Question

所以我之前已经问过这个问题，并且我收到了以下答案，这真的有帮助！但是，假设该商家在上午9:00 开始营业，并在下午5:00 PST时间（加利福尼亚州）关闭，他们将在周六关闭和星期天。

我如何调整以下内容？

另请注意，以下脚本会根据营业时间触发图像显示/显示/隐藏。因此，在太平洋标准时间上午9点时间，图像显示“我们打开”，然后在 5:00 时，图像会显示“我们已关闭”。谢谢你们，我希望我已经输入了足够的数据/信息来回答这个问题。

以下是参考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);
  }
});

已编辑/更新的脚本

$(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);
  }
});

Answer 1

将底部的几行改为此，目前有点混乱。

if (d.getDay() !== 0 && d.getDay() !== 6 && (d >= open && d < closed)) {
    setOpenStatus(true);
} else {
    setOpenStatus(false);
}

所以你了解情况，它说：如果它不是星期日(d.getDay() !== 0)或星期六(d.getDay() !== 6)，当前时间是在开幕时间(d >= open)之后和关闭时间之前(d < closed)，然后设置打开状态，否则设置为(else)，设置已关闭状态。

如何确定营业时间/天内显示的图像？

1 个答案: