31

我基本上是在尝试将 Unix 时间戳（time() 函数）转换为与过去和未来日期兼容的相对日期/时间。所以输出可能是：

2个星期前

1 小时 60 分钟前

15 分 54 秒前

10分15秒后

首先我尝试对此进行编码，但创建了一个无法维护的巨大功能，然后我在互联网上搜索了几个小时，但我只能找到只产生一部分时间的脚本（嗯：“1小时前”没有纪要）。

你有一个已经这样做的脚本吗？

4

10 回答 10

60

此功能为您提供“现在”和“特定时间戳”之间的“1 小时前”或“明天”之类的结果。

function time2str($ts)
{
    if(!ctype_digit($ts))
        $ts = strtotime($ts);

    $diff = time() - $ts;
    if($diff == 0)
        return 'now';
    elseif($diff > 0)
    {
        $day_diff = floor($diff / 86400);
        if($day_diff == 0)
        {
            if($diff < 60) return 'just now';
            if($diff < 120) return '1 minute ago';
            if($diff < 3600) return floor($diff / 60) . ' minutes ago';
            if($diff < 7200) return '1 hour ago';
            if($diff < 86400) return floor($diff / 3600) . ' hours ago';
        }
        if($day_diff == 1) return 'Yesterday';
        if($day_diff < 7) return $day_diff . ' days ago';
        if($day_diff < 31) return ceil($day_diff / 7) . ' weeks ago';
        if($day_diff < 60) return 'last month';
        return date('F Y', $ts);
    }
    else
    {
        $diff = abs($diff);
        $day_diff = floor($diff / 86400);
        if($day_diff == 0)
        {
            if($diff < 120) return 'in a minute';
            if($diff < 3600) return 'in ' . floor($diff / 60) . ' minutes';
            if($diff < 7200) return 'in an hour';
            if($diff < 86400) return 'in ' . floor($diff / 3600) . ' hours';
        }
        if($day_diff == 1) return 'Tomorrow';
        if($day_diff < 4) return date('l', $ts);
        if($day_diff < 7 + (7 - date('w'))) return 'next week';
        if(ceil($day_diff / 7) < 4) return 'in ' . ceil($day_diff / 7) . ' weeks';
        if(date('n', $ts) == date('n') + 1) return 'next month';
        return date('F Y', $ts);
    }
}
于 2010-04-22T12:01:39.627 回答
15 
function relativeTime($time) {

    $d[0] = array(1,"second");
    $d[1] = array(60,"minute");
    $d[2] = array(3600,"hour");
    $d[3] = array(86400,"day");
    $d[4] = array(604800,"week");
    $d[5] = array(2592000,"month");
    $d[6] = array(31104000,"year");

    $w = array();

    $return = "";
    $now = time();
    $diff = ($now-$time);
    $secondsLeft = $diff;

    for($i=6;$i>-1;$i--)
    {
         $w[$i] = intval($secondsLeft/$d[$i][0]);
         $secondsLeft -= ($w[$i]*$d[$i][0]);
         if($w[$i]!=0)
         {
            $return.= abs($w[$i]) . " " . $d[$i][1] . (($w[$i]>1)?'s':'') ." ";
         }

    }

    $return .= ($diff>0)?"ago":"left";
    return $return;
}

用法：

echo relativeTime((time()-256));
4 minutes 16 seconds ago
于 2011-09-20T15:35:30.883 回答
10

这是我写的。显示相对于今天日期的过去日期。

/**
 * @param $date integer of unixtimestamp format, not actual date type
 * @return string
 */
function zdateRelative($date)
{
    $now = time();
    $diff = $now - $date;

    if ($diff < 60){
        return sprintf($diff > 1 ? '%s seconds ago' : 'a second ago', $diff);
    }

    $diff = floor($diff/60);

    if ($diff < 60){
        return sprintf($diff > 1 ? '%s minutes ago' : 'one minute ago', $diff);
    }

    $diff = floor($diff/60);

    if ($diff < 24){
        return sprintf($diff > 1 ? '%s hours ago' : 'an hour ago', $diff);
    }

    $diff = floor($diff/24);

    if ($diff < 7){
        return sprintf($diff > 1 ? '%s days ago' : 'yesterday', $diff);
    }

    if ($diff < 30)
    {
        $diff = floor($diff / 7);

        return sprintf($diff > 1 ? '%s weeks ago' : 'one week ago', $diff);
    }

    $diff = floor($diff/30);

    if ($diff < 12){
        return sprintf($diff > 1 ? '%s months ago' : 'last month', $diff);
    }

    $diff = date('Y', $now) - date('Y', $date);

    return sprintf($diff > 1 ? '%s years ago' : 'last year', $diff);
}
于 2012-05-25T03:00:36.547 回答
5

我喜欢 xdebug 的 relativeTime函数。问题是我需要它有一些粒度。

换句话说，如果我决定，就停在几秒钟或几分钟。所以现在，

echo fTime(strtotime('-23 hours 5 minutes 55 seconds'),0);

会显示，

23 小时 5 分钟前

代替

23 小时 5 分 55 秒前

如果它达到更高的时间量之一，我还希望它不会在数组中降低。所以如果它显示年份，我只想显示年份和月份。所以现在，

echo fTime(strtotime('-1 year 2 months 3 weeks 4 days 16 hours 15 minutes 22 seconds'),0);

会显示

1 年零 2 个月前

代替

1 年 2 个月 3 周 4 天 16 小时 15 分 22 秒 前

以下代码更改做了我需要的。道具当然首先去 xdebug。希望其他人可能会发现它有用：

function fTime($time, $gran=-1) {

    $d[0] = array(1,"second");
    $d[1] = array(60,"minute");
    $d[2] = array(3600,"hour");
    $d[3] = array(86400,"day");
    $d[4] = array(604800,"week");
    $d[5] = array(2592000,"month");
    $d[6] = array(31104000,"year");

    $w = array();

    $return = "";
    $now = time();
    $diff = ($now-$time);
    $secondsLeft = $diff;
    $stopat = 0;
    for($i=6;$i>$gran;$i--)
    {
         $w[$i] = intval($secondsLeft/$d[$i][0]);
         $secondsLeft -= ($w[$i]*$d[$i][0]);
         if($w[$i]!=0)
         {
            $return.= abs($w[$i]) . " " . $d[$i][1] . (($w[$i]>1)?'s':'') ." ";
             switch ($i) {
                case 6: // shows years and months
                    if ($stopat==0) { $stopat=5; }
                    break;
                case 5: // shows months and weeks
                    if ($stopat==0) { $stopat=4; }
                    break;
                case 4: // shows weeks and days
                    if ($stopat==0) { $stopat=3; }
                    break;
                case 3: // shows days and hours
                    if ($stopat==0) { $stopat=2; }
                    break;
                case 2: // shows hours and minutes
                    if ($stopat==0) { $stopat=1; }
                    break;
                case 1: // shows minutes and seconds if granularity is not set higher
                    break;
             }
             if ($i===$stopat) { break 0; }
         }
    }

    $return .= ($diff>0)?"ago":"left";
    return $return;
}

马库斯

于 2013-05-05T00:02:21.840 回答
4

这是我过去使用的：

function zdateRelative($date)
{
  $diff = time() - $date;
  $periods[] = [60, 1, '%s seconds ago', 'a second ago'];
  $periods[] = [60*100, 60, '%s minutes ago', 'one minute ago'];
  $periods[] = [3600*70, 3600, '%s hours ago', 'an hour ago'];
  $periods[] = [3600*24*10, 3600*24, '%s days ago', 'yesterday'];
  $periods[] = [3600*24*30, 3600*24*7, '%s weeks ago', 'one week ago'];
  $periods[] = [3600*24*30*30, 3600*24*30, '%s months ago', 'last month'];
  $periods[] = [INF, 3600*24*265, '%s years ago', 'last year'];
  foreach ($periods as $period) {
    if ($diff > $period[0]) continue;
    $diff = floor($diff / $period[1]);
    return sprintf($diff > 1 ? $period[2] : $period[3], $diff);
  }
}
于 2017-02-24T19:40:58.180 回答
3

我需要一个给我如下结果，所以我自己写了。希望这会对某人有所帮助。

示例用法：

$datetime = "2014-08-13 12:52:48";  
echo getRelativeTime($datetime);    //10 hours ago  
echo getRelativeTime($datetime, 1); //10 hours ago  
echo getRelativeTime($datetime, 2); //10 hours and 50 minutes ago  
echo getRelativeTime($datetime, 3); //10 hours, 50 minutes and 50 seconds ago  
echo getRelativeTime($datetime, 4); //10 hours, 50 minutes and 50 seconds ago

代码：

public function getRelativeTime($datetime, $depth=1) {

    $units = array(
        "year"=>31104000,
        "month"=>2592000,
        "week"=>604800,
        "day"=>86400,
        "hour"=>3600,
        "minute"=>60,
        "second"=>1
    );

    $plural = "s";
    $conjugator = " and ";
    $separator = ", ";
    $suffix1 = " ago";
    $suffix2 = " left";
    $now = "now";
    $empty = "";

    # DO NOT EDIT BELOW

    $timediff = time()-strtotime($datetime);
    if ($timediff == 0) return $now;
    if ($depth < 1) return $empty;

    $max_depth = count($units);
    $remainder = abs($timediff);
    $output = "";
    $count_depth = 0;
    $fix_depth = true;

    foreach ($units as $unit=>$value) {
        if ($remainder>$value && $depth-->0) {
            if ($fix_depth) {
                $max_depth -= ++$count_depth;
                if ($depth>=$max_depth) $depth=$max_depth;
                $fix_depth = false;
            }
            $u = (int)($remainder/$value);
            $remainder %= $value;
            $pluralise = $u>1?$plural:$empty;
            $separate = $remainder==0||$depth==0?$empty:
                            ($depth==1?$conjugator:$separator);
            $output .= "{$u} {$unit}{$pluralise}{$separate}";
        }
        $count_depth++;
    }
    return $output.($timediff<0?$suffix2:$suffix1);
}
于 2014-08-13T23:55:27.900 回答
2

你可以通过 packagist 使用 Carbon，太棒了:) https://github.com/briannesbitt/Carbon#api-humandiff

于 2014-02-23T19:24:56.960 回答
0

为什么不撕掉drupal的做法-http: //api.drupal.org/api/drupal/includes%21common.inc/function/format_interval/7

<?php
function format_interval($interval, $granularity = 2, $langcode = NULL) {
  $units = array(
    '1 year|@count years' => 31536000, 
    '1 month|@count months' => 2592000, 
    '1 week|@count weeks' => 604800, 
    '1 day|@count days' => 86400, 
    '1 hour|@count hours' => 3600, 
    '1 min|@count min' => 60, 
    '1 sec|@count sec' => 1,
  );
  $output = '';
  foreach ($units as $key => $value) {
    $key = explode('|', $key);
    if ($interval >= $value) {
      $output .= ($output ? ' ' : '') . format_plural(floor($interval / $value), $key[0], $key[1], array(), array('langcode' => $langcode));
      $interval %= $value;
      $granularity--;
    }

    if ($granularity == 0) {
      break;
    }
  }
  return $output ? $output : t('0 sec', array(), array('langcode' => $langcode));
}
?>

您可能不需要替换 t() 并且您可以很容易地为 format_plural 做自己的事情，因为您（可能）不必支持多种语言。http://api.drupal.org/api/drupal/includes%21common.inc/function/format_plural/7

于 2012-12-10T03:18:32.493 回答
0

应该很容易适应不同的格式。这个简单的函数只适用于过去的时间戳。

// return relative date/time string from timestamp
// [n yrs] [n mos] [n days] h:i:s
function relative_time(int $time): string
{
        $dt = new DateTime();
        $dt->setTimestamp($time);
        $diff = (new DateTime())->diff($dt);

        $s = "";
        if ($diff->y) $s .= " {$diff->y} " . (($diff->y > 1) ? "yrs" : "yr");
        if ($diff->m) $s .= " {$diff->m} " . (($diff->m > 1) ? "mos" : "mo");
        if ($diff->d) $s .= " {$diff->d} " . (($diff->d > 1) ? "days" : "day");
        $s .= sprintf(" %02d:%02d:%02d", $diff->h, $diff->i, $diff->s);
        return trim($s);
}
于 2021-06-05T03:37:18.487 回答
0

PHP 8.0.0 现在通过IntlDateFormatter::format添加了IntlDateFormatter::RELATIVE_*常量，相对日期的实现非常弱。

这在撰写本文时可能不是超级有用，因为它只输出字符串“昨天”、“今天”和“明天”......然后它会退回到完整/长/中/短日期之外的任何内容那些界限。

最大的吸引力在于它是完全国际化的。所以使用其他语言环境，如de_DE,ko_KR或pa_IN会给你相关脚本中的翻译字符串......这可能是值得妥协的。

$formatter = new IntlDateFormatter(
    'en_US',
    IntlDateFormatter::RELATIVE_FULL,
    IntlDateFormatter::NONE,
    'America/Los_Angeles',
    IntlDateFormatter::GREGORIAN
);

echo $formatter->format( time() - 86400 );
echo $formatter->format( time() );
echo $formatter->format( time() + 86400 );

未来的读者也有可能使用具有更全面的相对时间格式的更新版本的 PHP。

于 2021-09-16T21:07:02.857 回答