PHP:从时间戳生成相对日期/时间

I'm basically trying to convert a Unix timestamp (the time() function) to a relative date/time that's both compatible with past and future date. So outputs could be:

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。所以输出可以是:

2 weeks ago

两个星期前

1 hour and 60 minutes ago

1小时60分钟前

15 minutes and 54 seconds ago

15分钟54秒前

after 10 minutes and 15 seconds

10分钟15秒后

First I tried to code this, but made a huge unmaintainable function, and then I searched the internet for a couple of hours, yet all I can find are scripts that produce only one part of the time (e.h: "1 hour ago" without the minutes).

首先，我尝试编写这个代码，但是做了一个巨大的不可维护的功能，然后我在互联网上搜索了几个小时，但是我所能找到的只是脚本，它们只产生了一部分时间(e)。h:“1小时前”没有记录。

Do you have a script that already does this?

您是否已经有了这样的脚本?

8 个解决方案

#1

This function gives you "1 hour ago" or "Tomorrow" like results between 'now' and 'specific timestamp'.

这个函数提供“1小时前”或“明天”之类的结果，介于“现在”和“特定时间戳”之间。

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);
    }
}

#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;
}

Usage:

用法:

echo relativeTime((time()-256));
4 minutes 16 seconds ago

#3

Here is what I've written. Displays a past date relative to today's date.

这是我写的。显示相对于今天的日期的过去日期。

/**
 * @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);
}

#4

I love the relativeTime function by xdebug. Problem is I needed it to have some granularity.

我喜欢xdebug的相对论时间函数。问题是我需要它有一些粒度。

In other words stop at seconds or minutes if I decide. So now,

换句话说，如果我决定的话，在几秒钟或几分钟内停止。现在,

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

would show,

将显示,

23 hours, 5 minutes ago

23小时，5分钟前

Instead of

而不是

23 hours, 5 minutes, 55 seconds ago

23小时，5分钟，55秒之前。

I also wanted it to NOT go lower in the array if it reached one of the higher time amounts. So if it shows years, I only want to show years and months. So now,

我还希望它不会在数组中降低，如果它达到一个更高的时间量。如果它显示的是年份，我只想显示年份和月份。现在,

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

Would show

将显示

1 year, 2 months ago

一年前，两个月前

Instead of

而不是

1 year, 2 months, 3 weeks, 4 days, 16 hours, 15 minutes, 22 seconds ago

1年，2个月，3周，4天，16小时，15分钟，22秒之前

The following code change did what I needed. Props go to xdebug first of course. Hopefully someone else might find it useful:

下面的代码改变了我所需要的。首先是xdebug。希望其他人能发现它有用:

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;
}

Marcus

马库斯

#5

I needed one to give me results as below, so I wrote my own. Hopefully, this will help somebody.

我需要一个这样的结果，所以我写了我自己的。希望这能对某些人有所帮助。

Example usage:

使用示例:

$datetime = "2014-08-13 12:52:48";  
echo getRelativeTime($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  

Code:

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);
}

#6

You can use Carbon via packagist, just amazing :) https://github.com/briannesbitt/Carbon#api-humandiff

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

#7

Here is what I use for past times:

以下是我过去使用的:

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);
  }
}

#8

Why not rip off the way that drupal does it - http://api.drupal.org/api/drupal/includes%21common.inc/function/format_interval/7

为什么不剽窃drupal的方式——http://api.drupal.org/api/drupal/includes%21common.inc/function/format_interval7

<?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));
}
?>

You probably don't need a replacement for t() and you could do your own thing for format_plural pretty easily as you (probably) don't have to support multiple languages. http://api.drupal.org/api/drupal/includes%21common.inc/function/format_plural/7

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