帮助处理重复的 JavaScript 倒计时和时区

Question

目标： 创建下一个可用直播的倒计时。

详细信息： 我们每周直播六次（太平洋标准时间）。
1. 周日上午 8:00
2. 周日上午 10:00
3. 周日中午 12:00
4. 周日下午 6:30
5. 周三晚上 7:00
6. 周六上午 10:00

我的方法： 最初，我检查今天是星期几以及时间，然后创建下一个流的倒计时。这对于我们的时区 (PST) 以及任何相对接近的时区来说效果很好。

问题： 首先检查日期打破了我对位于时区的人们的方法，这些时区使他们比我们早一天。例如，如果来自伦敦的某人在周四凌晨 2:30（格林威治标准时间）检查该网站，我的代码将开始倒计时到周六上午 10:00 直播，而不是周三晚上 7:00（太平洋标准时间），后者将于 30 点开始分钟

我确信我所做的事情可以得到清理和改进，请告诉我如何做。

我的代码：

// Countdown
function countDown( addDay, hour, min ) {
    var newCount = new Date(); 
    newCount = new Date(newCount.getFullYear(), newCount.getMonth(), newCount.getDate() + addDay, hour, min); 
    $('#service_countdown').countdown({until: $.countdown.UTCDate(-8, newCount)}); 
}

var day_ref, day, addDay, hour, min;
addDay = 0;
hour = 0;
min = 0;
day = {};

day_ref = new Date();
day = {"day" : day_ref.getDay(), "hour" : day_ref.getHours(), "min" : day_ref.getMinutes()};

// It's Sunday
if ( day.day === 0 ) {

    if ( day.hour < 8 ) {
          hour = 8;
        countDown(addDay, hour, min);
    } 
    else if ( day.hour < 10 ) {
          hour = 10;
        countDown(addDay, hour, min);
    } 
    else if ( day.hour < 12 ) {
          hour = 12;
        countDown(addDay, hour, min);
    }
    else if ( day.hour <= 18 && day.min < 30 ) {
          hour = 18;
           min = 30;
        countDown(addDay, hour, min);
    } 
    else {
        addDay = 3;
          hour = 19;
        countDown(addDay, hour, min);
    }
} 
// It's Monday
else if ( day.day === 1 ) {
    addDay = 2;
      hour = 19;
    countDown(addDay, hour, min);
} 
// It's Tuesday
else if ( day.day === 2 ) {
    addDay = 1;
      hour = 19;
    countDown(addDay, hour, min);
} 
// It's Wednesday
else if ( day.day === 3) {

    if ( day.hour < 19 ) {
          hour = 19;
        countDown(addDay, hour, min);
    } else {
        addDay = 3;
          hour = 10;
        countDown(addDay, hour, min);
    }
} 
// It's Thursday
else if ( day.day === 4 ) {
    addDay = 2;
      hour = 10;
    countDown(addDay, hour, min);
} 
// It's Friday
else if ( day.day === 5 ) {
    addDay = 1;
      hour = 10;
    countDown(addDay, hour, min);
} 
// All that's left is Saturday
else {

    if ( day.hour < 10 ) {
          hour = 10;
        countDown(addDay, hour, min);
    } else {
        addDay = 1;
          hour = 8;
        countDown(addDay, hour, min);
    }
}

提前感谢您的帮助！

Answer 1

我最终只是一直在计算服务器端。在 bob-the-destroyer 的帮助下，考虑到要求，工作 php 非常紧凑。

<?php

$schedule = array(
    'this Sunday 8am',
    'this Sunday 10am',
    'this Sunday 12pm',
    'this Sunday 6:30pm',
    'this Wednesday 7pm',
    'this Saturday 10am'
    );

$current_time = strtotime('now');
foreach ($schedule as &$val) {
    $val = strtotime($val);
    // fix schedule to next week if time resolved to the past
    if ($val - $current_time < 0) $val += 604800; 
    }
sort($schedule);
$countdown = $schedule[0] - $current_time;

?>

<script type="text/javascript">
    var myTime = <?php echo $countdown; // just personally prefer full tags ?>;
    $('#countdown').countdown({ until: myTime}); 
</script>