如果当前时间在餐厅的开放时间内,我会尝试计算。
这个问题在Stackoverflow上已经被问了很多,但我还没有找到一个可以解释我所遇到的问题的原因。此外,将很高兴看到一个更好的方式来做这件事的想法。
目前,如果当天关闭(在本例中是周日),或者如果是“周六”凌晨1点(所以从技术上讲是周日凌晨1点),它就会中断。我有一种感觉,我将不得不改变数据存储的方式,以解决午夜后的问题,但我正在尝试使用我现在拥有的东西。这是一个问题,因为大多数餐馆列出的某一天的营业时间是下午5点到凌晨2点,而不是下午5点到12点,中午12点到凌晨2点
不管怎么说,这就是我所拥有的。请告诉我做这件事的更好方法。
我有这样的时间存储:
$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';
}
}
}
发布于 2013-10-20 04:49:20
以下是我的面向对象的解决方案,基于PHP (从5.2版本开始可用)的使用:
<?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;
}
?>
正如您所看到的,构造函数构建了一个内部表单( DateTime对象的openings
数组),然后在isOpen
方法中通过简单的比较使用该表单,以检查在调用时餐厅是开着还是关着。
您还会注意到,我使用了DateTime:add方法来计算明天的日期,而不是在当前日期时间戳上加上86400 (24*60*60),以避免DST时间转换的问题。
概念验证:
<?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;
?>
它给出了以下结果:
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;至少一年不是两次……
发布于 2013-10-22 17:16:32
假设我们有另一个数组,而不是这样的数组,其中包含以下类型的条目:
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开门,我们必须做一些简单的事情,比如:
$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,格式为mon
、tue
、wed
等,并且有两个定义时间范围的字符串,例如8:30am
和3:15pm
,则如下函数将返回相应的时间段,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'=>$to_ts);
}
strtotime()
可以创造奇迹,对吧?请注意,如果时间段的结束时间比开始时间早,我们假设它指的是第二天,并重新计算它。
编辑:一开始,我天真地认为我可以通过添加一天的秒数来更正它。这并不完全准确,因为操作时间戳不会保留DST信息。因此,如果一个时隙包含一个白班(午夜)和一个DST班次,它将在一个小时内给出不准确的结果。再次使用strtotime()
,加上相同的参数,再加一天,就可以解决这个问题。
yaEDIT:修复了另一个错误(希望是最后一个):当餐厅在周日营业到午夜后,$to_time
应该会包装到本周的周一,同样的时间。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]);
}
}
这是a little phpfiddle to demonstrate this。
编辑:受到另一个答案中“简洁”讨论的启发,我想我应该给出我的“紧凑”版本。使用完全相同的逻辑,它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>";
然而,我自己仍然更喜欢原始版本。除了拥有函数的明显好处之外,$slots
数组还可以很好地缓存和重用,这使得相关计算比重新解析原始数据容易得多。
发布于 2013-10-22 18:38:02
这可能不是最有效的,但它应该能很好地解决你手头的问题:
$times = array(
'opening_hours_mon' => '9am - 8pm',
'opening_hours_tue' => '9am - 2am',
'opening_hours_wed' => '8:30am - 2am',
'opening_hours_thu' => '9am - 3pm',
'opening_hours_fri' => '8:30am - 11am',
'opening_hours_sat' => '9am - 3pm, 5pm - 2am',
'opening_hours_sun' => 'closed'
);
var_dump(is_open($times, strtotime('sun 1am'))); // true
这里是第一个函数,设计简单;它使用一个开放和关闭时间的网格,并确定给定的时间是否与任何一个范围匹配:
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 < $range[1]) {
return true;
}
}
return false;
}
这个函数构建实际的网格;如果一个范围结束在它相应的开始之前,它将创建两个范围,一个范围代表被跨越的每一天。
function get_time_grid($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;
}
代码中只有几个注释,但我希望您能跟随正在做的事情。如果你需要任何澄清,请告诉我。
https://stackoverflow.com/questions/14692506
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