How to check if DST (Daylight Saving Time) is in effect, and if so, the offset?

Question

This is a bit of my JS code for which this is needed:

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);

I want to get the datetime in "ago", but if the DST is in use then the dates are off by 1 hour. I don't know how to check if the DST is in effect or not.

How can I know when the daylight saving starts and ends?

Answer 1

This code uses the fact that getTimezoneOffset returns a greater value during Standard Time versus Daylight Saving Time (DST). Thus it determines the expected output during Standard Time, and it compares whether the output of the given date the same (Standard) or less (DST).

Note that getTimezoneOffset returns positive numbers of minutes for zones west of UTC, which are usually stated as negative hours (since they're "behind" UTC). For example, Los Angeles is UTC–8h Standard, UTC-7h DST. getTimezoneOffset returns 480 (positive 480 minutes) in December (winter, Standard Time), rather than -480. It returns negative numbers for the Eastern Hemisphere (such -600 for Sydney in winter, despite this being "ahead" (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!");
}

Answer 2

This answer is quite similar to the accepted answer, but doesn't override the Date prototype, and only uses one function call to check if Daylight Savings Time is in effect, rather than two.

---

The idea is that, since no country observes DST that lasts for seven months^[1], in an area that observes DST the offset from UTC time in January will be different to the one in July.

While Daylight Savings Time moves clocks forwards, JavaScript always returns a greater value during Standard Time. Therefore, getting the minimum offset between January and July will get the timezone offset during DST.

We then check if the dates timezone is equal to that minimum value. If it is, then we are in DST; otherwise we are not.

The following function uses this algorithm. It takes a date object, d, and returns true if daylight savings time is in effect for that date, and false if it is not:

function isDST(d) {
    let jan = new Date(d.getFullYear(), 0, 1).getTimezoneOffset();
    let jul = new Date(d.getFullYear(), 6, 1).getTimezoneOffset();
    return Math.max(jan, jul) !== d.getTimezoneOffset();    
}

Answer 3

Create two dates: one in June, one in January. Compare their getTimezoneOffset() values.

• if January offset > June offset, client is in northern hemisphere
• if January offset < June offset, client is in southern hemisphere
• if no difference, client timezone does not observe DST

Now check getTimezoneOffset() of the current date.

• if equal to June, northern hemisphere, then current time zone is DST (+1 hour)
• if equal to January, southern hemisphere, then current time zone is DST (+1 hour)

Answer 4

I was faced with this same problem today but since our daylight saving starts and stops at differing times from the USA (at least from my understanding), I used a slightly different route..

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);

Then you simply compare the current timezone offset with DST and nonDST to see which one matches.

Answer 5

The getTimezoneOffset() method in JavaScript, in a browser, returns the number of minutes offset from the 00:00 time zone. For example, America/New_York time zone in Daylight Savings (DST) returns the number 300. 300 minutes is 5 hours difference from zero. 300 minutes divided by 60 minutes is 5 hours. Every time zone is compared to the zero time zone, +00:00 / Etc/GMT / Greenwich time.

MDN Web Docs

The next thing that you must know, is that the offset has the opposite sign of the actual time zone.

Information about time zones is maintained by the Internet Assigned Numbers Authority (iana)

iana time zones

A nicely formatted table of Time Zones is supplied by joda.org

joda-time Time Zones

+00:00 or Etc/GMT is Greenwich time

All time zones are offset from +00:00 / "Etc/GMT" / Greenwich time

Daylight Savings Time is always an earlier time than the "regular" time in the summer. You set your clocks back in the fall season. ("Fall Back" slogan to remember what to do)

So, America/New_York time in Daylight Savings (winter) is one hour before the regular time. So, for example, what was normally 5 p.m. in the afternoon in New York city in the summer, is now 4 p.m. America/New_York time in Daylight Savings. The name "America/New_York" time is a "Long Format" time zone name. The east coast of the U.S typically calls their time zone Eastern Standard Time (EST)

If you want to compare today's time zone offset to the time zone offset of some other date, you need to know that mathematical sign (+/- "Positive / Negative") of the time zone offset is the opposite of the time zone.

Look at the time zone table at joda.org and find the time zone for "America/New_York" It will have a negative sign in front of the Standard Offset.

The earth rotates counter-clockwise on its axis. A person watch the sunrise in Greenwich sees the sunrise 5 hours before someone in New York City will see the sunrise. And someone on the West Coast of the U.S. will see the sunrise after someone on the East Coast of the U.S. sees the sunrise.

There's a reason why you need to know all of this. So that you'll be able to logically determine whether some JavaScript code is getting the DST status correctly or not, without needing to test every time zone at different times of the year.

Imagine that it's November in New York City, and the clocks have been set back an hour. In the summer in New York City, the offset is 240 minutes or 4 hours.

You can test this by creating a date that is in July and then getting the offset.

var July_Date = new Date(2017, 6, 1);
var july_Timezone_OffSet = July_Date.getTimezoneOffset();

console.log('july_Timezone_OffSet: ' + july_Timezone_OffSet)

What will print to the browser's developer tools console log?

Answer is: 240

So, now you can create a date in January and see what your browser returns for a time zone offset for the winter season.

var Jan_Date = new Date(2017, 0, 1);//Month is zero indexed - Jan is zero
var jan_Timezone_OffSet = Jan_Date.getTimezoneOffset();

console.log('jan_Timezone_OffSet: ' + jan_Timezone_OffSet)

Answer is: 300

Obviously 300 is bigger than 240. So, what does this mean? Should you write code that tests for the winter offset being bigger than the summer offset? Or the summer offset less than the winter offset? If there is a difference between the summer and winter time zone offsets, then you can assume that DST is being used for this time zone. But that doesn't tell you if today is using DST for the browsers time zone. So, you'll need to get the time zone offset for today.

var today = new Date();
var todaysTimeZone = today.getTimezoneOffset();

console.log('todaysTimeZone : ' + todaysTimeZone)

Answer is: ? - Depends on the time of year

If today's time zone offset and the summer time zone offset is the same, AND the summer and winter time zone offsets are different, then by logical deduction, today must be NOT be in DST.

Can you omit comparing the summer and winter time zone offsets, (To know if DST is used for this time zone) and just compare today's time zone offset to the summer TZ offset, and always get the correct answer?

today's TZ Offset !== Summer TZ Offset

Well, is today in the winter or summer? If you knew that then you could apply the following logic:

if ( it_is_winter && ( todays_TZ_Offset !== summer_TZ_Offset) {
  var are_We_In_DST = true;
}

But the problem is, that you don't know if today's date is in winter or summer. Every time zone can have its own rules for when DST starts and stops. You'd need to keep track of every time zone's rules for every time zone in the world. So, if there is a better and easier way then you might as well do it the better and easier way.

What we are left with, is that you need to know if this time zone uses DST, and then compare today's time zone offset with the summer time zone offset. That will always give you a reliable answer.

The final logic is:

if ( DST_Is_Used_In_This_Time_Zone && ( todays_TZ_Offset !== summer_TZ_Offset) {
  var are_We_In_DST = true;
}

Function to determine if the time zone in the browser uses DST:

function is_DST_Used_In_This_TimeZone() {
  var Jan_Date, jan_Timezone_OffSet, July_Date, july_Timezone_OffSet 
      offsetsNotEqual, thisYear, today;

  today = new Date();//Create a date object that is now
  thisYear = today.getFullYear();//Get the year as a number

  Jan_Date = new Date(thisYear, 0, 1);//Month is zero indexed - Jan is zero
  jan_Timezone_OffSet = Jan_Date.getTimezoneOffset();

  console.log('jan_Timezone_OffSet: ' + jan_Timezone_OffSet)

  July_Date = new Date(thisYear, 6, 1);
  july_Timezone_OffSet = July_Date.getTimezoneOffset();

  console.log('july_Timezone_OffSet: ' + july_Timezone_OffSet)

  offsetsNotEqual = july_Timezone_OffSet !== jan_Timezone_OffSet;//True if not equal

  console.log('offsetsNotEqual: ' + offsetsNotEqual);

  return offsetsNotEqual;//If the offsets are not equal for summer and
       //winter then the only possible reason is that DST is used for
       //this time zone
}

Answer 6

Based on Matt Johanson's comment on the solution provided by Sheldon Griffin I created the following code:

    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(); 
    };

It tries to get the best of all worlds taking into account all the comments and previously suggested answers and specifically it:

1) Caches the result for per year stdTimezoneOffset so that you don't need to recalculate it when testing multiple dates in the same year.

2) It does not assume that DST (if it exists at all) is necessarily in July, and will work even if it will at some point and some place be any month. However Performance-wise it will work faster if indeed July (or near by months) are indeed DST.

3) Worse case it will compare the getTimezoneOffset of the first of each month. [and do that Once per tested year].

The assumption it does still makes is that the if there is DST period is larger then a single month.

If someone wants to remove that assumption he can change loop into something more like whats in the solutin provided by Aaron Cole - but I would still jump half a year ahead and break out of the loop when two different offsets are found]

Answer 7

Future-Proof Solution That Works In All Time Zones

1. Let x be the expected number of milliseconds into the year of interest without factoring in daylight savings.
2. Let y be the number of milliseconds since the Epoch from the start of the year of the date of interest.
3. Let z be the number of milliseconds since the Epoch of the full date and time of interest
4. Let t be the subtraction of both x and y from z: z - y - x. This yields the offset due to DST.
5. If t is zero, then DST is not in effect. If t is not zero, then DST is in effect.

"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");

I believe that the above code snippet is superior to all other answers posted here for many reasons.

• 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.

However, if you are not preparing for over 2 DST periods, then the below code can be used to determine whether DST is in effect as a boolean.

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

Answer 8

The moment.js library provides an .isDst() method on its time objects.

moment#isDST checks if the current moment is in daylight saving time.

moment([2011, 2, 12]).isDST(); // false, March 12 2011 is not DST
moment([2011, 2, 14]).isDST(); // true, March 14 2011 is DST

Answer 9

I've found that using the Moment.js library with some of the concepts described here (comparing Jan to June) works very well.

This simple function will return whether the timezone that the user is in observes Daylight Saving Time:

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

A simple way to check that this works (on Windows) is to change your timezone to a non DST zone, for example Arizona will return false, whereas EST or PST will return true.

Answer 10

Use Moment.js (https://momentjs.com/)

moment().isDST(); will give you if Day light savings is observed.

Also it has helper function to calculate relative time for you. You don't need to do manual calculations e.g moment("20200105", "YYYYMMDD").fromNow();

Answer 11

Your're close but a little off. You never need to calculate your own time as it is a result of your own clock. It can detect if you are using daylight saving time in your location but not for a remote location produced by the offset:

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.

Answer 12

Update: After trying to use these functions in a custom datetime picker, I noticed that switching from March to April toggled the timezone as expected, since my zone toggles DST in March. Unexpectedly, it was toggling to the next time zone over instead of switching between Standard and Daylight in the same time zone.

Turns out that's because my original functions were always creating new Date() for the current time or the arbitrary fixed time in the past. Comparing that to the the relative times from March and April meant it would logically detect the DST toggle as switching time zones instead.

The solution was to pass the relative times into the utility functions, so all my comparisons were for the relative time instead of now or the arbitrary fixed time. Lost some of the compactness, but now the logic works as needed.

Updates to workflow:

• 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));
}

---

Expanding on the compact and cryptic solution from @nkitku to turn it into a set of reusable functions.

Workflow:

• 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

Since this version is meant to be compact, you could save even more space by stripping out extraneous whitespace. Though that's really a job for a minifier.

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) => {
    let z = std ? ns.std() : new Date().getTimezoneOffset(),
      zm = z % 60;
    return 'UTC' + (z > 0 ? '-' : '+') + (z / 60) + (zm ? ':' + zm : '');
  }
};

console.log(ns.std(), ns.is_dst(), ns.utc(), ns.utc(1));

Answer 13

With https://date-fns.org/v2.22.1/docs/Time-Zones can be solved with one line

new Date().getUTCHours() + getTimezoneOffset('Europe/Amsterdam') / 1000 / 60 / 60;

Answer 14

Assuming the DST is changed on 27th March, the following can be used for detection of DST:

//utcT1: UTC time in [ms] before DST (DST has no impact on UTC!)
const utcT1 = (new Date(Date.UTC(2023, 2, 26))).getTime();
//locT1: local time in [ms] before setting DST
const locT1 = (new Date(2023, 2, 26)).getTime();
//tz1: time zone before DST
const tz1 = (locT1 - utcT1)/1000/60;
//expected output -60 [min] for e.g. Berlin

//utcT2: UTC time in [ms] after DST
const utcT2 = (new Date(Date.UTC(2023, 2, 27))).getTime();
//locT2: local time in [ms] after DST
const locT2 = (new Date(2023, 2, 27)).getTime();
//tz2: time zone after DST
const tz2 = (locT2 - utcT2)/1000/60;
//expected output -120 [min] for e.g. Berlin

//I.e. the DST for the local time zone is 60 [min]

//Similar result can be obtain by using the getTimezoneOffset
console.log((new Date(2023, 2, 26)).getTimezoneOffset());//-60
console.log((new Date(2023, 2, 27)).getTimezoneOffset());//-120

Answer 15

I recently needed to create a date string with UTC and DST, and based on Sheldon's answer I put this together:

Date.prototype.getTimezone = function(showDST) {
    var jan = new Date(this.getFullYear(), 0, 1);
    var jul = new Date(this.getFullYear(), 6, 1);

    var utcOffset = new Date().getTimezoneOffset() / 60 * -1;
    var dstOffset = (jan.getTimezoneOffset() - jul.getTimezoneOffset()) / 60;

    var utc = "UTC" + utcOffset.getSign() + (utcOffset * 100).preFixed(1000);
    var dst = "DST" + dstOffset.getSign() + (dstOffset * 100).preFixed(1000);

    if (showDST) {
        return utc + " (" + dst + ")";
    }

    return utc;
}
Number.prototype.preFixed = function (preCeiling) {
    var num = parseInt(this, 10);
    if (preCeiling && num < preCeiling) {
        num = Math.abs(num);
        var numLength   = num.toString().length;
        var preCeilingLength = preCeiling.toString().length;
        var preOffset   = preCeilingLength - numLength;
        for (var i = 0; i < preOffset; i++) {
            num = "0" + num;
        }
    }
    return num;
}
Number.prototype.getSign = function () {
    var num  = parseInt(this, 10);
    var sign = "+";
    if (num < 0) {
        sign = "-";
    }
    return sign;
}

document.body.innerHTML += new Date().getTimezone() + "<br>";
document.body.innerHTML += new Date().getTimezone(true);

<p>Output for Turkey (UTC+0200) and currently in DST: &nbsp; UTC+0300 (DST+0100)</p>
<hr>

Answer 16

Is there an issue using the Date.toString().indexOf('Daylight Time') > -1

"" + new Date()

Sat Jan 01 100050 00:00:00 GMT-0500 (Eastern Standard Time)

"" + new Date(...)

Sun May 01 100033 00:00:00 GMT-0400 (Eastern Daylight Time)

This seems compatible with all browsers.