如何确定在营业时间/天期间显示的图像？ [英] How to determine the image that is displayed during business hours/days?
问题描述
所以我已经问过这个问题，我收到以下作为答案，真正帮助！但是，假设该商家在上午9:00 开启，并在下午5:00 PST时间（加利福尼亚州）关闭，周六休息
请注意，下面的脚本触发了基于营业时间显示/显示/隐藏的图像。因此，在早上9:00太平洋标准时间时，图片显示为我们打开，在下午5:00 ，图片将显示为我们已关闭。感谢各位，我希望我已经输入足够的数据/信息，为您回答这个问题。
以下是小提琴的参考资料
$（window）.load（function（）{
//正在使用
//中央标准时间（-0500 UTC）
// UTC的开放时间为16，第二天的关闭时间为0
var d = new Date ），
open = new Date（），
closed = new Date（）;
//打开
的静态设置UTC日期open.setUTCHours（16） ;
open.setUTCMinutes（0）;
open.setUTCSeconds（0）;
open.setUTCMilliseconds（0）;
//静态设置关闭的UTC日期
closed.setUTCDate（d.getUTCDate（）+ 1）; // UTC时间旋转回0，添加一天
closed.setUTCHours（0）; // UTC小时为0
close.setUTCMinutes（0）;
closed.setUTCSeconds（0）;
closed.setUTCMilliseconds（0）;
//调试
console.log date：+ d）;
console.log（store open time in user's timezone：+ open）;
console.log（store close time in user's timezone：+ closed）;
console.log（d> open）; //用户的时间大于开放时间
console.log（d< closed）; //是用户的时间少于关闭时间
//（你不必回家...）
//测试商店开放？
if（d> open&& d< closed）{
setOpenStatus（true）;
} else {
setOpenStatus（false）;
}
function setOpenStatus（isOpen）{
$（'＃opend'）。toggle（isOpen）;
$（'＃closed'）。toggle（！isOpen）;
}
}）;
已编辑/更新的脚本
$（window）.load（function b $ b //将您的营业时间转换为UTC，例如，使用
//中央标准时间（-0500 UTC）
// UTC时间为16小时，营业时间为0第二天
var d = new Date（），
open = new Date（），
closed = new Date（）;
//静态设置UTC打开
的日期open.setUTCHours（16）;
open.setUTCMinutes（0）;
open.setUTCSeconds（0）;
open.setUTCMilliseconds（0）;
//静态设置关闭的UTC日期
closed.setUTCDate（d.getUTCDate（）+ 1）; // UTC时间旋转回0，添加一天
closed.setUTCHours 0）; // UTC小时为0
closed.setUTCMinutes（0）;
closed.setUTCSeconds（0）;
closed.setUTCMilliseconds（0）;
// Debugging
console.log（user's date：+ d）;
console.log（store open time in user's timezone：+ open）;
console.log（store close time in user's timezone：+ closed）;
console.log（d> open）; //用户的时间大于开放时间
console.log（d< closed）; //是用户的时间少于关闭时间
//（你不必回家...）
//测试商店开放？
if（d> open&& d< closed）{
setOpenStatus（true）;
}
if（d.getDay（）！== 0& d.getDay（）！== 6&&（d> open&& d< closed ））
else {
setOpenStatus（false）;
}
function setOpenStatus（isOpen）{
$（'＃opend'）。toggle（isOpen）;
$（'＃closed'）。toggle（！isOpen）;
}
}）;
将底部几行改为此，目前有点混乱。
if（d.getDay == 0&& d.getDay（）！== 6&&（d> = open& d< closed））{
setOpenStatus（true）;
} else {
setOpenStatus（false）;
}
所以你理解条件，它说：如果不是星期日（d.getDay（）！== 0）或星期六（d.getDay（）！== 6）当前时间是在开始时间（d> =打开）之后和在结束时间之前（d <闭合），然后设置打开状态，否则（else），设置关闭状态。
So I already asked this question before and I received the following as an answer which truly helps! However lets say the business opens at 9:00 AM and closes at 5:00 PM PST time (California) and they are closed on Saturday and Sunday.
How can I adjust that below?
Also keep in mind that the script below is triggering an image to show/display/hide based on the business hours. So at 9:00 AM PST time the image say's 'We're Open' and at 5:00 PM the image then goes to 'We're Closed'. Thanks guys I hope I have entered enough data/information for you to answer this question.
Here is a reference Fiddle.
$(window).load(function(){
// Translate your hours to UTC, example here is using
// Central Standard Time (-0500 UTC)
// Opening hour in UTC is 16, Closing hour is 0 the next day
var d = new Date(),
open = new Date(),
closed = new Date();
// Statically set UTC date for open
open.setUTCHours(16);
open.setUTCMinutes(0);
open.setUTCSeconds(0);
open.setUTCMilliseconds(0);
// Statically Set UTC date for closing
closed.setUTCDate(d.getUTCDate()+1); // UTC time rotates back to 0, add a day
closed.setUTCHours(0); // UTC hours is 0
closed.setUTCMinutes(0);
closed.setUTCSeconds(0);
closed.setUTCMilliseconds(0);
// Debugging
console.log("user's date:" + d);
console.log("store open time in user's timezone:" + open);
console.log("store close time in user's timezone:" + closed);
console.log(d > open); // user's time is greater than opening time
console.log(d < closed); // is user's time less than closing time
// (you don't have to go home...)
// Test for store open?
if (d > open && d < closed) {
setOpenStatus(true);
} else {
setOpenStatus(false);
}
function setOpenStatus(isOpen) {
$('#opend').toggle(isOpen);
$('#closed').toggle(!isOpen);
}
});
EDITED/UPDATED SCRIPT
$(window).load(function(){
// Translate your hours to UTC, example here is using
// Central Standard Time (-0500 UTC)
// Opening hour in UTC is 16, Closing hour is 0 the next day
var d = new Date(),
open = new Date(),
closed = new Date();
// Statically set UTC date for open
open.setUTCHours(16);
open.setUTCMinutes(0);
open.setUTCSeconds(0);
open.setUTCMilliseconds(0);
// Statically Set UTC date for closing
closed.setUTCDate(d.getUTCDate()+1); // UTC time rotates back to 0, add a day
closed.setUTCHours(0); // UTC hours is 0
closed.setUTCMinutes(0);
closed.setUTCSeconds(0);
closed.setUTCMilliseconds(0);
// Debugging
console.log("user's date:" + d);
console.log("store open time in user's timezone:" + open);
console.log("store close time in user's timezone:" + closed);
console.log(d > open); // user's time is greater than opening time
console.log(d < closed); // is user's time less than closing time
// (you don't have to go home...)
// Test for store open?
if (d > open && d < closed) {
setOpenStatus(true);
}
if (d.getDay() !== 0 && d.getDay() !== 6 && (d > open && d < closed))
else {
setOpenStatus(false);
}
function setOpenStatus(isOpen) {
$('#opend').toggle(isOpen);
$('#closed').toggle(!isOpen);
}
});
Change the bottom few lines to this, it is a little muddled at present.
if (d.getDay() !== 0 && d.getDay() !== 6 && (d >= open && d < closed)) {
setOpenStatus(true);
} else {
setOpenStatus(false);
}
So you understand the condition, it says: If it is not Sunday (d.getDay() !== 0) or Saturday (d.getDay() !== 6) and the current time is after or at the opening time (d >= open) and before the closing time (d < closed), then set the open status, otherwise (else), set closed status.
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