日期方法
目标:一旦时间达到两天或更短,就会切换到小时而不是天。
问题:通过更改日期来检查代码时,它仍然只显示剩余的天数,而不显示小时。我是不是错过了什么?
var countDownDate = new Date("Nov 15, 2020 11:59:59").getTime();
var x = setInterval(function() {
var now = new Date().getTime();
var distance = countDownDate - now;
var days = Math.floor(distance / (1000 * 60 * 60 * 24));
var hours = Math.floor((distance % (1000 * 60 * 60 * 24)) / (1000 * 60 * 60));
if (distance < 2) {
clearInterval(x);
document.getElementById("aep-countdown-date").innerHTML = hours + " hours left";
}
else if (distance < 0) {
clearInterval(x);
document.getElementById("aep-countdown-date").innerHTML = "The time has ended.";
}
else {
clearInterval(x);
document.getElementById("aep-countdown-date").innerHTML = days + " days left";
}
});
getTime
返回:
自Unix Epoch以来的毫秒数*。
您的代码countDownDate - now
返回从现在到目标日期的毫秒数。
然而,您的比较distance < 2
将distance
(以毫秒为单位(视为以天为单位。
你可能想要
ms in seconds ------------------+
seconds in hour ---------+ |
hours in day ------+ | |
days -+ | | |
| | | |
if (distance < 2 * 24 * 3600 * 1000) {
/* ... */
}
在前两个响应的帮助下,我找到了答案。我添加了";2x";更改为hours变量,并在if语句中将该变量更改为days,并取消setInterval,因为它不需要(我只需要days和hours,而不是毫秒,并且它可以不断启动(:
var countDownDate = new Date("Nov 15, 2020 11:59:59").getTime();
$(function() {
var now = new Date().getTime();
var distance = countDownDate - now;
var days = Math.floor(distance / (1000 * 60 * 60 * 24));
var hours = Math.floor(
(2 * (distance % (1000 * 60 * 60 * 24))) / (1000 * 60 * 60)
);
document.getElementById("aep-countdown-date").innerHTML =
days + " days left.";
if (days < 0) {
document.getElementById("aep-countdown-date").innerHTML = "Time has ended.";
}
if (days < 2) {
document.getElementById("aep-countdown-date").innerHTML =
hours + " hours left.";
}
});