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

更新: 在尝试在自定义日期时间选择器中使用这些函数后,我注意到从3月切换到4月会像预期的那样切换时区,因为我的区域在3月切换夏令时。出乎意料的是,它正在切换到下一个时区,而不是在同一时区的标准时区和夏令时时区之间切换。

原来,这是因为我原来的函数总是为当前时间或过去的任意固定时间创建新的Date()。将其与3月和4月的相对时间进行比较,意味着它将在逻辑上检测到夏令时切换为切换时区。

解决办法是将相对时间传递到效用函数中,所以我所有的比较都是相对时间,而不是现在或任意固定的时间。失去了一些紧凑性,但现在逻辑可以根据需要工作。

更新工作流程:

t parameter defaults to new Date() For fixed time, pass in an existing Date For current time, pass in null or nothing std() updated to use t.setMonth(v); to change the month for fixed times .getTimezoneOffset() cannot chain to .setMonth(), so we need to swap from one-line notation to use closures ({}), terminators (;), and return console.log() example loops through each month (0 to 11) The fixed date object needs to be cloned using the same timestamp (let ts = +t;) The + before the Date type casts it to a number with the Unix timestamp Date() also accepts Unix timestamps to create fixed times If we don't clone it, each call would pass around the same Date object with the months set to 6, which defeats the purpose Ok, we're not actually cloning, just creating a new object using the same settings; same difference ;)

let ns = { std: (t = new Date()) => Math.max(...[0, 6].map(v => { t.setMonth(v); return t.getTimezoneOffset(); })), is_dst: (t = new Date()) => t.getTimezoneOffset() < ns.std(t), utc: (t, std = 0) => { t = t || new Date(); let z = std ? ns.std(t) : t.getTimezoneOffset(), zm = z % 60; return 'UTC' + (z > 0 ? '-' : '+') + (z / 60) + (zm ? ':' + zm : ''); } }; //current time only console.log(ns.std(), ns.is_dst(), ns.utc(), ns.utc(null, 1)); //iterate each month let t = new Date(2021,0,1); for (let i = 0; i < 12; i++) { t.setMonth(i); let ts = +t; console.log(t.toDateString().split(" ")[1], ns.std(new Date(ts)), ns.is_dst(new Date(ts)), ns.utc(new Date(ts)), ns.utc(new Date(ts), 1)); }


扩展来自@nkitku的紧凑而神秘的解决方案,将其转换为一组可重用的函数。

工作流程:

All functions are scoped in a namespace ns so they don't conflict with other functions in the code that may have the same name Namespacing also allows for compact function notation; std: ()=>Math.max(), is equivalent to function std(){ return Math.max(); } std() returns the timezone offset in Standard Time [0, 6] sets up a comparison of a month without DST and a month with DST 0 for January, since Date.setMonth() is zero-indexed 6 for July Apparently, Standard Time is not in January for everyone, so we have to check both January and July ...[] converts the Array of months to a Set so we can apply the map() function Raw arrays cannot run map() map() runs a set of variables on the same function and returns an array of results Create a new Date object with year, month, day The year (95 in the example) is arbitrary since the year isn't important for this calculation The month plugs in our values [0, 6] as a variable v The day (1 in the example) is also arbitrary Logically we could have created a new Date(), then .setMonth(v), but using the arbitrary numbers is more compact and faster Now that we have the dates, getTimezoneOffset() returns the offsets for each month and pushes them to the results array Math.max() finds the largest value from the results, which will be the Standard Time offset is_dst() checks if it is currently Daylight Savings Time new Date().getTimezoneOffset() gets the current offset, with or without DST ns.std() gets the offset in Standard Time If the current offset is lower, then it's DST utc() returns a string in UTC notation The std parameter defaults to off z = std ? ns.std() : new Date().getTimezoneOffset() sets the time to DST or standard based on the flag zm = z % 60 captures minutes since some zones use 30 minutes for example (z > 0 ? '-' : '+') assigns the correct sign per UTC notation; positive offset values are shown as negative offsets in the notation (z / 60) captures the hours in single-digit format per the notation, so no need to .toString().padStart(2,'0)` for double-digit format (zm ? ':' + zm : '') appends minutes if they exist for the timezone

由于这个版本是紧凑的,您可以通过去掉多余的空白来节省更多的空间。不过这真的是一个迷你机的工作。

std:()=>Math.max(...[0,6].map(v=>new Date(95,v,1).getTimezoneOffset())),

Const ns = { std: () => Math.max(…(0, 6)。map(v => new Date(95, v, 1).getTimezoneOffset())), is_dst: () => new Date().getTimezoneOffset() < ns.std(), Utc: (std = 0) => { 让z = STD ?ns.std(): new Date().getTimezoneOffset(), Zm = z % 60; 返回'UTC' + (z > 0 ?'-': '+') + (z / 60) + (zm ?':' + zm: "); } }; ns.is_dst console.log (ns.std () (), ns.utc (), ns.utc (1));

其他回答

我今天也遇到了同样的问题,但由于我们的夏令时开始和结束的时间与美国不同(至少从我的理解来看),我使用了稍微不同的路线。

var arr = [];
for (var i = 0; i < 365; i++) {
 var d = new Date();
 d.setDate(i);
 newoffset = d.getTimezoneOffset();
 arr.push(newoffset);
}
DST = Math.min.apply(null, arr);
nonDST = Math.max.apply(null, arr);

然后,您只需将当前时区偏移量与DST和非DST进行比较,以查看哪一个匹配。

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.

创建两个日期:一个在六月,一个在一月。比较它们的getTimezoneOffset()值。

如果一月抵消>六月抵消,客户端在北半球 如果1月偏移量< 6月偏移量,则客户在南半球 如果没有差异,则客户端时区不遵守夏令时

现在检查当前日期的getTimezoneOffset()。

如果等于北半球的6月,则当前时区为夏时制(+1小时) 如果等于南半球的1月,则当前时区为夏时制(+1小时)

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。

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;
}