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？
问问题
24092 次
10 回答
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