基本上,我想一个Unix时间戳(时间()函数)转换为相对日期/时间这都与过去和未来日期兼容。 所以输出可以是:
2个星期前
1小时和60分钟前
15分钟54秒前
10分钟,并在15秒后
首先,我试图代码这一点,但作出了巨大的不可维护的功能,然后我在网上搜索了几个小时,但所有我能找到的是产生的时间只是其中的一部分(EH脚本:“1小时前”无纪要)。
你有一个脚本,已经这样做了?
基本上,我想一个Unix时间戳(时间()函数)转换为相对日期/时间这都与过去和未来日期兼容。 所以输出可以是:
2个星期前
1小时和60分钟前
15分钟54秒前
10分钟,并在15秒后
首先,我试图代码这一点,但作出了巨大的不可维护的功能,然后我在网上搜索了几个小时,但所有我能找到的是产生的时间只是其中的一部分(EH脚本:“1小时前”无纪要)。
你有一个脚本,已经这样做了?
该功能为您提供了“1小时前”或“明天”像“现在”和“特定的时间戳”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', $ts);
}
}
function relativeTime($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
这是我写的东西。 显示过去相对于今天的日期日期。
/**
* @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', $diff);
}
我喜欢用的XDebug的RELATIVETIME功能。 问题是我需要它有一定的粒度。
换句话说,在数秒或数分钟停止,如果我决定。 所以,现在,
echo fTime(strtotime('-23 hours 5 minutes 55 seconds'),0);
将显示,
23小时,5分钟前
代替
23小时,5分钟,55秒前
我也希望它不要去下数组中,如果它达到更高的时间量之一。 所以,如果它显示了几年,我只想显示年和月。 所以,现在,
echo fTime(strtotime('-1 year 2 months 3 weeks 4 days 16 hours 15 minutes 22 seconds'),0);
将显示
1年,2个月前
代替
1年,2月,3周4天,16小时,15分钟后,22秒前
下面的代码改变做了我所需要的。 道具去Xdebug的第一期培训班。 希望别人可能会发现它有用:
function fTime($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
我需要一个给我如下的结果,所以我写我自己。 希望这将帮助别人。
实例:
$datetime = "2014-08-13 12:52:48";
echo getRelativeTime($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
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);
}
您可以通过使用packagist碳,仅仅是惊人:) https://github.com/briannesbitt/Carbon#api-humandiff
以下是我用过去的时间:
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], $diff);
}
}
这Drupal的做它为什么不撕掉的方式- http://api.drupal.org/api/drupal/includes%21common.inc/function/format_interval/7
<?php
function format_interval($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));
}
?>
你也许并不需要在t()的替代,你可以做你自己的事情format_plural很容易地为你(可能)不具备支持多国语言。 http://api.drupal.org/api/drupal/includes%21common.inc/function/format_plural/7