PHP如何查找自日期时间以来经过的时间？

Question

如何查找自日期时间戳以来经过的时间2010-04-28 17:25:43,最终输出文本应该像xx Minutes Ago/xx Days Ago

Answer 1

大多数答案似乎都集中在将日期从字符串转换为时间.看起来你大部分时间都在考虑将日期纳入'5天前'的格式等等.对吧？

这就是我要这样做的方式:

$time = strtotime('2010-04-28 17:25:43');

echo 'event happened '.humanTiming($time).' ago';

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':'');
    }

}

我没有测试过,但它应该工作.

结果看起来像

event happened 4 days ago

要么

event happened 1 minute ago

干杯

Answer 2

想要分享php功能,这导致语法正确的Facebook像人类可读的时间格式.

例:

echo get_time_ago(strtotime('now'));

结果:

不到1分钟前

function get_time_ago($time_stamp)
{
    $time_difference = strtotime('now') - $time_stamp;

    if ($time_difference >= 60 * 60 * 24 * 365.242199)
    {
        /*
         * 60 seconds/minute * 60 minutes/hour * 24 hours/day * 365.242199 days/year
         * This means that the time difference is 1 year or more
         */
        return get_time_ago_string($time_stamp, 60 * 60 * 24 * 365.242199, 'year');
    }
    elseif ($time_difference >= 60 * 60 * 24 * 30.4368499)
    {
        /*
         * 60 seconds/minute * 60 minutes/hour * 24 hours/day * 30.4368499 days/month
         * This means that the time difference is 1 month or more
         */
        return get_time_ago_string($time_stamp, 60 * 60 * 24 * 30.4368499, 'month');
    }
    elseif ($time_difference >= 60 * 60 * 24 * 7)
    {
        /*
         * 60 seconds/minute * 60 minutes/hour * 24 hours/day * 7 days/week
         * This means that the time difference is 1 week or more
         */
        return get_time_ago_string($time_stamp, 60 * 60 * 24 * 7, 'week');
    }
    elseif ($time_difference >= 60 * 60 * 24)
    {
        /*
         * 60 seconds/minute * 60 minutes/hour * 24 hours/day
         * This means that the time difference is 1 day or more
         */
        return get_time_ago_string($time_stamp, 60 * 60 * 24, 'day');
    }
    elseif ($time_difference >= 60 * 60)
    {
        /*
         * 60 seconds/minute * 60 minutes/hour
         * This means that the time difference is 1 hour or more
         */
        return get_time_ago_string($time_stamp, 60 * 60, 'hour');
    }
    else
    {
        /*
         * 60 seconds/minute
         * This means that the time difference is a matter of minutes
         */
        return get_time_ago_string($time_stamp, 60, 'minute');
    }
}

function get_time_ago_string($time_stamp, $divisor, $time_unit)
{
    $time_difference = strtotime("now") - $time_stamp;
    $time_units      = floor($time_difference / $divisor);

    settype($time_units, 'string');

    if ($time_units === '0')
    {
        return 'less than 1 ' . $time_unit . ' ago';
    }
    elseif ($time_units === '1')
    {
        return '1 ' . $time_unit . ' ago';
    }
    else
    {
        /*
         * More than "1" $time_unit. This is the "plural" message.
         */
        // TODO: This pluralizes the time unit, which is done by adding "s" at the end; this will not work for i18n!
        return $time_units . ' ' . $time_unit . 's ago';
    }
}

Answer 3

我想我有一个应该做你想做的功能:

function time2string($timeline) {
    $periods = array('day' => 86400, 'hour' => 3600, 'minute' => 60, 'second' => 1);

    foreach($periods AS $name => $seconds){
        $num = floor($timeline / $seconds);
        $timeline -= ($num * $seconds);
        $ret .= $num.' '.$name.(($num > 1) ? 's' : '').' ';
    }

    return trim($ret);
}

只需将其应用于之间的区别time(),strtotime('2010-04-28 17:25:43')如下:

print time2string(time()-strtotime('2010-04-28 17:25:43')).' ago';

Answer 4

为了改进@arnorhs 的答案，我添加了获得更精确结果的能力，因此如果您想要例如自用户加入以来的年、月、日和小时。

我添加了一个新参数，允许您指定希望返回的精度点数。

function get_friendly_time_ago($distant_timestamp, $max_units = 3) {
    $i = 0;
    $time = time() - $distant_timestamp; // to get the time since that moment
    $tokens = [
        31536000 => 'year',
        2592000 => 'month',
        604800 => 'week',
        86400 => 'day',
        3600 => 'hour',
        60 => 'minute',
        1 => 'second'
    ];

    $responses = [];
    while ($i < $max_units && $time > 0) {
        foreach ($tokens as $unit => $text) {
            if ($time < $unit) {
                continue;
            }
            $i++;
            $numberOfUnits = floor($time / $unit);

            $responses[] = $numberOfUnits . ' ' . $text . (($numberOfUnits > 1) ? 's' : '');
            $time -= ($unit * $numberOfUnits);
            break;
        }
    }

    if (!empty($responses)) {
        return implode(', ', $responses) . ' ago';
    }

    return 'Just now';
}

Answer 5

如果您使用 php Datetime 类，您可以使用：

function time_ago(Datetime $date) {
  $time_ago = '';

  $diff = $date->diff(new Datetime('now'));


  if (($t = $diff->format("%m")) > 0)
    $time_ago = $t . ' months';
  else if (($t = $diff->format("%d")) > 0)
    $time_ago = $t . ' days';
  else if (($t = $diff->format("%H")) > 0)
    $time_ago = $t . ' hours';
  else
    $time_ago = 'minutes';

  return $time_ago . ' ago (' . $date->format('M j, Y') . ')';
}