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在 PHP 中将时间戳转换为前时间,例如 1 天前、2 天前...

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我正在尝试转换格式为 2009-09-12 20:57:19 的时间戳,并使用 PHP 将其转换为 3 minutes ago 之类的东西。

我找到了一个有用的脚本来执行此操作,但我认为它正在寻找一种不同的格式来用作时间变量。我要修改以使用这种格式的脚本是:

function _ago($tm,$rcs = 0) {
    $cur_tm = time(); 
    $dif = $cur_tm-$tm;
    $pds = array('second','minute','hour','day','week','month','year','decade');
    $lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);

    for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]);
        $no = floor($no);
        if($no <> 1)
            $pds[$v] .='s';
        $x = sprintf("%d %s ",$no,$pds[$v]);
        if(($rcs == 1)&&($v >= 1)&&(($cur_tm-$_tm) > 0))
            $x .= time_ago($_tm);
        return $x;
    }

我认为在前几行中,脚本试图做一些看起来像这样的事情(不同的日期格式数学):

$dif = 1252809479 - 2009-09-12 20:57:19;

我将如何将我的时间戳转换为那种(unix?)格式?


m
mahen3d

使用示例:

echo time_elapsed_string('2013-05-01 00:22:35');
echo time_elapsed_string('@1367367755'); # timestamp input
echo time_elapsed_string('2013-05-01 00:22:35', true);

输入可以是任何 supported date and time format

输出 :

4 months ago
4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago

功能 :

function time_elapsed_string($datetime, $full = false) {
    $now = new DateTime;
    $ago = new DateTime($datetime);
    $diff = $now->diff($ago);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'year',
        'm' => 'month',
        'w' => 'week',
        'd' => 'day',
        'h' => 'hour',
        'i' => 'minute',
        's' => 'second',
    );
    foreach ($string as $k => &$v) {
        if ($diff->$k) {
            $v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
        } else {
            unset($string[$k]);
        }
    }

    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . ' ago' : 'just now';
}

星期部分很好,但总体而言,该函数应该更灵活($full 应该是一个字符串输入,以根据需要过滤输出)。例如time_elapsed_string($datetime, $format = "ymw")。 PS平面版:stackoverflow.com/a/5010169/318765
关于我的最后一条评论:将 $full = false 更改为 $level = 7 并将 if (!$full) $string = array_slice($string, 0, 1); 更改为 $string = array_slice($string, 0, $level); 并调用 time_elapsed_string($datetime, 2) 以仅获取两个最高日期字符串。我认为这应该满足所有需求。
@mgutt:ofc 此功能不适用于任何情况下的用户想要的场景;但这是一个很好的起点,您可以通过最少的修复获得所需的东西,就像您已经展示的那样......
我在这条线 $diff->w = floor($diff->d / 7); 得到 Unknown property (w)
要在 PHP5.3 及更低版本中修复此问题 Unknown property (w),请将 $diff 从对象转换为数组并相应地调整其余代码。我在此处发布了修复:stackoverflow.com/a/32723846/235633
P
Pablo
function time_elapsed_string($ptime)
{
    $etime = time() - $ptime;

    if ($etime < 1)
    {
        return '0 seconds';
    }

    $a = array( 365 * 24 * 60 * 60  =>  'year',
                 30 * 24 * 60 * 60  =>  'month',
                      24 * 60 * 60  =>  'day',
                           60 * 60  =>  'hour',
                                60  =>  'minute',
                                 1  =>  'second'
                );
    $a_plural = array( 'year'   => 'years',
                       'month'  => 'months',
                       'day'    => 'days',
                       'hour'   => 'hours',
                       'minute' => 'minutes',
                       'second' => 'seconds'
                );

    foreach ($a as $secs => $str)
    {
        $d = $etime / $secs;
        if ($d >= 1)
        {
            $r = round($d);
            return $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ago';
        }
    }
}

这不是一个好的解决方案,因为它使用 30days for month12x30days for year,因此它将返回低于年 <= 1978 年的无效年数。Example 它返回 39 年,但应该是 38 . 它在 1970 年以后的几年里也行不通。
可怕的解决方案。为什么这有 42 个赞成票和选定的答案?从什么时候开始每个月有30天?
@wassimboy,你能告诉我你为什么要否决我的回答吗?如果我的回答不够好,您可以写评论来纠正它而无需投反对票。
您的答案不好,因为它每月计算 30 天,并非所有月份都有 30 天。阅读上面的其他评论。
这是一个返回一般短语的函数,例如“这条评论大约是多久以前发表的”?如上所述,它并不精确并且有一些缺点,但对于最近的日期(比如过去 30 年或更短),它让我们不精确的人类知道事件发生在多长时间之前。非常适合我的应用程序。
H
H6.
$time_elapsed = timeAgo($time_ago); //The argument $time_ago is in timestamp (Y-m-d H:i:s)format.

//Function definition

function timeAgo($time_ago)
{
    $time_ago = strtotime($time_ago);
    $cur_time   = time();
    $time_elapsed   = $cur_time - $time_ago;
    $seconds    = $time_elapsed ;
    $minutes    = round($time_elapsed / 60 );
    $hours      = round($time_elapsed / 3600);
    $days       = round($time_elapsed / 86400 );
    $weeks      = round($time_elapsed / 604800);
    $months     = round($time_elapsed / 2600640 );
    $years      = round($time_elapsed / 31207680 );
    // Seconds
    if($seconds <= 60){
        return "just now";
    }
    //Minutes
    else if($minutes <=60){
        if($minutes==1){
            return "one minute ago";
        }
        else{
            return "$minutes minutes ago";
        }
    }
    //Hours
    else if($hours <=24){
        if($hours==1){
            return "an hour ago";
        }else{
            return "$hours hrs ago";
        }
    }
    //Days
    else if($days <= 7){
        if($days==1){
            return "yesterday";
        }else{
            return "$days days ago";
        }
    }
    //Weeks
    else if($weeks <= 4.3){
        if($weeks==1){
            return "a week ago";
        }else{
            return "$weeks weeks ago";
        }
    }
    //Months
    else if($months <=12){
        if($months==1){
            return "a month ago";
        }else{
            return "$months months ago";
        }
    }
    //Years
    else{
        if($years==1){
            return "one year ago";
        }else{
            return "$years years ago";
        }
    }
}

K
Krzysztof Duszczyk

我不知道为什么还没有人提到碳。

https://github.com/briannesbitt/Carbon

这实际上是 php dateTime 的扩展(已在此处使用),它具有:diffForHumans 方法。所以你需要做的就是:

$dt = Carbon::parse('2012-9-5 23:26:11.123789');
echo $dt->diffForHumans();

更多示例:http://carbon.nesbot.com/docs/#api-humandiff

此解决方案的优点:

它适用于未来的日期,并将在 2 个月内返回类似的东西。

您可以使用本地化来获取其他语言,并且多元化工作正常

如果您将开始将 Carbon 用于其他事情,那么使用日期将变得前所未有的简单。


P
Panama Jack

这实际上是我找到的更好的解决方案。使用 jQuery,但它完美地工作。此外,它会自动刷新,类似于 SO 和 Facebook 的方式,因此您无需刷新页面即可查看更新。

该插件将读取您在 <time> 标记中的 datetime 属性并为您填写。

e.g. "4 minutes ago" or "about 1 day ago

http://timeago.yarp.com/


完美工作并在 IST 时间内显示所需的结果
如何在此插件中使用 php 时间戳 2020-12-19 13:12:58
m
mgutt

我发现如下丑陋的结果:

1年2个月0天0小时53分1秒

因此,我实现了一个尊重复数的函数,删除空值,并且可以选择缩短输出:

function since($timestamp, $level=6) {
    global $lang;
    $date = new DateTime();
    $date->setTimestamp($timestamp);
    $date = $date->diff(new DateTime());
    // build array
    $since = array_combine(array('year', 'month', 'day', 'hour', 'minute', 'second'), explode(',', $date->format('%y,%m,%d,%h,%i,%s')));
    // remove empty date values
    $since = array_filter($since);
    // output only the first x date values
    $since = array_slice($since, 0, $level);
    // build string
    $last_key = key(array_slice($since, -1, 1, true));
    $string = '';
    foreach ($since as $key => $val) {
        // separator
        if ($string) {
            $string .= $key != $last_key ? ', ' : ' ' . $lang['and'] . ' ';
        }
        // set plural
        $key .= $val > 1 ? 's' : '';
        // add date value
        $string .= $val . ' ' . $lang[ $key ];
    }
    return $string;
}

看起来好多了:

1年2个月53分1秒

可以选择使用 $level = 2 来缩短它,如下所示:

1年2个月

如果您只需要英文版,请删除 $lang 部分,或编辑此翻译以满足您的需要:

$lang = array(
    'second' => 'Sekunde',
    'seconds' => 'Sekunden',
    'minute' => 'Minute',
    'minutes' => 'Minuten',
    'hour' => 'Stunde',
    'hours' => 'Stunden',
    'day' => 'Tag',
    'days' => 'Tage',
    'month' => 'Monat',
    'months' => 'Monate',
    'year' => 'Jahr',
    'years' => 'Jahre',
    'and' => 'und',
);

总之,这个对我很有效。虽然它没有显示去或以前,但双向工作。简单修复: if ($date->invert > 0) { $ending = " to go"; } else { $ending = "以前";只需将 $ending 添加到 $string。
在这里使用 json 函数会对性能造成很大影响,而且完全没有必要。这不是一个好的解决方案。
@zombat 感谢您的反馈。我更改了针对 explodearray_combine 的行。
为什么是global?请阅读一些 www.PhpTheRightWay.com。
b
brycejl
function humanTiming ($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':'');
            }

        }

echo humanTiming( strtotime($mytimestring) );

这真的很棒。感谢你的分享。但是在某些情况下,该函数返回 $numberOfUnits 为空,它只输出“之前”。貌似是因为值小于1秒。在那种情况下,如果您可以将其默认为“刚刚”之类的内容,那将是完美的。
B
Bobb Fwed

我稍微修改了原始功能(在我看来更有用,或更合乎逻辑)。

// display "X time" ago, $rcs is precision depth
function time_ago ($tm, $rcs = 0) {
  $cur_tm = time(); 
  $dif = $cur_tm - $tm;
  $pds = array('second','minute','hour','day','week','month','year','decade');
  $lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);

  for ($v = count($lngh) - 1; ($v >= 0) && (($no = $dif / $lngh[$v]) <= 1); $v--);
    if ($v < 0)
      $v = 0;
  $_tm = $cur_tm - ($dif % $lngh[$v]);

  $no = ($rcs ? floor($no) : round($no)); // if last denomination, round

  if ($no != 1)
    $pds[$v] .= 's';
  $x = $no . ' ' . $pds[$v];

  if (($rcs > 0) && ($v >= 1))
    $x .= ' ' . $this->time_ago($_tm, $rcs - 1);

  return $x;
}

任何使用基于固定日历的数学的函数都存在根本缺陷。使用 Date 对象,不要对时间进行数学运算。
@chris-baker 当大多数人使用这些类型的函数时,不需要精确到秒。该功能在短时间内是准确的,而在很长一段时间内,接近就足够了。
A
Ahmad ghoneim

我做了这个,它工作得很好,它适用于像 1470919932 这样的 unix 时间戳或像 16-08-11 14:53:30 这样的格式化时间

function timeAgo($time_ago) {
    $time_ago =  strtotime($time_ago) ? strtotime($time_ago) : $time_ago;
    $time  = time() - $time_ago;

switch($time):
// seconds
case $time <= 60;
return 'lessthan a minute ago';
// minutes
case $time >= 60 && $time < 3600;
return (round($time/60) == 1) ? 'a minute' : round($time/60).' minutes ago';
// hours
case $time >= 3600 && $time < 86400;
return (round($time/3600) == 1) ? 'a hour ago' : round($time/3600).' hours ago';
// days
case $time >= 86400 && $time < 604800;
return (round($time/86400) == 1) ? 'a day ago' : round($time/86400).' days ago';
// weeks
case $time >= 604800 && $time < 2600640;
return (round($time/604800) == 1) ? 'a week ago' : round($time/604800).' weeks ago';
// months
case $time >= 2600640 && $time < 31207680;
return (round($time/2600640) == 1) ? 'a month ago' : round($time/2600640).' months ago';
// years
case $time >= 31207680;
return (round($time/31207680) == 1) ? 'a year ago' : round($time/31207680).' years ago' ;

endswitch;
}

?>

即使 $time 将评估为 true,那不应该是 switch (true) 吗?
C
Community

只是为了提供另一种选择......

虽然我更喜欢发布 here 的 DateTime 方法,但我不喜欢它显示 0 年等的事实。

/* 
 * Returns a string stating how long ago this happened
 */

private function timeElapsedString($ptime){
    $diff = time() - $ptime;
    $calc_times = array();
    $timeleft   = array();

    // Prepare array, depending on the output we want to get.
    $calc_times[] = array('Year',   'Years',   31557600);
    $calc_times[] = array('Month',  'Months',  2592000);
    $calc_times[] = array('Day',    'Days',    86400);
    $calc_times[] = array('Hour',   'Hours',   3600);
    $calc_times[] = array('Minute', 'Minutes', 60);
    $calc_times[] = array('Second', 'Seconds', 1);

    foreach ($calc_times AS $timedata){
        list($time_sing, $time_plur, $offset) = $timedata;

        if ($diff >= $offset){
            $left = floor($diff / $offset);
            $diff -= ($left * $offset);
            $timeleft[] = "{$left} " . ($left == 1 ? $time_sing : $time_plur);
        }
    }

    return $timeleft ? (time() > $ptime ? null : '-') . implode(' ', $timeleft) : 0;
}

V
Vicky Salunkhe

我通常用它来找出 currentpassed datetime stamp 之间的区别

输出

//If difference is greater than 7 days
7 June 2019

// if difference is greater than 24 hours and less than 7 days
1 days ago
6 days ago

1 hour ago
23 hours ago

1 minute ago
58 minutes ago

1 second ago
20 seconds ago

代码

//return current date time
function getCurrentDateTime(){
    //date_default_timezone_set("Asia/Calcutta");
    return date("Y-m-d H:i:s");
}
function getDateString($date){
    $dateArray = date_parse_from_format('Y/m/d', $date);
    $monthName = DateTime::createFromFormat('!m', $dateArray['month'])->format('F');
    return $dateArray['day'] . " " . $monthName  . " " . $dateArray['year'];
}

function getDateTimeDifferenceString($datetime){
    $currentDateTime = new DateTime(getCurrentDateTime());
    $passedDateTime = new DateTime($datetime);
    $interval = $currentDateTime->diff($passedDateTime);
    //$elapsed = $interval->format('%y years %m months %a days %h hours %i minutes %s seconds');
    $day = $interval->format('%a');
    $hour = $interval->format('%h');
    $min = $interval->format('%i');
    $seconds = $interval->format('%s');

    if($day > 7)
        return getDateString($datetime);
    else if($day >= 1 && $day <= 7 ){
        if($day == 1) return $day . " day ago";
        return $day . " days ago";
    }else if($hour >= 1 && $hour <= 24){
        if($hour == 1) return $hour . " hour ago";
        return $hour . " hours ago";
    }else if($min >= 1 && $min <= 60){
        if($min == 1) return $min . " minute ago";
        return $min . " minutes ago";
    }else if($seconds >= 1 && $seconds <= 60){
        if($seconds == 1) return $seconds . " second ago";
        return $seconds . " seconds ago";
    }
}

谢谢你。我实际上将它转换为 C# 代码。
未定义函数 getCurrentDateTime() 仅供参考
@gfivehost 添加了功能代码,你现在可以查看它。
我真的很喜欢这个片段。但 DateTime 不支持语言环境设置来翻译日期时间格式,如日期或月份的名称。例如,在 getDateString() 格式中('F')只有英文。在前 7 天内切换到 IntlDateFormatter 和 gettext() 用于自定义输出将为 i18n 解决此问题
A
Abbbas khan

它可以帮助您检查

   function calculate_time_span($seconds)
{  
 $year = floor($seconds /31556926);
$months = floor($seconds /2629743);
$week=floor($seconds /604800);
$day = floor($seconds /86400); 
$hours = floor($seconds / 3600);
 $mins = floor(($seconds - ($hours*3600)) / 60); 
$secs = floor($seconds % 60);
 if($seconds < 60) $time = $secs." seconds ago";
 else if($seconds < 3600 ) $time =($mins==1)?$mins."now":$mins." mins ago";
 else if($seconds < 86400) $time = ($hours==1)?$hours." hour ago":$hours." hours ago";
 else if($seconds < 604800) $time = ($day==1)?$day." day ago":$day." days ago";
 else if($seconds < 2629743) $time = ($week==1)?$week." week ago":$week." weeks ago";
 else if($seconds < 31556926) $time =($months==1)? $months." month ago":$months." months ago";
 else $time = ($year==1)? $year." year ago":$year." years ago";
return $time; 
}  
  $seconds = time() - strtotime($post->post_date); 
echo calculate_time_span($seconds); 

M
Monzur

试试这个,我从我的旧代码中找到它,它显示了正确的结果

function ago($datefrom, $dateto = -1) {
    // Defaults and assume if 0 is passed in that
    // its an error rather than the epoch

    if ($datefrom == 0) {
        return "A long time ago";
    }
    if ($dateto == -1) {
        $dateto = time();
    }

    // Make the entered date into Unix timestamp from MySQL datetime field

    $datefrom = strtotime($datefrom);

    // Calculate the difference in seconds betweeen
    // the two timestamps

    $difference = $dateto - $datefrom;

    // Based on the interval, determine the
    // number of units between the two dates
    // From this point on, you would be hard
    // pushed telling the difference between
    // this function and DateDiff. If the $datediff
    // returned is 1, be sure to return the singular
    // of the unit, e.g. 'day' rather 'days'

    switch (true) {
        // If difference is less than 60 seconds,
        // seconds is a good interval of choice
        case(strtotime('-1 min', $dateto) < $datefrom):
            $datediff = $difference;
            $res = ($datediff == 1) ? $datediff . ' second' : $datediff . ' seconds';
            break;
        // If difference is between 60 seconds and
        // 60 minutes, minutes is a good interval
        case(strtotime('-1 hour', $dateto) < $datefrom):
            $datediff = floor($difference / 60);
            $res = ($datediff == 1) ? $datediff . ' minute' : $datediff . ' minutes';
            break;
        // If difference is between 1 hour and 24 hours
        // hours is a good interval
        case(strtotime('-1 day', $dateto) < $datefrom):
            $datediff = floor($difference / 60 / 60);
            $res = ($datediff == 1) ? $datediff . ' hour' : $datediff . ' hours';
            break;
        // If difference is between 1 day and 7 days
        // days is a good interval                
        case(strtotime('-1 week', $dateto) < $datefrom):
            $day_difference = 1;
            while (strtotime('-' . $day_difference . ' day', $dateto) >= $datefrom) {
                $day_difference++;
            }

            $datediff = $day_difference;
            $res = ($datediff == 1) ? 'yesterday' : $datediff . ' days';
            break;
        // If difference is between 1 week and 30 days
        // weeks is a good interval            
        case(strtotime('-1 month', $dateto) < $datefrom):
            $week_difference = 1;
            while (strtotime('-' . $week_difference . ' week', $dateto) >= $datefrom) {
                $week_difference++;
            }

            $datediff = $week_difference;
            $res = ($datediff == 1) ? 'last week' : $datediff . ' weeks';
            break;
        // If difference is between 30 days and 365 days
        // months is a good interval, again, the same thing
        // applies, if the 29th February happens to exist
        // between your 2 dates, the function will return
        // the 'incorrect' value for a day
        case(strtotime('-1 year', $dateto) < $datefrom):
            $months_difference = 1;
            while (strtotime('-' . $months_difference . ' month', $dateto) >= $datefrom) {
                $months_difference++;
            }

            $datediff = $months_difference;
            $res = ($datediff == 1) ? $datediff . ' month' : $datediff . ' months';

            break;
        // If difference is greater than or equal to 365
        // days, return year. This will be incorrect if
        // for example, you call the function on the 28th April
        // 2008 passing in 29th April 2007. It will return
        // 1 year ago when in actual fact (yawn!) not quite
        // a year has gone by
        case(strtotime('-1 year', $dateto) >= $datefrom):
            $year_difference = 1;
            while (strtotime('-' . $year_difference . ' year', $dateto) >= $datefrom) {
                $year_difference++;
            }

            $datediff = $year_difference;
            $res = ($datediff == 1) ? $datediff . ' year' : $datediff . ' years';
            break;
    }
    return $res;
}

示例:echo ago('2020-06-03 00:14:21 AM');

输出:6 days


D
Dazza

要直接回答问题...您可以使用...

strtotime()

https://www.php.net/manual/en/function.strtotime.php

$dif = time() - strtotime("2009-09-12 20:57:19");

例如:

echo round(((( time() - strtotime("2021-08-01 21:57:50") )/60)/60)/24).' day(s) ago';

结果:1 天前


r
rdpcoder

我知道这里有几个答案,但这就是我想出的。这仅根据我回答的原始问题处理 MySQL DATETIME 值。数组 $a 需要一些工作。我欢迎就如何改进提出意见。调用为:

echo time_elapsed_string('2014-11-14 09:42:28');

function time_elapsed_string($ptime)
{
    // Past time as MySQL DATETIME value
    $ptime = strtotime($ptime);

    // Current time as MySQL DATETIME value
    $csqltime = date('Y-m-d H:i:s');

    // Current time as Unix timestamp
    $ctime = strtotime($csqltime); 

    // Elapsed time
    $etime = $ctime - $ptime;

    // If no elapsed time, return 0
    if ($etime < 1){
        return '0 seconds';
    }

    $a = array( 365 * 24 * 60 * 60  =>  'year',
                 30 * 24 * 60 * 60  =>  'month',
                      24 * 60 * 60  =>  'day',
                           60 * 60  =>  'hour',
                                60  =>  'minute',
                                 1  =>  'second'
    );

    $a_plural = array( 'year'   => 'years',
                       'month'  => 'months',
                       'day'    => 'days',
                       'hour'   => 'hours',
                       'minute' => 'minutes',
                       'second' => 'seconds'
    );

    foreach ($a as $secs => $str){
        // Divide elapsed time by seconds
        $d = $etime / $secs;
        if ($d >= 1){
            // Round to the next lowest integer 
            $r = floor($d);
            // Calculate time to remove from elapsed time
            $rtime = $r * $secs;
            // Recalculate and store elapsed time for next loop
            if(($etime - $rtime)  < 0){
                $etime -= ($r - 1) * $secs;
            }
            else{
                $etime -= $rtime;
            }
            // Create string to return
            $estring = $estring . $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ';
        }
    }
    return $estring . ' ago';
}

R
Ruberandinda Patience

我试过了,对我来说效果很好

$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-10');
$difference = $datetime1->diff($datetime2);
echo formatOutput($difference);

function formatOutput($diff){
    /* function to return the highrst defference fount */
    if(!is_object($diff)){
        return;
    }

    if($diff->y > 0){
        return $diff->y .(" year".($diff->y > 1?"s":"")." ago");
    }

    if($diff->m > 0){
        return $diff->m .(" month".($diff->m > 1?"s":"")." ago");
    }

    if($diff->d > 0){
        return $diff->d .(" day".($diff->d > 1?"s":"")." ago");
    }

    if($diff->h > 0){
        return $diff->h .(" hour".($diff->h > 1?"s":"")." ago");
    }

    if($diff->i > 0){
        return $diff->i .(" minute".($diff->i > 1?"s":"")." ago");
    }

    if($diff->s > 0){
        return $diff->s .(" second".($diff->s > 1?"s":"")." ago");
    }
}

检查此链接以获取参考 here

谢谢!玩得开心。


d
drtechno

这就是我一起去的。它是 Abbbas khan 帖子的修改版本:

<?php

  function calculate_time_span($post_time)
  {  
  $seconds = time() - strtotime($post);
  $year = floor($seconds /31556926);
  $months = floor($seconds /2629743);
  $week=floor($seconds /604800);
  $day = floor($seconds /86400); 
  $hours = floor($seconds / 3600);
  $mins = floor(($seconds - ($hours*3600)) / 60); 
  $secs = floor($seconds % 60);
  if($seconds < 60) $time = $secs." seconds ago";
  else if($seconds < 3600 ) $time =($mins==1)?$mins."now":$mins." mins ago";
  else if($seconds < 86400) $time = ($hours==1)?$hours." hour ago":$hours." hours ago";
  else if($seconds < 604800) $time = ($day==1)?$day." day ago":$day." days ago";
  else if($seconds < 2629743) $time = ($week==1)?$week." week ago":$week." weeks ago";
  else if($seconds < 31556926) $time =($months==1)? $months." month ago":$months." months ago";
  else $time = ($year==1)? $year." year ago":$year." years ago";
  return $time; 
  }  



 // uses
 // $post_time="2017-12-05 02:05:12";
 // echo calculate_time_span($post_time); 

F
Frank Forte

这里的许多解决方案都没有考虑四舍五入。例如:

事件发生在两天前的下午 3 点。如果您在下午 2 点检查,它将显示一天前。如果您在下午 4 点检查,它将显示两天前。

如果您使用的是 unix 时间,这会有所帮助:

// how long since event has passed in seconds
$secs = time() - $time_ago;

// how many seconds in a day
$sec_per_day = 60*60*24;

// days elapsed
$days_elapsed = floor($secs / $sec_per_day);

// how many seconds passed today
$today_seconds = date('G')*3600 + date('i') * 60 + date('s');

// how many seconds passed in the final day calculation
$remain_seconds = $secs % $sec_per_day;

if($today_seconds < $remain_seconds)
{
    $days_elapsed++;
}

echo 'The event was '.$days_ago.' days ago.';

如果您担心闰秒和夏令时,这并不完美。


L
LukeWarm74

您必须获取时间戳的每一部分,并将其转换为 Unix 时间。例如时间戳,2009-09-12 20:57:19。

(((2008-1970)*365)+(8*30)+12)*24+20 会给你一个粗略估计自 1970 年 1 月 1 日以来的小时数。

取那个数字,乘以 60 再加上 57 得到分钟。

把它乘以 60 再加上 19。

然而,这会非常粗略且不准确地转换它。

你有什么理由不能从正常的 Unix 时间开始?


在 sql 表中存储为 unix 时间更好吗?我目前在时间戳列(可以更改为 unix)上使用 mysqls 自动时间戳更新。我只是在学习什么更好?
确实。我相信 mySQL 表的默认值是您引用的类型,但 Unix 时间更实用。您始终可以将其存储为 int。
您的数据库应该具有将日期转换为 UNIX 格式的功能。在 mysql 中,您使用 UNIX_TIMESTAMP()。哦,您通常应该将日期存储为 DATETIME 而不是 INT,以便您可以使用 sql 函数进行日期操作。
你永远不应该按时使用数学。你假设一个固定的日历,它不存在。使用 php 中提供的 Date 对象来处理...日期。
M
Mourad Karoudi

某些语言显示时间之前存在一些问题,例如在阿拉伯语中,显示日期需要 3 种格式。我在我的项目中使用这个功能希望他们可以帮助某人(任何建议或改进我都会很感激:))

/**
 *
 * @param   string $date1 
 * @param   string $date2 the date that you want to compare with $date1
 * @param   int $level  
 * @param   bool $absolute  
 */

function app_date_diff( $date1, $date2, $level = 3, $absolute = false ) {

    $date1 = date_create($date1);   
    $date2 = date_create($date2);
    $diff = date_diff( $date1, $date2, $absolute );

    $d = [
        'invert' => $diff->invert
    ];  

    $diffs = [
        'y' => $diff->y, 
        'm' => $diff->m, 
        'd' => $diff->d
    ];

    $level_reached = 0;

    foreach($diffs as $k=>$v) {

        if($level_reached >= $level) {
            break;
        }

        if($v > 0) {
            $d[$k] = $v;
            $level_reached++;
        }

    }

    return  $d;

}

/**
 * 
 */

function date_timestring( $periods, $format = 'latin', $separator = ',' ) {

    $formats = [
        'latin' => [
            'y' => ['year','years'],
            'm' => ['month','months'],
            'd' => ['day','days']
        ],
        'arabic' => [
            'y' => ['سنة','سنتين','سنوات'],
            'm' => ['شهر','شهرين','شهور'],
            'd' => ['يوم','يومين','أيام']
        ]
    ];

    $formats = $formats[$format];

    $string = [];

    foreach($periods as $period=>$value) {

        if(!isset($formats[$period])) {
            continue;
        }

        $string[$period] = $value.' ';
        if($format == 'arabic') {
            if($value == 2) {
                $string[$period] = $formats[$period][1];
            }elseif($value > 2 && $value <= 10) {
                $string[$period] .= $formats[$period][2];
            }else{
                $string[$period] .= $formats[$period][0];
            }

        }elseif($format == 'latin') {
            $string[$period] .= ($value > 1) ? $formats[$period][1] : $formats[$period][0];
        }

    }

    return implode($separator, $string);


}

function timeago( $date ) {

    $today = date('Y-m-d h:i:s');

    $diff = app_date_diff($date,$today,2);

    if($diff['invert'] == 1) {
        return '';
    }

    unset($diff[0]);

    $date_timestring = date_timestring($diff,'latin');

    return 'About '.$date_timestring;

}

$date1 = date('Y-m-d');
$date2 = '2018-05-14';

$diff = timeago($date2);
echo $diff;

S
Stefanov.sm

如果您使用的是 PostgreSQL,那么它将为您完成这项工作:

const DT_SQL = <<<SQL
WITH lapse AS (SELECT (?::timestamp(0) - now()::timestamp(0))::text t)
SELECT CASE
  WHEN (select t from lapse) ~ '^\s*-' THEN replace((select t from lapse), '-', '') ||' ago' 
  ELSE (select t from lapse) END;
SQL;

function timeSpanText($ts, $conn)
// $ts: date-time string, $conn: PostgreSQL PDO connection
{
 return $conn -> prepare(DT_SQL) -> execute([ts]) -> fetchColumn();
}

M
Maarten

我想要支持单数和复数的荷兰语版本。仅在末尾添加一个“s”是不够的,我们使用了完全不同的词,所以我重写了这篇文章的最佳答案。

这将导致:

2 jaren 1 maand 2 weken 1 dag 1 minuten 2 seconden

或者

1 jaar 2 maanden 1 week 2 dagen 1 minuut 1 seconde

    public function getTimeAgo($full = false){

    $now = new \DateTime;
    $ago = new \DateTime($this->datetime());
    $diff = $now->diff($ago);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'jaren',
        'm' => 'maanden',
        'w' => 'weken',
        'd' => 'dagen',
        'h' => 'uren',
        'i' => 'minuten',
        's' => 'seconden',
    );
    $singleString = array(
        'y' => 'jaar',
        'm' => 'maand',
        'w' => 'week',
        'd' => 'dag',
        'h' => 'uur',
        'i' => 'minuut',
        's' => 'seconde',
    );
    // M.O. 2022-02-11 I rewrote this function to support dutch singles and plurals. Added some docs for next programmer to break his brain :)
    // For each possible notation, if corresponding value of current key is true (>1) otherwise remove its key/value from array
    // If the value from current key is 1, use value from $singleString array. Otherwise use value from $string array
    foreach ($string as $k => &$v) {
        if ($diff->$k) {
            if($diff->$k == 1){
                $v = $diff->$k . ' ' . $singleString[$k];
            } else {
                $v = $diff->$k . ' ' . $v;
            }
        } else {
            if($diff->$k == 1){
                unset($singleString[$k]);
            } else {
                unset($string[$k]);
            }
        }
    }

    // If $full = true, print all values.
    // Values have already been filtered with foreach removing keys that contain a 0 as value
    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . '' : 'zojuist';
}

您可能应该先对其进行测试,因为我不是那么好的程序员:)


M
Mehreen Jamil
$time_ago = ' ';
$time = time() - $time; // to get the time since that moment
$tokens = array (
31536000 => 'year',2592000 => 'month',604800 => 'week',86400 => 'day',3600 => 'hour',
60  => 'minute',1 => 'second');
foreach ($tokens as $unit => $text) {
if ($time < $unit)continue;
$numberOfUnits = floor($time / $unit);
$time_ago = ' '.$time_ago. $numberOfUnits.' '.$text.(($numberOfUnits>1)?'s':'').'  ';
$time = $time % $unit;}echo $time_ago;

D
Dineshaws

这是我的解决方案,请根据您的要求检查和修改

function getHowLongAgo($date, $display = array('Year', 'Month', 'Day', 'Hour', 'Minute', 'Second'), $ago = '') {
        date_default_timezone_set('Australia/Sydney');
        $timestamp = strtotime($date);
        $timestamp = (int) $timestamp;
        $current_time = time();
        $diff = $current_time - $timestamp;

        //intervals in seconds
        $intervals = array(
            'year' => 31556926, 'month' => 2629744, 'week' => 604800, 'day' => 86400, 'hour' => 3600, 'minute' => 60
        );

        //now we just find the difference
        if ($diff == 0) {
            return ' Just now ';
        }

        if ($diff < 60) {
            return $diff == 1 ? $diff . ' second ago ' : $diff . ' seconds ago ';
        }

        if ($diff >= 60 && $diff < $intervals['hour']) {
            $diff = floor($diff / $intervals['minute']);
            return $diff == 1 ? $diff . ' minute ago ' : $diff . ' minutes ago ';
        }

        if ($diff >= $intervals['hour'] && $diff < $intervals['day']) {
            $diff = floor($diff / $intervals['hour']);
            return $diff == 1 ? $diff . ' hour ago ' : $diff . ' hours ago ';
        }

        if ($diff >= $intervals['day'] && $diff < $intervals['week']) {
            $diff = floor($diff / $intervals['day']);
            return $diff == 1 ? $diff . ' day ago ' : $diff . ' days ago ';
        }

        if ($diff >= $intervals['week'] && $diff < $intervals['month']) {
            $diff = floor($diff / $intervals['week']);
            return $diff == 1 ? $diff . ' week ago ' : $diff . ' weeks ago ';
        }

        if ($diff >= $intervals['month'] && $diff < $intervals['year']) {
            $diff = floor($diff / $intervals['month']);
            return $diff == 1 ? $diff . ' month ago ' : $diff . ' months ago ';
        }

        if ($diff >= $intervals['year']) {
            $diff = floor($diff / $intervals['year']);
            return $diff == 1 ? $diff . ' year ago ' : $diff . ' years ago ';
        }
    }

谢谢


R
Ramesh
# This function prints the difference between two php datetime objects
# in a more human readable form
# inputs should be like strtotime($date)
function humanizeDateDiffference($now,$otherDate=null,$offset=null){
    if($otherDate != null){
        $offset = $now - $otherDate;
    }
    if($offset != null){
        $deltaS = $offset%60;
        $offset /= 60;
        $deltaM = $offset%60;
        $offset /= 60;
        $deltaH = $offset%24;
        $offset /= 24;
        $deltaD = ($offset > 1)?ceil($offset):$offset;      
    } else{
        throw new Exception("Must supply otherdate or offset (from now)");
    }
    if($deltaD > 1){
        if($deltaD > 365){
            $years = ceil($deltaD/365);
            if($years ==1){
                return "last year"; 
            } else{
                return "<br>$years years ago";
            }   
        }
        if($deltaD > 6){
            return date('d-M',strtotime("$deltaD days ago"));
        }       
        return "$deltaD days ago";
    }
    if($deltaD == 1){
        return "Yesterday";
    }
    if($deltaH == 1){
        return "last hour";
    }
    if($deltaM == 1){
        return "last minute";
    }
    if($deltaH > 0){
        return $deltaH." hours ago";
    }
    if($deltaM > 0){
        return $deltaM." minutes ago";
    }
    else{
        return "few seconds ago";
    }
}

s
syrkull

此功能不适用于英语。我把这些话翻译成英文。这需要在用于英语之前进行更多修复。

function ago($d) {
$ts = time() - strtotime(str_replace("-","/",$d));

        if($ts>315360000) $val = round($ts/31536000,0).' year';
        else if($ts>94608000) $val = round($ts/31536000,0).' years';
        else if($ts>63072000) $val = ' two years';
        else if($ts>31536000) $val = ' a year';

        else if($ts>24192000) $val = round($ts/2419200,0).' month';
        else if($ts>7257600) $val = round($ts/2419200,0).' months';
        else if($ts>4838400) $val = ' two months';
        else if($ts>2419200) $val = ' a month';


        else if($ts>6048000) $val = round($ts/604800,0).' week';
        else if($ts>1814400) $val = round($ts/604800,0).' weeks';
        else if($ts>1209600) $val = ' two weeks';
        else if($ts>604800) $val = ' a week';

        else if($ts>864000) $val = round($ts/86400,0).' day';
        else if($ts>259200) $val = round($ts/86400,0).' days';
        else if($ts>172800) $val = ' two days';
        else if($ts>86400) $val = ' a day';

        else if($ts>36000) $val = round($ts/3600,0).' year';
        else if($ts>10800) $val = round($ts/3600,0).' years';
        else if($ts>7200) $val = ' two years';
        else if($ts>3600) $val = ' a year';

        else if($ts>600) $val = round($ts/60,0).' minute';
        else if($ts>180) $val = round($ts/60,0).' minutes';
        else if($ts>120) $val = ' two minutes';
        else if($ts>60) $val = ' a minute';

        else if($ts>10) $val = round($ts,0).' second';
        else if($ts>2) $val = round($ts,0).' seconds';
        else if($ts>1) $val = ' two seconds';
        else $val = $ts.' a second';


        return $val;
    }

L
Limitless isa

用于:

echo elapsed_time('2016-05-09 17:00:00'); // 18 saat 8 dakika önce yazıldı.

功能:

function elapsed_time($time){// Nekadar zaman geçmiş

        $diff = time() - strtotime($time); 

        $sec = $diff;
        $min = floor($diff/60);
        $hour = floor($diff/(60*60));
        $hour_min = floor($min - ($hour*60));
        $day = floor($diff/(60*60*24));
        $day_hour = floor($hour - ($day*24));
        $week = floor($diff/(60*60*24*7));
        $mon = floor($diff/(60*60*24*7*4));
        $year = floor($diff/(60*60*24*7*4*12));

        //difference calculate to string
        if($sec < (60*5)){
            return 'şimdi yazıldı.';
        }elseif($min < 60){
            return 'biraz önce yazıldı.';
        }elseif($hour < 24){
            return $hour.' saat '.$hour_min.' dakika önce yazıldı.';
        }elseif($day < 7){
            if($day_hour!=0){$day_hour=$day_hour.' saat ';}else{$day_hour='';}
            return $day.' gün '.$day_hour.'önce yazıldı.';
        }elseif($week < 4){
            return $week.' hafta önce yazıldı.';
        }elseif($mon < 12){
            return $mon.' ay önce yazıldı.';
        }else{
            return $year.' yıl önce yazıldı.';
        }
    }

一个月多于 28 天,一年多于 52 周。
M
Mr Sorbose

从上面稍微修改的答案:

  $commentTime = strtotime($whatever)
  $today       = strtotime('today');
  $yesterday   = strtotime('yesterday');
  $todaysHours = strtotime('now') - strtotime('today');

private function timeElapsedString(
    $commentTime,
    $todaysHours,
    $today,
    $yesterday
) {
    $tokens = array(
        31536000 => 'year',
        2592000 => 'month',
        604800 => 'week',
        86400 => 'day',
        3600 => 'hour',
        60 => 'minute',
        1 => 'second'
    );
    $time = time() - $commentTime;
    $time = ($time < 1) ? 1 : $time;
    if ($commentTime >= $today || $commentTime < $yesterday) {
        foreach ($tokens as $unit => $text) {
            if ($time < $unit) {
                continue;
            }
            if ($text == 'day') {
                $numberOfUnits = floor(($time - $todaysHours) / $unit) + 1;
            } else {
                $numberOfUnits = floor(($time)/ $unit);
            }
            return $numberOfUnits . ' ' . $text . (($numberOfUnits > 1) ? 's' : '') . ' ago';
        }
    } else {
        return 'Yesterday';
    }
}

d
demongolem

以下是一个非常简单且极其有效的解决方案。

function timeElapsed($originalTime){

        $timeElapsed=time()-$originalTime;

        /*
          You can change the values of the following 2 variables 
          based on your opinion. For 100% accuracy, you can call
          php's cal_days_in_month() and do some additional coding
          using the values you get for each month. After all the
          coding, your final answer will be approximately equal to
          mine. That is why it is okay to simply use the average
          values below.
        */
        $averageNumbDaysPerMonth=(365.242/12);
        $averageNumbWeeksPerMonth=($averageNumbDaysPerMonth/7);

        $time1=(((($timeElapsed/60)/60)/24)/365.242);
        $time2=floor($time1);//Years
        $time3=($time1-$time2)*(365.242);
        $time4=($time3/$averageNumbDaysPerMonth);
        $time5=floor($time4);//Months
        $time6=($time4-$time5)*$averageNumbWeeksPerMonth;
        $time7=floor($time6);//Weeks
        $time8=($time6-$time7)*7;
        $time9=floor($time8);//Days
        $time10=($time8-$time9)*24;
        $time11=floor($time10);//Hours
        $time12=($time10-$time11)*60;
        $time13=floor($time12);//Minutes
        $time14=($time12-$time13)*60;
        $time15=round($time14);//Seconds

        $timeElapsed=$time2 . 'yrs ' . $time5 . 'months ' . $time7 . 
                     'weeks ' . $time9 .  'days ' . $time11 . 'hrs '
                     . $time13 . 'mins and ' . $time15 . 'secs.';

        return $timeElapsed;

}

回声 timeElapsed(1201570814);

样本输出:

6 年 4 个月 3 周 4 天 12 小时 40 分 36 秒。


C
Ciaran

这是我前段时间构建的通知模块的解决方案。它返回类似于 Facebook 的通知下拉列表的输出(例如 1 天前、刚刚等)。

public function getTimeDifference($time) {
    //Let's set the current time
    $currentTime = date('Y-m-d H:i:s');
    $toTime = strtotime($currentTime);

    //And the time the notification was set
    $fromTime = strtotime($time);

    //Now calc the difference between the two
    $timeDiff = floor(abs($toTime - $fromTime) / 60);

    //Now we need find out whether or not the time difference needs to be in
    //minutes, hours, or days
    if ($timeDiff < 2) {
        $timeDiff = "Just now";
    } elseif ($timeDiff > 2 && $timeDiff < 60) {
        $timeDiff = floor(abs($timeDiff)) . " minutes ago";
    } elseif ($timeDiff > 60 && $timeDiff < 120) {
        $timeDiff = floor(abs($timeDiff / 60)) . " hour ago";
    } elseif ($timeDiff < 1440) {
        $timeDiff = floor(abs($timeDiff / 60)) . " hours ago";
    } elseif ($timeDiff > 1440 && $timeDiff < 2880) {
        $timeDiff = floor(abs($timeDiff / 1440)) . " day ago";
    } elseif ($timeDiff > 2880) {
        $timeDiff = floor(abs($timeDiff / 1440)) . " days ago";
    }

    return $timeDiff;
}

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