这是我的一点JS代码,这是需要的:

var secDiff = Math.abs(Math.round((utc_date-this.premiere_date)/1000));
this.years = this.calculateUnit(secDiff,(86400*365));
this.days = this.calculateUnit(secDiff-(this.years*(86400*365)),86400);
this.hours = this.calculateUnit((secDiff-(this.years*(86400*365))-(this.days*86400)),3600);
this.minutes = this.calculateUnit((secDiff-(this.years*(86400*365))-(this.days*86400)-(this.hours*3600)),60);
this.seconds = this.calculateUnit((secDiff-(this.years*(86400*365))-(this.days*86400)-(this.hours*3600)-(this.minutes*60)),1);

我想在“前”得到日期时间,但如果DST正在使用,那么日期是1小时。我不知道如何检查夏令时是否有效。

我怎样才能知道夏令时何时开始和结束?


当前回答

我发现使用Moment.js库和这里描述的一些概念(比较Jan和June)效果非常好。

这个简单的函数将返回用户所在的时区是否遵守日光节约时间:

function HasDST() {
    return moment([2017, 1, 1]).isDST() != moment([2017, 6, 1]).isDST();
}

检查它是否有效的一个简单方法(在Windows上)是将您的时区更改为非夏令时区域,例如亚利桑那州将返回false,而EST或PST将返回true。

其他回答

这段代码使用了这样一个事实,即getTimezoneOffset在标准时间与日光节约时间(DST)期间返回更大的值。因此,它确定标准时间内的预期输出,并比较给定日期的输出是否相同(标准)或更少(DST)。

注意,getTimezoneOffset对于UTC以西的区域返回正数分钟,通常表示为负小时(因为它们“落后”UTC)。例如,洛杉矶是UTC-8h标准,UTC-7h夏令时。getTimezoneOffset在12月(冬季,标准时间)返回480(正480分钟),而不是-480。它返回东半球的负数(例如冬季悉尼的-600,尽管这是“超前”(UTC+10h)。

Date.prototype.stdTimezoneOffset = function () {
    var jan = new Date(this.getFullYear(), 0, 1);
    var jul = new Date(this.getFullYear(), 6, 1);
    return Math.max(jan.getTimezoneOffset(), jul.getTimezoneOffset());
}

Date.prototype.isDstObserved = function () {
    return this.getTimezoneOffset() < this.stdTimezoneOffset();
}

var today = new Date();
if (today.isDstObserved()) { 
    alert ("Daylight saving time!");
}

基于Matt Johanson对Sheldon Griffin提供的解决方案的评论,我创建了以下代码:

    Date.prototype.stdTimezoneOffset = function() {
        var fy=this.getFullYear();
        if (!Date.prototype.stdTimezoneOffset.cache.hasOwnProperty(fy)) {

            var maxOffset = new Date(fy, 0, 1).getTimezoneOffset();
            var monthsTestOrder=[6,7,5,8,4,9,3,10,2,11,1];

            for(var mi=0;mi<12;mi++) {
                var offset=new Date(fy, monthsTestOrder[mi], 1).getTimezoneOffset();
                if (offset!=maxOffset) { 
                    maxOffset=Math.max(maxOffset,offset);
                    break;
                }
            }
            Date.prototype.stdTimezoneOffset.cache[fy]=maxOffset;
        }
        return Date.prototype.stdTimezoneOffset.cache[fy];
    };

    Date.prototype.stdTimezoneOffset.cache={};

    Date.prototype.isDST = function() {
        return this.getTimezoneOffset() < this.stdTimezoneOffset(); 
    };

考虑到所有的评论和之前建议的答案,它试图得到所有世界的最好结果,特别是:

1)缓存每年stdTimezoneOffset的结果,这样当您在同一年测试多个日期时就不需要重新计算它。

2)它没有假设夏令时(如果它存在的话)一定是在7月, 即使在任何一个月的某个时间某个地点,它也会起作用。 但是性能方面,如果七月(或临近月份)确实是DST,它会工作得更快。

3)在更糟糕的情况下,它将比较每个月的第一个getTimezoneOffset。[并且每年测试一次]。

它所做的假设仍然是,如果有夏令时周期比一个月大。

如果有人想要消除这种假设,他可以把循环变成更像Aaron Cole提供的溶液中的东西-但我仍然会提前半年,当发现两个不同的偏移时跳出循环]

你很接近了,但是有点差。你永远不需要计算你自己的时间,因为它是你自己的时钟的结果。它可以检测您是否在您的位置使用日光节约时间,但不能检测由偏移量产生的远程位置:

newDateWithOffset = new Date(utc + (3600000*(offset)));

This will still be wrong and off an hour if they are in DST. You need for a remote time account if they are currently inside their DST or not and adjust accordingly. try calculating this and change your clock to - lets say 2/1/2015 and reset the clock back an hour as if outside DST. Then calculate for an offset for a place that should still be 2 hours behind. It will show an hour ahead of the two hour window. You would still need to account for the hour and adjust. I did it for NY and Denver and always go the incorrect (hour ahead) in Denver.

面向未来的解决方案,适用于所有时区

设x为在不考虑夏时制的情况下进入利息年的预期毫秒数。 设y为从感兴趣日期的年份开始到Epoch的毫秒数。 设z为自感兴趣的完整日期和时间的Epoch以来的毫秒数 设t是z减去x和y: z - y - x。这就得到了由于夏令时而产生的偏移量。 如果t为零,则DST不生效。如果t不为零,则DST生效。

"use strict"; function dstOffsetAtDate(dateInput) { var fullYear = dateInput.getFullYear()|0; // "Leap Years are any year that can be exactly divided by 4 (2012, 2016, etc) // except if it can be exactly divided by 100, then it isn't (2100,2200,etc) // except if it can be exactly divided by 400, then it is (2000, 2400)" // (https://www.mathsisfun.com/leap-years.html). var isLeapYear = ((fullYear & 3) | (fullYear/100 & 3)) === 0 ? 1 : 0; // (fullYear & 3) = (fullYear % 4), but faster //Alternative:var isLeapYear=(new Date(currentYear,1,29,12)).getDate()===29?1:0 var fullMonth = dateInput.getMonth()|0; return ( // 1. We know what the time since the Epoch really is (+dateInput) // same as the dateInput.getTime() method // 2. We know what the time since the Epoch at the start of the year is - (+new Date(fullYear, 0)) // day defaults to 1 if not explicitly zeroed // 3. Now, subtract what we would expect the time to be if daylight savings // did not exist. This yields the time-offset due to daylight savings. - (( (( // Calculate the day of the year in the Gregorian calendar // The code below works based upon the facts of signed right shifts // • (x) >> n: shifts n and fills in the n highest bits with 0s // • (-x) >> n: shifts n and fills in the n highest bits with 1s // (This assumes that x is a positive integer) -1 + // first day in the year is day 1 (31 & ((-fullMonth) >> 4)) + // January // (-11)>>4 = -1 ((28 + isLeapYear) & ((1-fullMonth) >> 4)) + // February (31 & ((2-fullMonth) >> 4)) + // March (30 & ((3-fullMonth) >> 4)) + // April (31 & ((4-fullMonth) >> 4)) + // May (30 & ((5-fullMonth) >> 4)) + // June (31 & ((6-fullMonth) >> 4)) + // July (31 & ((7-fullMonth) >> 4)) + // August (30 & ((8-fullMonth) >> 4)) + // September (31 & ((9-fullMonth) >> 4)) + // October (30 & ((10-fullMonth) >> 4)) + // November // There are no months past December: the year rolls into the next. // Thus, fullMonth is 0-based, so it will never be 12 in Javascript (dateInput.getDate()|0) // get day of the month )&0xffff) * 24 * 60 // 24 hours in a day, 60 minutes in an hour + (dateInput.getHours()&0xff) * 60 // 60 minutes in an hour + (dateInput.getMinutes()&0xff) )|0) * 60 * 1000 // 60 seconds in a minute * 1000 milliseconds in a second - (dateInput.getSeconds()&0xff) * 1000 // 1000 milliseconds in a second - dateInput.getMilliseconds() ); } // Demonstration: var date = new Date(2100, 0, 1) for (var i=0; i<12; i=i+1|0, date.setMonth(date.getMonth()+1|0)) console.log(date.getMonth()+":\t"+dstOffsetAtDate(date)/60/60/1000+"h\t"+date); date = new Date(1900, 0, 1); for (var i=0; i<12; i=i+1|0, date.setMonth(date.getMonth()+1|0)) console.log(date.getMonth()+":\t"+dstOffsetAtDate(date)/60/60/1000+"h\t"+date); // Performance Benchmark: console.time("Speed of processing 16384 dates"); for (var i=0,month=date.getMonth()|0; i<16384; i=i+1|0) date.setMonth(month=month+1+(dstOffsetAtDate(date)|0)|0); console.timeEnd("Speed of processing 16384 dates");

我相信上面的代码片段优于这里发布的所有其他答案,原因有很多。

This answer works in all time zones, even Antarctica/Casey. Daylight savings is very much subject to change. It might be that 20 years from now, some country might have 3 DST periods instead of the normal 2. This code handles that case by returning the DST offset in milliseconds, not just whether DST is in effect or not in effect. The size of the months of the year and the way that Leap Years work fits perfectly into keeping our time on track with the sun. Heck, it works so perfectly that all we ever do is just adjust mere seconds here and there. Our current system of leap years has been in effect since February 24th, 1582, and will likely stay in effect for the foreseeable future. This code works in timezones that do not use DST. This code works in historic times before when DST was implemented (such as the 1900s). This code is maximally integer-optimized and should give you no problem if called in a tight loop. After running the code snippet above, scroll down to the bottom of the output to see the performance benchmark. My computer is able to process 16384 dates in 29ms on FireFox.

但是,如果您没有为超过2个DST周期做准备,那么可以使用下面的代码来确定DST是否作为布尔值有效。

function isDaylightSavingsInEffect(dateInput) {
    // To satisfy the original question
    return dstOffsetAtDate(dateInput) !== 0;
}

使用Moment.js (https://momentjs.com/)

.isDST时刻()();如果日光节约被观察,将会给你。

还有辅助功能,为您计算相对时间。你不需要手动计算 例如moment("20200105", "YYYYMMDD").fromNow();