a year';

        else if($ts>24192000) $val = round($ts/2419200,0).' month';
        else if($ts>7257600) $val = round($ts/2419200,0).' months';
        else if($ts>4838400) $val = ' two months';
        else if($ts>2419200) $val = ' a month';


        else if($ts>6048000) $val = round($ts/604800,0).' week';
        else if($ts>1814400) $val = round($ts/604800,0).' weeks';
        else if($ts>1209600) $val = ' two weeks';
        else if($ts>604800) $val = ' a week';

        else if($ts>864000) $val = round($ts/86400,0).' day';
        else if($ts>259200) $val = round($ts/86400,0).' days';
        else if($ts>172800) $val = ' two days';
        else if($ts>86400) $val = ' a day';

        else if($ts>36000) $val = round($ts/3600,0).' year';
        else if($ts>10800) $val = round($ts/3600,0).' years';
        else if($ts>7200) $val = ' two years';
        else if($ts>3600) $val = ' a year';

        else if($ts>600) $val = round($ts/60,0).' minute';
        else if($ts>180) $val = round($ts/60,0).' minutes';
        else if($ts>120) $val = ' two minutes';
        else if($ts>60) $val = ' a minute';

        else if($ts>10) $val = round($ts,0).' second';
        else if($ts>2) $val = round($ts,0).' seconds';
        else if($ts>1) $val = ' two seconds';
        else $val = $ts.' a second';


        return $val;
    }