PHP:从时间戳生成相对日期/时间

Question

我基本上试图将Unix时间戳(time()函数)转换为兼容过去和未来日期的相对日期/时间.所以产出可能是:

2个星期前

1小时60分钟前

15分54秒前

10分15秒后

首先我尝试编写这个代码,但是创建了一个巨大的无法维护的功能,然后我在互联网上搜索了几个小时,但我能找到的只是一部分时间的脚本(呃:"1小时前"没有纪要).

你有一个已经这样做的脚本吗？

Answer 1

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

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);
    }
}

Answer 2

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

Answer 3

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);
}

Answer 4

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,请在秒或分钟停止.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分钟,55秒前

如果达到较高时间量之一,我也希望它不会在数组中降低.因此,如果它显示多年,我只想显示年和月.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,2个月前

代替

1年,2个月,3周,4天,16小时,15分钟,22秒之前

以下代码更改执行了我需要的操作.道具首先去xdebug.希望其他人可能会发现它很有用:

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

Answer 5

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);
  }
}