如何查找自日期时间戳以来经过的时间2010-04-28 17:25:43，最终输出文本应为xx Minutes Ago/xx Days Ago
问问题
130798 次
17 回答
250
大多数答案似乎都集中在将日期从字符串转换为时间。看来您主要是在考虑将日期转换为“5 天前”格式等。对吗？
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.(($numberOfUnits>1)?'s':'');
}
}
我还没有测试过，但它应该可以工作。
结果看起来像
event happened 4 days ago
或者
event happened 1 minute ago
干杯
于 2010-05-26T19:39:34.013 回答
16
想要分享 php 函数，这会导致语法正确的 Facebook 像人类可读的时间格式。
例子：
echo get_time_ago(strtotime('now'));
结果：
不到 1 分钟前
function get_time_ago($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 . ' ' . $time_unit . 's ago';
}
}
于 2011-03-30T02:35:58.833 回答
12
我想我有一个功能应该做你想做的事：
function time2string($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);
}
time()只需将其应用于和之间的差异strtotime('2010-04-28 17:25:43')：
print time2string(time()-strtotime('2010-04-28 17:25:43')).' ago';
于 2010-05-26T19:34:23.410 回答
7
为了改进@arnorhs 的答案，我添加了获得更精确结果的能力，因此如果您想要几年、几个月、几天和几小时，例如自用户加入以来。
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';
}
于 2014-11-23T14:47:28.423 回答
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') . ')';
}
于 2013-04-23T18:11:18.413 回答
3
请注意，大多数数学计算的示例都有日期的硬性限制，2038-01-18并且不适用于虚构的日期。
由于缺乏基于示例的示例DateTime，DateInterval我想提供一个多功能功能，以满足 OP 的需求和其他想要复合经过时间段的需求，例如1 month 2 days ago. 连同一堆其他用例，例如显示日期而不是经过时间的限制，或者过滤掉经过时间结果的部分。
此外，大多数示例都假设过去是从当前时间开始的，下面的函数允许用所需的结束日期覆盖它。
/**
* multi-purpose function to calculate the time elapsed between $start and optional $end
* @param string|null $start the date string to start calculation
* @param string|null $end the date string to end calculation
* @param string $suffix the suffix string to include in the calculated string
* @param string $format the format of the resulting date if limit is reached or no periods were found
* @param string $separator the separator between periods to use when filter is not true
* @param null|string $limit date string to stop calculations on and display the date if reached - ex: 1 month
* @param bool|array $filter false to display all periods, true to display first period matching the minimum, or array of periods to display ['year', 'month']
* @param int $minimum the minimum value needed to include a period
* @return string
*/
function elapsedTimeString($start, $end = null, $limit = null, $filter = true, $suffix = 'ago', $format = 'Y-m-d', $separator = ' ', $minimum = 1)
{
$dates = (object) array(
'start' => new DateTime($start ? : 'now'),
'end' => new DateTime($end ? : 'now'),
'intervals' => array('y' => 'year', 'm' => 'month', 'd' => 'day', 'h' => 'hour', 'i' => 'minute', 's' => 'second'),
'periods' => array()
);
$elapsed = (object) array(
'interval' => $dates->start->diff($dates->end),
'unknown' => 'unknown'
);
if ($elapsed->interval->invert === 1) {
return trim('0 seconds ' . $suffix);
}
if (false === empty($limit)) {
$dates->limit = new DateTime($limit);
if (date_create()->add($elapsed->interval) > $dates->limit) {
return $dates->start->format($format) ? : $elapsed->unknown;
}
}
if (true === is_array($filter)) {
$dates->intervals = array_intersect($dates->intervals, $filter);
$filter = false;
}
foreach ($dates->intervals as $period => $name) {
$value = $elapsed->interval->$period;
if ($value >= $minimum) {
$dates->periods[] = vsprintf('%1$s %2$s%3$s', array($value, $name, ($value !== 1 ? 's' : '')));
if (true === $filter) {
break;
}
}
}
if (false === empty($dates->periods)) {
return trim(vsprintf('%1$s %2$s', array(implode($separator, $dates->periods), $suffix)));
}
return $dates->start->format($format) ? : $elapsed->unknown;
}
需要注意的一点 - 提供的过滤器值的检索间隔不会延续到下一个时期。过滤器仅显示所提供周期的结果值，并且不会重新计算周期以仅显示所需的过滤器总数。
用法
由于 OP 需要显示最高时段（截至 2015-02-24）。
echo elapsedTimeString('2010-04-26');
/** 4 years ago */
显示复合期间并提供自定义结束日期（请注意缺少提供的时间和虚构的日期）。
echo elapsedTimeString('1920-01-01', '2500-02-24', null, false);
/** 580 years 1 month 23 days ago */
显示过滤周期的结果（数组的顺序无关紧要）。
echo elapsedTimeString('2010-05-26', '2012-02-24', null, ['month', 'year']);
/** 1 year 8 months ago */
如果达到限制，则以提供的格式（默认 Ymd）显示开始日期。
echo elapsedTimeString('2010-05-26', '2012-02-24', '1 year');
/** 2010-05-26 */
还有很多其他用例。它也可以很容易地适应用于开始、结束或限制参数的 unix 时间戳和/或 DateInterval 对象。
于 2015-02-24T21:52:40.797 回答
2
适用于任何版本的 PHP 的一个选项是执行已经建议的操作，如下所示：
$eventTime = '2010-04-28 17:25:43';
$age = time() - strtotime($eventTime);
这将以秒为单位给出您的年龄。从那里，您可以随心所欲地显示它。
然而，这种方法的一个问题是它不会考虑 DST 导致的时间偏移。如果这不是一个问题，那就去吧。否则，您可能需要在 DateTime 类中使用 diff() 方法。不幸的是，如果您至少使用 PHP 5.3，这只是一个选项。
于 2010-05-26T19:13:50.447 回答
2
我喜欢 Mithun 的代码，但我对其进行了一些调整以使其给出更合理的答案。
function getTimeSince($eventTime)
{
$totaldelay = time() - strtotime($eventTime);
if($totaldelay <= 0)
{
return '';
}
else
{
$first = '';
$marker = 0;
if($years=floor($totaldelay/31536000))
{
$totaldelay = $totaldelay % 31536000;
$plural = '';
if ($years > 1) $plural='s';
$interval = $years." year".$plural;
$timesince = $timesince.$first.$interval;
if ($marker) return $timesince;
$marker = 1;
$first = ", ";
}
if($months=floor($totaldelay/2628000))
{
$totaldelay = $totaldelay % 2628000;
$plural = '';
if ($months > 1) $plural='s';
$interval = $months." month".$plural;
$timesince = $timesince.$first.$interval;
if ($marker) return $timesince;
$marker = 1;
$first = ", ";
}
if($days=floor($totaldelay/86400))
{
$totaldelay = $totaldelay % 86400;
$plural = '';
if ($days > 1) $plural='s';
$interval = $days." day".$plural;
$timesince = $timesince.$first.$interval;
if ($marker) return $timesince;
$marker = 1;
$first = ", ";
}
if ($marker) return $timesince;
if($hours=floor($totaldelay/3600))
{
$totaldelay = $totaldelay % 3600;
$plural = '';
if ($hours > 1) $plural='s';
$interval = $hours." hour".$plural;
$timesince = $timesince.$first.$interval;
if ($marker) return $timesince;
$marker = 1;
$first = ", ";
}
if($minutes=floor($totaldelay/60))
{
$totaldelay = $totaldelay % 60;
$plural = '';
if ($minutes > 1) $plural='s';
$interval = $minutes." minute".$plural;
$timesince = $timesince.$first.$interval;
if ($marker) return $timesince;
$first = ", ";
}
if($seconds=floor($totaldelay/1))
{
$totaldelay = $totaldelay % 1;
$plural = '';
if ($seconds > 1) $plural='s';
$interval = $seconds." second".$plural;
$timesince = $timesince.$first.$interval;
}
return $timesince;
}
}
于 2014-04-20T12:29:06.450 回答
1
将 [saved_date] 转换为时间戳。获取当前时间戳。
当前时间戳 - [saved_date] 时间戳。
然后你可以用 date(); 格式化它。
您通常可以使用 strtotime() 函数将大多数日期格式转换为时间戳。
于 2010-05-26T19:01:05.857 回答
1
使用这个，你可以得到
$previousDate = '2013-7-26 17:01:10';
$startdate = new DateTime($previousDate);
$endDate = new DateTime('now');
$interval = $endDate->diff($startdate);
echo$interval->format('%y years, %m months, %d days');
于 2014-12-16T11:18:49.420 回答
1
尝试以下回购之一：
https://github.com/salavert/time-ago-in-words
https://github.com/jimmiw/php-time-ago
我刚开始使用后者，可以解决问题，但是当所讨论的日期太远时，没有stackoverflow风格的回退到确切的日期，也不支持未来的日期 - 而且API有点时髦，但至少它工作看似完美无瑕，并得到维护......
于 2015-08-19T09:20:58.767 回答
0
为了找出经过的时间，我通常使用time()而不是date()格式化时间戳。然后得到后一个值和前一个值之间的差异并相应地格式化。 time()不同的是，它不是替代品，date()但在计算经过时间时完全有帮助。
例子：
time()看起来像这样的值1274467343每秒递增。因此，您可以拥有$erlierTimewith value1274467343和$latterTimewith value 1274467500，然后只需执行$latterTime - $erlierTime以秒为单位的时间。
于 2010-05-26T19:00:31.657 回答
0
自己写的
function getElapsedTime($eventTime)
{
$totaldelay = time() - strtotime($eventTime);
if($totaldelay <= 0)
{
return '';
}
else
{
if($days=floor($totaldelay/86400))
{
$totaldelay = $totaldelay % 86400;
return $days.' days ago.';
}
if($hours=floor($totaldelay/3600))
{
$totaldelay = $totaldelay % 3600;
return $hours.' hours ago.';
}
if($minutes=floor($totaldelay/60))
{
$totaldelay = $totaldelay % 60;
return $minutes.' minutes ago.';
}
if($seconds=floor($totaldelay/1))
{
$totaldelay = $totaldelay % 1;
return $seconds.' seconds ago.';
}
}
}
于 2010-05-31T06:13:14.577 回答
0
在这里，我使用自定义函数来查找自日期时间以来经过的时间。
echo Datetodays('2013-7-26 17:01:10');
功能日期（$d）{
$date_start = $d;
$date_end = date('Ymd H:i:s');
定义（'第二'，1）；
定义（'分钟'，秒 * 60）；
定义（'小时'，分钟 * 60）；
定义（'天'，小时* 24）；
定义（'星期'，天 * 7）；
$t1 = strtotime($date_start);
$t2 = strtotime($date_end);
如果（$t1 > $t2）{
$差异 = $t1 - $t2;
} 别的 {
$差异 = $t2 - $t1;
}
//echo "
".$date_end." ".$date_start." ".$diffrence;
$results['major'] = array(); // 整数表示日期时间关系中的较大数字
$results1 = 数组（）；
$字符串 = '';
$results['major']['weeks'] = floor($diffrence / WEEK);
$results['major']['days'] = floor($diffrence / DAY);
$results['major']['hours'] = floor($diffrence / HOUR);
$results['major']['minutes'] = floor($diffrence / MINUTE);
$results['major']['seconds'] = floor($diffrence / SECOND);
//print_r($results);
// 逻辑：
// 第 1 步：取主要结果并将其转换为原始秒数（将减去差异的秒数）
// 例如：$result = ($results['major']['weeks']*WEEK)
// 第 2 步：从差值（总时间）中减去较小的数字（结果）
// 例如：$minor_result = $difference - $result
// 第 3 步：以秒为单位获取结果时间并将其转换为次要格式
// 例如：地板（$minor_result/DAY）
$results1['weeks'] = floor($diffrence / WEEK);
$results1['days'] = floor((($diffrence - ($results['major']['weeks'] * WEEK)) / DAY));
$results1['hours'] = floor((($diffrence - ($results['major']['days'] * DAY)) / HOUR));
$results1['minutes'] = floor((($diffrence - ($results['major']['hours'] * HOUR)) / MINUTE));
$results1['seconds'] = floor((($diffrence - ($results['major']['minutes'] * MINUTE)) / SECOND));
//print_r($results1);
if ($results1['weeks'] != 0 && $results1['days'] == 0) {
if ($results1['weeks'] == 1) {
$string = $results1['weeks'] 。' 一周前';
} 别的 {
if ($results1['weeks'] == 2) {
$string = $results1['weeks'] 。' 几周前';
} 别的 {
$string = '2 周前';
}
}
} elseif ($results1['weeks'] != 0 && $results1['days'] != 0) {
if ($results1['weeks'] == 1) {
$string = $results1['weeks'] 。' 一周前';
} 别的 {
if ($results1['weeks'] == 2) {
$string = $results1['weeks'] 。' 几周前';
} 别的 {
$string = '2 周前';
}
}
} elseif ($results1['weeks'] == 0 && $results1['days'] != 0) {
if ($results1['days'] == 1) {
$string = $results1['days'] 。'一天前';
} 别的 {
$string = $results1['days'] 。' 几天前';
}
} elseif ($results1['days'] != 0 && $results1['hours'] != 0) {
$string = $results1['days'] 。'天和'。$results1['小时'] 。'几小时前';
} elseif ($results1['days'] == 0 && $results1['hours'] != 0) {
if ($results1['hours'] == 1) {
$string = $results1['hours'] 。' 一个小时前';
} 别的 {
$string = $results1['hours'] 。'几小时前';
}
} elseif ($results1['hours'] != 0 && $results1['minutes'] != 0) {
$string = $results1['hours'] 。'小时和'。$results1['分钟'] 。' 几分钟前';
} elseif ($results1['hours'] == 0 && $results1['minutes'] != 0) {
if ($results1['minutes'] == 1) {
$string = $results1['分钟'] 。'分钟前';
} 别的 {
$string = $results1['分钟'] 。' 几分钟前';
}
} elseif ($results1['minutes'] != 0 && $results1['seconds'] != 0) {
$string = $results1['分钟'] 。'分钟和'。$results1['seconds'] 。'几秒钟前';
} elseif ($results1['minutes'] == 0 && $results1['seconds'] != 0) {
if ($results1['seconds'] == 1) {
$string = $results1['seconds'] 。'第二之前';
} 别的 {
$string = $results1['seconds'] 。'几秒钟前';
}
}
返回$字符串；
}
?>
于 2013-07-26T13:24:24.647 回答
0
您可以直接从 WordPress 核心文件中获取此功能，看看这里
http://core.trac.wordpress.org/browser/tags/3.6/wp-includes/formatting.php#L2121
function human_time_diff( $from, $to = '' ) {
if ( empty( $to ) )
$to = time();
$diff = (int) abs( $to - $from );
if ( $diff < HOUR_IN_SECONDS ) {
$mins = round( $diff / MINUTE_IN_SECONDS );
if ( $mins <= 1 )
$mins = 1;
/* translators: min=minute */
$since = sprintf( _n( '%s min', '%s mins', $mins ), $mins );
} elseif ( $diff < DAY_IN_SECONDS && $diff >= HOUR_IN_SECONDS ) {
$hours = round( $diff / HOUR_IN_SECONDS );
if ( $hours <= 1 )
$hours = 1;
$since = sprintf( _n( '%s hour', '%s hours', $hours ), $hours );
} elseif ( $diff < WEEK_IN_SECONDS && $diff >= DAY_IN_SECONDS ) {
$days = round( $diff / DAY_IN_SECONDS );
if ( $days <= 1 )
$days = 1;
$since = sprintf( _n( '%s day', '%s days', $days ), $days );
} elseif ( $diff < 30 * DAY_IN_SECONDS && $diff >= WEEK_IN_SECONDS ) {
$weeks = round( $diff / WEEK_IN_SECONDS );
if ( $weeks <= 1 )
$weeks = 1;
$since = sprintf( _n( '%s week', '%s weeks', $weeks ), $weeks );
} elseif ( $diff < YEAR_IN_SECONDS && $diff >= 30 * DAY_IN_SECONDS ) {
$months = round( $diff / ( 30 * DAY_IN_SECONDS ) );
if ( $months <= 1 )
$months = 1;
$since = sprintf( _n( '%s month', '%s months', $months ), $months );
} elseif ( $diff >= YEAR_IN_SECONDS ) {
$years = round( $diff / YEAR_IN_SECONDS );
if ( $years <= 1 )
$years = 1;
$since = sprintf( _n( '%s year', '%s years', $years ), $years );
}
return $since;
}
于 2013-09-02T07:54:51.760 回答
0
arnorhs 对功能“humanTiming”的即兴创作。它将计算时间字符串到人类可读文本版本的“完全拉伸”翻译。例如说它像“1周2天1小时28分14秒”
function humantime ($oldtime, $newtime = null, $returnarray = false) {
if(!$newtime) $newtime = time();
$time = $newtime - $oldtime; // to get the time since that moment
$tokens = array (
31536000 => 'year',
2592000 => 'month',
604800 => 'week',
86400 => 'day',
3600 => 'hour',
60 => 'minute',
1 => 'second'
);
$htarray = array();
foreach ($tokens as $unit => $text) {
if ($time < $unit) continue;
$numberOfUnits = floor($time / $unit);
$htarray[$text] = $numberOfUnits.' '.$text.(($numberOfUnits>1)?'s':'');
$time = $time - ( $unit * $numberOfUnits );
}
if($returnarray) return $htarray;
return implode(' ', $htarray);
}
于 2015-03-07T10:21:52.340 回答
0
最近不得不这样做 - 希望这对某人有所帮助。它不能满足所有可能性，但满足了我对项目的需求。
https://github.com/duncanheron/twitter_date_format
https://github.com/duncanheron/twitter_date_format/blob/master/twitter_date_format.php
于 2015-05-06T13:00:26.100 回答