javascript 如何在Javascript中计算自午夜以来的毫秒数
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How to calculate ms since midnight in Javascript
提问by Olga
What is the best way to calculate the time passed since (last) midnight in ms?
计算自（最后）午夜以来经过的时间（以毫秒为单位）的最佳方法是什么？
回答by RobG
A bunch of answers so here another:
一堆答案，所以这里是另一个：
var d = new Date(), e = new Date(d);
var msSinceMidnight = e - d.setHours(0,0,0,0);
As a function:
作为一个函数：
function getMsSinceMidnight(d) {
var e = new Date(d);
return d - e.setHours(0,0,0,0);
}
alert(getMsSinceMidnight(new Date()));
回答by Niet the Dark Absol
Create a new date using the current day/month/year, and get the difference.
使用当前日/月/年创建一个新日期，并获取差异。
var now = new Date(),
then = new Date(
now.getFullYear(),
now.getMonth(),
now.getDate(),
0,0,0),
diff = now.getTime() - then.getTime(); // difference in milliseconds
回答by Dan
Many answers except RobG's (recommended answer), Kolink'sand Lai'sare wrong here
除了RobG 的（推荐答案）、Kolink和Lai之外的许多答案在这里都是错误的
Let's look closer together
让我们一起仔细看看
First mistake
第一个错误
OptimusCrimeand Andrew D.answers:
As Malasugested, if the daylight saving correction was applied the nearest midnight, we get incorrect value. Let's debug:
正如Mala所说，如果在最近的午夜应用夏令时校正，我们会得到错误的值。让我们调试：
- Suppose it's last Sunday of March
- The time is fixed at 2 am.
- If we see 10 am on the clock, there's actually 11 hours passed from midnight
- But instead we count
10 * 60 * 60 * 1000ms - The trick is played when midnight happens in different DST state then current
- 假设它是三月的最后一个星期日
- 时间固定在凌晨2点。
- 如果我们看到时钟上是上午 10 点，那么实际上距离午夜已经过去了 11 个小时
- 但相反，我们计算
10 * 60 * 60 * 1000ms - 当午夜发生在不同的 DST 状态然后当前
Second mistake
第二个错误
kennebeck'sanswer:
肯尼贝克的回答：
As RobGwrote, the clock can tick if you get the system time twice. We can even appear in different dates sometimes. You can reproduce this in a loop:
正如RobG所写，如果您两次获得系统时间，时钟就会滴答作响。我们有时甚至会出现在不同的日期。您可以在循环中重现此内容：
for (var i = 0; true; i++) {
if ((new Date()).getTime() - (new Date()).getTime()) {
alert(i); // console.log(i); // for me it's about a 1000
break;
}
}
Thirdis my personal pitfall you could possibly experience
第三是我个人的陷阱，你可能会遇到
Consider the following code:
考虑以下代码：
var d = new Date(),
msSinceMidnight = d - d.setHours(0,0,0,0);
msSinceMidnightis always 0as the object is changed during computation before the substraction operation
msSinceMidnight总是0因为在减法运算之前的计算过程中对象发生了变化
At last, this code works:
最后，此代码有效：
var d = new Date(),
msSinceMidnight = d.getTime() - d.setHours(0,0,0,0);
回答by kennebec
Simpler to write, if you don't mind creating two dates.
如果您不介意创建两个日期，则编写起来更简单。
var msSinceMidnight= new Date()-new Date().setHours(0,0,0,0);
回答by Andrew D.
var d=new Date();
// offset from midnight in Greenwich timezone
var msFromMidnightInGMT=d%86400000;
// offset from midnight in locale timezone
var msFromMidnightLocale=(d.getTime()-d.getTimezoneOffset()*60000)%86400000;
回答by Lai Xin Chu
var today = new Date();
var d = new Date(today.getFullYear(), today.getMonth(), today.getDate(), 0, 0, 0, 0);
var difference = today.getTime() - d.getTime();
回答by OptimusCrime
Seconds since midnight would simply be to display the time, but instead of using hours:minutes:seconds, everything is converted into seconds.
从午夜开始的秒数只是显示时间，而不是使用小时：分钟：秒，所有内容都转换为秒。
I think this should do it:
我认为应该这样做：
var now = new Date();
var hours = now.getHours()*(60*60);
var minutes = now.getMinutes()*60;
var seconds = now.getSeconds();
var secSinceMidnight = hours+minutes+seconds;
