开放时间

我有一个脚本，根据星期几和一天中的时间显示不同的div。 他运作良好。 我需要修改它。 我想每天设置不同的开放时间。 我怎样才能做到这一点？ 请帮忙。

例

星期一开放时间为8:00至16:00 周二开放时间为8:00至16:00 周三开放时间为8:00至18:00 周四开放时间为8:00至18:00 周五开放时间为8:00至16:00 周六关闭 周日开放时间为8:00至13:00

我的剧本：

var d = new Date();
var dayOfWeek = d.getDay();
var hour = d.getHours();
var mins = d.getMinutes();
var status = 'open';

if (dayOfWeek !== 6 && dayOfWeek !== 0 && hour >= 9 && hour <= 15){
    if (hour=='9' && mins < '00'){
        status = 'closed';
    }else if (hour=='15' && mins > '30'){
        status = 'closed';
    }else{
        status = 'open';
    }
}else{
    status = 'closed';
}

if (status=='open') {
    $('.hours').show();
    $('.closed').hide();
}else{
    $('.hours').hide();
    $('.closed').show();
} 
  
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="hours">We are OPEN</div>
<div class="closed">We are CLOSED</div> 
  
 
I have a script that shows different div depending on the day of the week and the time of day. He works well. I need to modify it. I want to set different opening hours each day. How can I do this? Please help.
 
Example
 
Monday open from 8:00 to 16:00 Tuesday open from 8:00 to 16:00 Wednesday open from 8:00 to 18:00 Thursday open from 8:00 to 18:00 Friday open from 8:00 to 16:00 Saturday close Sunday open from 8:00 to 13:00
 
My script:
 
 
 
  
  
var d = new Date();
var dayOfWeek = d.getDay();
var hour = d.getHours();
var mins = d.getMinutes();
var status = 'open';

if (dayOfWeek !== 6 && dayOfWeek !== 0 && hour >= 9 && hour <= 15){
    if (hour=='9' && mins < '00'){
        status = 'closed';
    }else if (hour=='15' && mins > '30'){
        status = 'closed';
    }else{
        status = 'open';
    }
}else{
    status = 'closed';
}

if (status=='open') {
    $('.hours').show();
    $('.closed').hide();
}else{
    $('.hours').hide();
    $('.closed').show();
} 
  
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="hours">We are OPEN</div>
<div class="closed">We are CLOSED</div> 
  
 

                
最满意答案
                
                    
                        
 您可以构造一个数组来表示一周中的每一天，并在其中包含打开/关闭小时。 
 
 另外，我会更改你的“status”变量来存储一个布尔值，而不是一个字符串，因为它有效地回答了“是或否”的问题 - 为什么不把它称为“打开”并使其值为真或假？ 挑剔，但这种方式更正确。 字符串比较很笨拙。 
 
var openHours = [
    {
        openHour: 8,
        openMinute: 0,
        closeHour: 13,
        closeMinute: 0
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: -1,
        openMinute: -1,
        closeHour: -1,
        closeMinute: -1,
    }
];

var d = new Date();
var dayOfWeek = d.getDay();
var hour = d.getHours();
var mins = d.getMinutes();
var open = true;
var todayHours = openHours[dayOfWeek];

if (hour >= todayHours.openHour && hour <= todayHours.closeHour) {
    if ((hour==todayHours.openHour && mins < todayHours.openMinute) || (hour==todayHours.closeHour && mins > todayHours.closeMinute)) {
        open = false;
    } else {
        open = true;
    }
} else {
    open = false;
}

if (open) {
    $('.hours').show();
    $('.closed').hide();
} else {
    $('.hours').hide();
    $('.closed').show();
}
 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="hours">We are OPEN</div>
<div class="closed">We are CLOSED</div>
You can construct an array to represent each day of the week, and include the open/close hour within it. 
 
Also, I'd change your "status" variable to store a boolean, not a string, since it's effectively answering a "yes or no" question - why not just call it "open" and make its values true or false? Nit-picky, but this way is more correct. String comparisons are clunky.
 
var openHours = [
    {
        openHour: 8,
        openMinute: 0,
        closeHour: 13,
        closeMinute: 0
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: 9,
        openMinute: 0,
        closeHour: 17,
        closeMinute: 30,
    },
    {
        openHour: -1,
        openMinute: -1,
        closeHour: -1,
        closeMinute: -1,
    }
];

var d = new Date();
var dayOfWeek = d.getDay();
var hour = d.getHours();
var mins = d.getMinutes();
var open = true;
var todayHours = openHours[dayOfWeek];

if (hour >= todayHours.openHour && hour <= todayHours.closeHour) {
    if ((hour==todayHours.openHour && mins < todayHours.openMinute) || (hour==todayHours.closeHour && mins > todayHours.closeMinute)) {
        open = false;
    } else {
        open = true;
    }
} else {
    open = false;
}

if (open) {
    $('.hours').show();
    $('.closed').hide();
} else {
    $('.hours').hide();
    $('.closed').show();
}
 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="hours">We are OPEN</div>
<div class="closed">We are CLOSED</div>