当前时间是否在两个给定时间之间？

Question

我正在尝试计算当前时间是否在餐厅的营业时间内。

这个问题在 Stackoverflow 上被问了很多，但我还没有找到一个可以解释我遇到的问题的问题。另外，很高兴看到有更好的方法来做到这一点。

目前，如果当天关闭（本例中为星期日）或“星期六”凌晨 1 点（从技术上讲是星期天早上 1 点），它会中断。我有一种感觉，我必须改变数据的存储方式，以解决午夜之后的问题，但我正在尝试使用我现在拥有的东西。这是一个问题，因为大多数餐厅将特定日期的营业时间列为下午 5 点至凌晨 2 点，而不是下午 5 点至凌晨 12 点、凌晨 12 点至凌晨 2 点。

无论如何，这就是我所拥有的。请告诉我一个更好的方法。

我的时间是这样存储的：

$times = array(
    'opening_hours_mon' => '9am - 8pm',
    'opening_hours_tue' => '9am - 2am',
    'opening_hours_wed' => '8:30am - 2am',
    'opening_hours_thu' => '5:30pm - 2am',
    'opening_hours_fri' => '8:30am - 11am',
    'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
    'opening_hours_sun' => 'closed'
);

这是我现在使用的代码：

// Get the right key for today
$status = 'open';
$now = (int) current_time( 'timestamp' );
$day = strtolower( date('D', $now) );
$string = 'opening_hours_'.$day;

$times = $meta[$string][0]; // This should be a stirng like '6:00am - 2:00am' or even '6:00am - 11:00am, 1:00pm to 11:00pm'.

// Does it contain a '-', if not assume it's closed.
$pos = strpos($times, '-');
if ($pos === false) {       
    $status = 'closed';
} else {

    // Maybe a day has multiple opening times?
    $seating_times = explode(',', $times);
    foreach( $seating_times as $time ) {

        $chunks = explode('-', $time);
        $open_time = strtotime($chunks[0]);
        $close_time = strtotime($chunks[1]);

        // Calculate if now is between range of open and closed
        if(($open_time <= $now) && ($now <= $close_time)) {
            $status = 'open';
            break;
        } else {
            $status = 'closed';             
        }

    }

}

注意：current_time('timestamp',0) is a WordPress function.

Answer 1

这是我的面向对象的解决方案，基于 PHP DateTime class 的使用（从 5.2 版本开始可用）：

<?php 

class Restaurant {
    private $cw;
    private $times = array();
    private $openings = array();

    public function __construct(array $times) {
        $this->times = $times;
        $this->setTimes(date("w") ? "this" : "last");
        //print_r($this->openings);       // Debug
    }

    public function setTimes($cw) {
        $this->cw = $cw;
        foreach ($this->times as $key => $val) {
            $t = array();
            $buf = strtok($val, ' -,');
            for ($n = 0; $buf !== FALSE; $n++) {
                try {
                    $d = new DateTime($buf);
                    $d->setTimestamp(strtotime(substr($key, -3)." {$this->cw} week {$buf}"));
                    if ($n && ($d < $t[$n-1])) {
                        $d->add(new DateInterval('P1D'));
                    }
                    $t[] = $d;
                } catch (Exception $e) {
                    break;
                }
                $buf = strtok(' -,');
            }
            if ($n % 2) {
                throw new Exception("Invalid opening time: {$val}");
            } else {
                $this->openings[substr($key, -3)] = $t;
            }
        }
    }

    public function isOpen() {
        $cw = date("w") ? "this" : "last";
        if ($cw != $this->cw) {
            $this->setTimes($cw);
        }
        $d = new DateTime('now');
        foreach ($this->openings as $wd => $t) {
            $n = count($t);
            for ($i = 0; $i < $n; $i += 2) {
                if (($d >= $t[$i]) && ($d <= $t[$i+1])) {
                    return(TRUE);
                }
            }
        }
        return(FALSE);
    }
}

$times = array(
    'opening_hours_mon' => '9am - 8pm',
    'opening_hours_tue' => '9am - 2am',
    'opening_hours_wed' => '8:30am - 2am',
    'opening_hours_thu' => '9am - 3pm',
    'opening_hours_fri' => '8:30am - 11am',
    'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
    'opening_hours_sun' => 'closed'
);

try {
    $r = new Restaurant($times);
    $status = $r->isOpen() ? 'open' : 'closed';
    echo "status=".$status.PHP_EOL;
} catch (Exception $e) {
    echo $e->getMessage().PHP_EOL;
}

?>

如您所见，构造函数构建了一个内部表单（DateTime 对象的openings 数组），然后将其与isOpen 方法中的简单比较一起使用，以检查是否在调用餐厅时已打开或关闭。

您还会注意到，我使用 DateTime:add 方法来计算明天的日期，而不是在当前日期时间戳上添加 86400 (24*60*60)，以避免出现 DST 时间偏移问题。
概念证明：

<?php

ini_set("date.timezone", "Europe/Rome");
echo "date.timezone = ".ini_get("date.timezone").PHP_EOL;

$d1 = strtotime("2013-10-27 00:00:00");
$d2 = strtotime("2013-10-28 00:00:00");
// Expected: 86400, Result: 90000
echo "Test #1: ".($d2 - $d1).PHP_EOL;
// Expected: 2013-10-28 00:00:00, Result: 2013-10-27 23:00:00
echo "Test #2: ".date("Y-m-d H:i:s", $d1 + 86400).PHP_EOL;

$d1 = strtotime("2014-03-30 00:00:00");
$d2 = strtotime("2014-03-31 00:00:00");
// Expected: 86400, Result: 82800
echo "Test #3: ".($d2 - $d1).PHP_EOL;
// Expected: 2014-03-30 00:00:00, Result: 2014-03-29 23:00:00
echo "Test #4: ".date("Y-m-d H:i:s", $d2 - 86400).PHP_EOL;

?>

结果如下：

date.timezone = Europe/Rome
Test #1: 90000
Test #2: 2013-10-27 23:00:00
Test #3: 82800
Test #4: 2014-03-29 23:00:00

因此，似乎一天并不总是有 86400 秒；一年至少两次……

Answer 2

假设我们有另一个具有以下类型条目的数组，而不是这样的数组：

Array ( [from] => 1382335200 [to] => 1382374800 )

from 和 to 值是时间戳，通过将数组的信息投影到当前（运行）周来计算。

然后，为了检查餐厅现在是否营业，我们必须做一些简单的事情：

$slots=..... /* calculate time slots array */
$status='closed';
$rightnow=time();
foreach($slots as $slot)
  if($rightnow<=$slot['to'])
    {
    if($rightnow>=$slot['from']) $status='open';
    break;
    }
echo "The restaurant is <strong>$status</strong> right now<br>";

给定一个工作日，形式为mon、tue、wed等和两个定义时间范围的字符串，例如8:30am和3:15pm，下面的函数会返回对应的时间插槽，如上所述：

function get_time_slot($weekday,$fromtime,$totime)
  {
  $from_ts=strtotime("this week $weekday $fromtime");
  $to_ts=strtotime("this week $weekday $totime");
  if($to_ts<$from_ts)
    {
    $to_ts=strtotime("this week $weekday +1 day $totime");
    if($to_ts>strtotime("next week midnight")) 
      $to_ts=strtotime("this week mon $totime");
    }
  return array('from'=>$from_ts,'to'=>$to_ts);
  }

strtotime()可以创造奇迹吧？请注意，如果时间段的结束时间早于开始时间，我们假设它指的是第二天，我们会重新计算它。

编辑：起初，我天真地以为我会通过增加一天的秒数来纠正它。这并不完全准确，因为操作时间戳不会保留 DST 信息。因此，如果一个时间段包括白班（午夜）和还有一个 DST 班，它会给出一个小时的不准确结果。再次使用strtotime()，加上相同的参数加上一天，就可以解决问题了。

yaEDIT： 修复了另一个错误（希望是最后一个错误）：当餐厅在周日营业到午夜之后，$to_time 应该会在本周的星期一同一时间结束。 呼！

现在，为了转换您的数组，您需要执行以下操作：

$slots=array();
foreach($times as $key=>$entry)
  {
  list(,,$dow)=explode('_',$key);
  foreach(explode(',',$entry) as $d)
    {
    $arr=explode('-',$d);
    if(count($arr)==2) $slots[]=get_time_slot($dow,$arr[0],$arr[1]);
    }
  }

这里是a little phpfiddle to demonstrate this。

---

编辑：受另一个答案中“简洁”讨论的启发，我想我会给出我的“紧凑”版本。使用完全相同的逻辑，可以归结为以下内容：

$status='closed';
$rightnow=time();
foreach($times as $key=>$entry)
  {
  list(,,$dow)=explode('_',$key);
  foreach(explode(',',$entry) as $d)
    if(count($arr=explode('-',$d))==2)
      {
      $from_ts=strtotime("this week $dow {$arr[0]}");
      $to_ts=strtotime("this week $dow {$arr[1]}");
      if($to_ts<$from_ts) $to_ts=strtotime("this week $dow +1 day {$arr[1]}");
        {
        $to_ts=strtotime("this week $dow +1 day {$arr[1]}");
        if($to_ts>strtotime("next week midnight")) 
          $to_ts=strtotime("this week mon {$arr[1]}");
        }
      if($rightnow<=$to_ts)
        {
        if($rightnow>=$from_ts) $status='open';
        break 2; // break both loops
        }
      }
  }
echo "<hr>The restaurant is <strong>$status</strong> right now<br>";

不过，我自己还是更喜欢原版。除了拥有函数的明显好处之外，$slots 数组还可以很好地被缓存和重复使用，这使得相关计算比重新解析原始数据容易得多。

Answer 3

这可能不是最有效的，但它应该可以很好地解决您手头的问题：

$times = array(
    'opening_hours_mon' => '9am - 8pm',
    'opening_hours_tue' => '9am - 2am',
    'opening_hours_wed' => '8:30am - 2am',
    'opening_hours_thu' => '9am - 3pm',
    'opening_hours_fri' => '8:30am - 11am',
    'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
    'opening_hours_sun' => 'closed'
);

var_dump(is_open($times, strtotime('sun 1am'))); // true

这是第一个功能，设计简单；它使用开盘和关盘时间网格，并确定给定时间是否与任何范围匹配：

function is_open($times, $now)
{
    $today = strtotime('today', $now);

    $grid = get_time_grid($times);
    $today_name = strtolower(date('D', $today));
    $today_seconds = $now - $today;

    foreach ($grid[$today_name] as $range) {
        if ($today_seconds >= $range[0] && $today_seconds < $range[1]) {
            return true;
        }
    }

    return false;
}

这个函数构建实际的网格；如果范围结束出现在其相应的开始之前，它将创建两个范围，一个用于跨越的每一天。

function get_time_grid($times)
{
    static $next_day = array(
        'mon' => 'tue', 'tue' => 'wed', 'wed' => 'thu',
        'thu' => 'fri', 'fri' => 'sat', 'sat' => 'sun',
        'sun' => 'mon'
    );
    static $time_r = '(\d{1,2}(?::\d{2})?(?:am|pm))';

    $today = strtotime('today');
    $grid = [];

    foreach ($times as $key => $schedule) {
        $day_name = substr($key, -3);
        // match all time ranges, skips "closed"
        preg_match_all("/$time_r - $time_r/", $schedule, $slots, PREG_SET_ORDER);
        foreach ($slots as $slot) {
            $from = strtotime($slot[1], $today) - $today;
            $to = strtotime($slot[2], $today) - $today;

            if ($to < $from) { // spans two days
                $grid[$day_name][] = [$from, 86400];
                $grid[$next_day[$day_name]][] = [0, $to];
            } else { // normal range
                $grid[$day_name][] = [$from, $to];
            }
        }
    }

    return $grid;
}

代码中只有几个cmets，但我希望你能关注正在做的事情。如果您需要任何说明，请告诉我。

Answer 4

我过去做过类似的事情，但采取了完全不同的方法。我将营业时间存储在单独的表格中。

CREATE TABLE `Openinghours`
(
    `OpeninghoursID` int, // serial
    `RestaurantID` int, // foreign key
    `Dayofweek` int, // day of week : 0 (for Sunday) through 6 (for Saturday)
    `Opentime` int, // time of day when restaurant opens (in seconds)
    `Closetime` int // time of day when restaurant closes (in seconds)
);

如果餐厅每天有多个营业时间，您只需添加 2 条记录（或需要更多 ID）。使用这样的表的好处是您可以简单地查询以查看哪些餐厅营业。

$day = date('w');
$now = time()-strtotime("00:00");
$query = "Select `RestaurantID` from `Openinghours` where `Dayofweek` = ".$day." and `Opentime` <= ".$now." and `Closetime` > ".$now;

使用这种系统的另一个好处是，您可以调整查询以获得不同的结果，例如：哪些餐厅现在营业并至少再营业一个小时（没有必要在关闭前几分钟去餐厅)

$day = date('w');
$now = time()-strtotime("00:00");
$query = "Select `RestaurantID` from `Openinghours` where `Dayofweek` = ".$day." and `Opentime` <= ".$now." and `Closetime` > ".($now+3600);

当然，它需要重新格式化您当前的数据，但它带来了不错的功能。

Answer 5

这是另一个无需重新格式化数据的解决方案。

$times = array(
    'opening_hours_mon' => '9am - 8pm',
    'opening_hours_tue' => '9am - 2am',
    'opening_hours_wed' => '8:30am - 2am',
    'opening_hours_thu' => '5:30pm - 2am',
    'opening_hours_fri' => '8:30am - 11am',
    'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
    'opening_hours_sun' => 'closed'
);

function compileHours($times, $timestamp) {
    $times = $times['opening_hours_'.strtolower(date('D',$timestamp))];
    if(!strpos($times, '-')) return array();
    $hours = explode(",", $times);
    $hours = array_map('explode', array_pad(array(),count($hours),'-'), $hours);
    $hours = array_map('array_map', array_pad(array(),count($hours),'strtotime'), $hours, array_pad(array(),count($hours),array_pad(array(),2,$timestamp)));
    end($hours);
    if ($hours[key($hours)][0] > $hours[key($hours)][1]) $hours[key($hours)][1] = strtotime('+1 day', $hours[key($hours)][1]);
    return $hours;
}

function isOpen($now, $times) {
    $open = 0; // time until closing in seconds or 0 if closed
    // merge opening hours of today and the day before
    $hours = array_merge(compileHours($times, strtotime('yesterday',$now)),compileHours($times, $now)); 

    foreach ($hours as $h) {
        if ($now >= $h[0] and $now < $h[1]) {
            $open = $h[1] - $now;
            return $open;
        } 
    }
    return $open;
}

$now = strtotime('7:59pm');
$open = isOpen($now, $times);

if ($open == 0) {
    echo "Is closed";
} else {
    echo "Is open. Will close in ".ceil($open/60)." minutes";
}

?>

我进行了几次测试，它似乎按预期工作，考虑了我能想到的所有方面。如果你发现这个有问题，请告诉我。 我知道这种方法看起来有点讨厌，但我只想使用简单的函数（除了 array_map 的棘手部分）并尽可能短。

Answer 6

周计时数组

$times = array(
    'opening_hours_mon' => '9am - 8pm',
    'opening_hours_tue' => '9am - 2am',
    'opening_hours_wed' => '8:30am - 2am',
    'opening_hours_thu' => '9am - 3pm',
    'opening_hours_fri' => '8:30am - 11am',
    'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
    'opening_hours_sun' => 'closed'
);

用法示例：

ini_set( "date.timezone", "Pacific/Auckland" ); // Make sure correct timezone is set

echo ( isOpen( $times ) ? 'Open' : 'Closed' );    

echo ( isOpen( $times ,"10am" ) ? 'Open' : 'Closed' );

函数定义：

/*
 *  First parameter : Weeks timings as array
 *  Second parameter : Time to check as string
 *  Return value : boolean
 */
function isOpen( $times ,$timeToCheck = 'now' )
{
    $timeToCheckAsUnixTimestamp = strtotime( $timeToCheck );

    $yesterdayTimes = $todayTimes = '';
    //  Find yesterday's times and today's times
    foreach( $times as $day => $timeRange )
    {
        $yesterdayTimes = ( stripos( $day ,date( "D" ,time() - 60 * 60 * 24 ) ) !== false ? $timeRange : $yesterdayTimes );
        $todayTimes = ( stripos( $day ,date( "D" ) ) !== false ? $timeRange : $todayTimes );
    }
    //  Handle closed
    if( strcasecmp( $todayTimes ,'closed' ) == 0 ) return false;
    if( strcasecmp( $yesterdayTimes ,'closed' ) == 0 ) $yesterdayTimes = '12am - 12am';
    //  Process and check with yesterday's timings
    foreach( explode( ',' ,$yesterdayTimes ) as $timeRanges )
    {
        list( $from ,$to ) = explode( '-' ,$timeRanges );
        list( $fromAsUnixTimestamp ,$toAsUnixTimestamp ) = array( strtotime( $from .' previous day' ) ,strtotime( $to .' previous day'  ) );
        $toAsUnixTimestamp = ( $toAsUnixTimestamp < $fromAsUnixTimestamp ? strtotime( $to ) : $toAsUnixTimestamp );
        if( $fromAsUnixTimestamp <= $timeToCheckAsUnixTimestamp and $timeToCheckAsUnixTimestamp <= $toAsUnixTimestamp ) return true;
    }
    //  Process and check with today's timings
    foreach( explode( ',' ,$todayTimes ) as $timeRanges )
    {
        list( $from ,$to ) = explode( '-' ,$timeRanges );
        list( $fromAsUnixTimestamp ,$toAsUnixTimestamp ) = array( strtotime( $from ) ,strtotime( $to ) );
        $toAsUnixTimestamp = ( $toAsUnixTimestamp < $fromAsUnixTimestamp ? strtotime( $to .' next day' ) : $toAsUnixTimestamp );
        if( $fromAsUnixTimestamp <= $timeToCheckAsUnixTimestamp and $timeToCheckAsUnixTimestamp <= $toAsUnixTimestamp ) return true;
    }
    return false;
}

Answer 7

如果您为此使用数据库，为什么不为此使用 datetime。

示例：

sunday 14:28, saturday 1:28

您可以拆分这两部分并在字符串时间（第 2 部分）中比较它们。您可以使用 strtotime 将字符串时间转换为时间戳并进行比较。

示例：

$date = "sunday 14:28"; 
echo $stamp = strtotime($date);

输出：

1360492200

喜欢这段代码：

$Time="sunday  14:28 , saturday 1:28";
$tA=explode(",",$Time);
$start=strtotime($tA[0]);
$end=strtotime($tA[1]);
$now=time();
if($now>$start and $now<$end){
   echo "is open";
}else{
   echo "is close";
}

但是您在更新它们时遇到问题，您可以这样做。

Answer 8

不改变时间存储格式的解决方案

<?php
    $times = array(
        'opening_hours_mon' => '9am - 8pm',
        'opening_hours_tue' => '5pm - 2am',
        'opening_hours_wed' => '8:30am - 2am',
        'opening_hours_thu' => '9am - 3pm',
        'opening_hours_fri' => '8:30am - 11am',
        'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
        'opening_hours_sun' => 'closed'
    );

    //$time_go = '13:00';
    $time_go = date('H:i');

    //$day_go = 1; //monday
    $day_go = (date('N') - 1);

    if(Are_they_open($time_go, $day_go, $times)){
        echo 'jep';
    }
    else{
        echo 'nope';
    }

    function Are_they_open($time_go, $day_go, $times){
        // the magic
        $times = array_values($times);
        $day_go = explode(',', $times[$day_go]);
        $time_go = hh_mm_toSeconds($time_go);

        foreach($day_go as $time){

            if((!$time) || ($time == 'closed')){
                return false;
            }

            $time = explode(' - ', $time);
            $time_open = hh_mm_toSeconds(date('H:i', strtotime($time[0])));
            $time_close = hh_mm_toSeconds(date('H:i', strtotime($time[1])));

            if(($time_go > $time_open) && ($time_go < $time_close)){
                return true;
            }
            elseif (($time_open > $time_close) || ($time_go > $time_open)){
                return true;
            }

        }

        return false;
    }

    function hh_mm_toSeconds($str_time){
        sscanf($str_time, "%d:%d", $hours, $minutes);
        return ($hours * 3600) + ($minutes * 60);
    }
?>

更改时间格式的解决方案

$times = array(
    1 => array(
        array('07:00', '17:00')
    ),
    2 => array(
        array('07:00', '14:30'),
        array('15:00', '20:00')
    ),
    3 => array(
        array('07:00', '17:00')
    ),
    4 => false, //closed
    5 => array(
        array('07:00', '17:00'),
        array('20:00', '24:00')
    ),
    6 => array(
        array('00:00', '03:00'),
        array('07:00', '17:00'),
        array('20:00', '24:00')
    ),
    7 => array(
        array('00:00', '03:00')
    ),
);

Answer 9

这是我的解决方案：

输入数据：

$meta = array(
   'opening_hours_mon' => '9am - 8pm',
   'opening_hours_tue' => '9am - 2am',
   'opening_hours_wed' => '8:30am - 2am',
   'opening_hours_thu' => '9am - 3pm',
   'opening_hours_fri' => '8:30am - 11am',
   'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
   'opening_hours_sun' => 'closed'

);

current_time('timestamp')（如作者所说）WordPress 中time() 的模拟
及解决方案：

    $now = (int) current_time( 'timestamp' );
    $day = strtolower(date('D', $now));
    $yesterday = strtolower(date('D', strtotime('-1 day')));
    $days = array(
        'yesterday' => $meta['opening_hours_'.$yesterday],
        'today' => $meta['opening_hours_'.$day],
    );
    $status = false;
    foreach($days as $when=>$times)
    {
        $parts = explode(',',$times);
        foreach($parts as $p)
        {
            if($p == 'closed')
                break;
            else{
                list($b,$e) = explode('-',$p);
                $b = strtotime("$when $b");
                $e = strtotime("$when $e");
                if($b > $e)
                    $e += strtotime("$when $e +1 day");;
                if($b <= $now && $now <= $e)
                {
                    $status =true;
                    break;
                }
            }
        }
    }

用于测试：
您可以将前 3 行更改为以下内容：

$now = (int) strtotime('today 3:00am');
$day = strtolower(date('D', $now));
$yesterday = strtolower(date('D', strtotime('yesterday 3:00am')));

Answer 10

将您的数组作为参数传递给此函数，您将在当前时间为打开和关闭为 false。这是简单直接的功能。我只检查今天的开放时间，如果有必要的话，我只检查昨天的开放时间，而无需在整个工作日中进行不必要的循环。也许它可以改进一点，但它的工作原理并不复杂。

function isOpen($times) {
    $times = array_values($times); //we will use numeric indexes
    $now = new DateTime();
    $day = $now->format('N'); 
    $day--; //days are counted 1 to 7 so we decrement it to match indexes
    $period = $times[$day];
    if($period!='closed') {
        $opening = explode('-', $period);
        $open = new DateTime($opening[0]);
        $close = new DateTime($opening[1]);
        if($close<$open) {
            //it means today we close after midnight, it is tomorrow
            $close->add(new DateInterval('P1D'));
        }
        if($open<$now && $now<$close) {
            //we are open
            return true;
        }
    }
    if($period=='closed' || $now<$open) {
        //now we check if we still open since yesterday
        $day = $day==0 ? 6 : $day-1;
        $period = $times[$day];
        if($period=='closed') return false;
        $opening = explode(' - ', $period);
        $open = new DateTime($opening[0]);
        $close = new DateTime($opening[1]);
        if($close<$open) {
        //it means yesterday we closed after midnight
            if($now<$close) {
                //we are before closing time
                return true;
            }
        }
    }
    return false;
}

Answer 11

也许我没有完全理解这一点，但是通过一个基本的循环，也许这可以工作：

if( ($date('D') == "3" ) )
{
    $open = "8";
    $close = "17";
    $now = time();

    if( ($now > $open) and ($now < $close) )
    {
        echo "open";
    }
    else
    {
        echo "closed";
    }
}

也许不是最好的解决方法，因为这不包括假期等，并且需要一些条件语句，但我认为这可以工作。嗯，不是最佳的，但一直对我有用。