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My date objects in JavaScript are always represented by UTC +2 because of where I am located. Hence like this
Mon Sep 28 10:00:00 UTC+0200 2009
Problem is doing a JSON.stringify
converts the above date to
2009-09-28T08:00:00Z (notice 2 hours missing i.e. 8 instead of 10)
What I need is for the date and time to be honoured but it's not, hence it should be
2009-09-28T10:00:00Z (this is how it should be)
Basically I use this:
var jsonData = JSON.stringify(jsonObject);
I tried passing a replacer parameter (second parameter on stringify) but the problem is that the value has already been processed.
I also tried using toString()
and toUTCString()
on the date object, but these don't give me what I want either..
Can anyone help me?
My date objects in JavaScript are always represented by UTC +2 because of where I am located. Hence like this
Mon Sep 28 10:00:00 UTC+0200 2009
Problem is doing a JSON.stringify
converts the above date to
2009-09-28T08:00:00Z (notice 2 hours missing i.e. 8 instead of 10)
What I need is for the date and time to be honoured but it's not, hence it should be
2009-09-28T10:00:00Z (this is how it should be)
Basically I use this:
var jsonData = JSON.stringify(jsonObject);
I tried passing a replacer parameter (second parameter on stringify) but the problem is that the value has already been processed.
I also tried using toString()
and toUTCString()
on the date object, but these don't give me what I want either..
Can anyone help me?
Share Improve this question edited Dec 3, 2014 at 10:41 Rahil Wazir 10.1k11 gold badges44 silver badges65 bronze badges asked Sep 28, 2009 at 11:03 mark smithmark smith 20.9k47 gold badges136 silver badges190 bronze badges 2 |21 Answers
Reset to default 79Recently I have run into the same issue. And it was resolved using the following code:
x = new Date();
let hoursDiff = x.getHours() - x.getTimezoneOffset() / 60;
let minutesDiff = (x.getHours() - x.getTimezoneOffset()) % 60;
x.setHours(hoursDiff);
x.setMinutes(minutesDiff);
JSON uses the Date.prototype.toISOString
function which does not represent local time -- it represents time in unmodified UTC -- if you look at your date output you can see you're at UTC+2 hours, which is why the JSON string changes by two hours, but if this allows the same time to be represented correctly across multiple time zones.
date.toJSON() prints the UTC-Date into a String formatted (So adds the offset with it when converts it to JSON format).
date = new Date();
new Date(date.getTime() - (date.getTimezoneOffset() * 60000)).toJSON();
Just for the record, remember that the last "Z" in "2009-09-28T08:00:00Z" means that the time is indeed in UTC.
See http://en.wikipedia.org/wiki/ISO_8601 for details.
Out-of-the-box solution to force JSON.stringify
ignore timezones:
- Pure javascript (based on Anatoliy answer):
// Before: JSON.stringify apply timezone offset
const date = new Date();
let string = JSON.stringify(date);
console.log(string);
// After: JSON.stringify keeps date as-is!
Date.prototype.toJSON = function(){
const hoursDiff = this.getHours() - this.getTimezoneOffset() / 60;
this.setHours(hoursDiff);
return this.toISOString();
};
string = JSON.stringify(date);
console.log(string);
Using moment + moment-timezone libraries:
const date = new Date();
let string = JSON.stringify(date);
console.log(string);
Date.prototype.toJSON = function(){
return moment(this).format("YYYY-MM-DDTHH:mm:ss:ms");;
};
string = JSON.stringify(date);
console.log(string);
<html>
<header>
<script src="https://momentjs.com/downloads/moment.min.js"></script>
<script src="https://momentjs.com/downloads/moment-timezone-with-data-10-year-range.min.js"></script>
</header>
</html>
Here is another answer (and personally I think it's more appropriate)
var currentDate = new Date();
currentDate = JSON.stringify(currentDate);
// Now currentDate is in a different format... oh gosh what do we do...
currentDate = new Date(JSON.parse(currentDate));
// Now currentDate is back to its original form :)
you can use moment.js to format with local time:
Date.prototype.toISOString = function () {
return moment(this).format("YYYY-MM-DDTHH:mm:ss");
};
I'm a little late but you can always overwrite the toJson function in case of a Date using Prototype like so:
Date.prototype.toJSON = function(){
return Util.getDateTimeString(this);
};
In my case, Util.getDateTimeString(this) return a string like this: "2017-01-19T00:00:00Z"
I run into this a bit working with legacy stuff where they only work on east coast US and don't store dates in UTC, it's all EST. I have to filter on the dates based on user input in the browser so must pass the date in local time in JSON format.
Just to elaborate on this solution already posted - this is what I use:
// Could be picked by user in date picker - local JS date
date = new Date();
// Create new Date from milliseconds of user input date (date.getTime() returns milliseconds)
// Subtract milliseconds that will be offset by toJSON before calling it
new Date(date.getTime() - (date.getTimezoneOffset() * 60000)).toJSON();
So my understanding is this will go ahead and subtract time (in milliseconds (hence 60000) from the starting date based on the timezone offset (returns minutes) - in anticipation for the addition of time toJSON() is going to add.
JavaScript normally convert local timezone to UTC .
date = new Date();
date.setMinutes(date.getMinutes()-date.getTimezoneOffset())
JSON.stringify(date)
Usually you want dates to be presented to each user in his own local time-
that is why we use GMT (UTC).
Use Date.parse(jsondatestring) to get the local time string,
unless you want your local time shown to each visitor.
In that case, use Anatoly's method.
Got around this issue by using the moment.js
library (the non-timezone version).
var newMinDate = moment(datePicker.selectedDates[0]);
var newMaxDate = moment(datePicker.selectedDates[1]);
// Define the data to ask the server for
var dataToGet = {"ArduinoDeviceIdentifier":"Temperatures",
"StartDate":newMinDate.format('YYYY-MM-DD HH:mm'),
"EndDate":newMaxDate.format('YYYY-MM-DD HH:mm')
};
alert(JSON.stringify(dataToGet));
I was using the flatpickr.min.js
library. The time of the resulting JSON object created matches the local time provided but the date picker.
Here is something really neat and simple (atleast I believe so :)) and requires no manipulation of date to be cloned or overloading any of browser's native functions like toJSON (reference: How to JSON stringify a javascript Date and preserve timezone, courtsy Shawson)
Pass a replacer function to JSON.stringify that stringifies stuff to your heart's content!!! This way you don't have to do hour and minute diffs or any other manipulations.
I have put in console.logs to see intermediate results so it is clear what is going on and how recursion is working. That reveals something worthy of notice: value param to replacer is already converted to ISO date format :). Use this[key] to work with original data.
var replacer = function(key, value)
{
var returnVal = value;
if(this[key] instanceof Date)
{
console.log("replacer called with key - ", key, " value - ", value, this[key]);
returnVal = this[key].toString();
/* Above line does not strictly speaking clone the date as in the cloned object
* it is a string in same format as the original but not a Date object. I tried
* multiple things but was unable to cause a Date object being created in the
* clone.
* Please Heeeeelp someone here!
returnVal = new Date(JSON.parse(JSON.stringify(this[key]))); //OR
returnVal = new Date(this[key]); //OR
returnVal = this[key]; //careful, returning original obj so may have potential side effect
*/
}
console.log("returning value: ", returnVal);
/* if undefined is returned, the key is not at all added to the new object(i.e. clone),
* so return null. null !== undefined but both are falsy and can be used as such*/
return this[key] === undefined ? null : returnVal;
};
ab = {prop1: "p1", prop2: [1, "str2", {p1: "p1inner", p2: undefined, p3: null, p4date: new Date()}]};
var abstr = JSON.stringify(ab, replacer);
var abcloned = JSON.parse(abstr);
console.log("ab is: ", ab);
console.log("abcloned is: ", abcloned);
/* abcloned is:
* {
"prop1": "p1",
"prop2": [
1,
"str2",
{
"p1": "p1inner",
"p2": null,
"p3": null,
"p4date": "Tue Jun 11 2019 18:47:50 GMT+0530 (India Standard Time)"
}
]
}
Note p4date is string not Date object but format and timezone are completely preserved.
*/
I ran into the same problem. The way I resolvet it was:
var currentTime = new Date();
Console.log(currentTime); //Return: Wed Sep 15 13:52:09 GMT-05:00 2021
Console.log(JSON.stringify(currentTime)); //Return: "2021-09-15T18:52:09.891Z"
var currentTimeFixed = new Date(currentTime.setHours(currentTime.getHours() - (currentTime.getUTCHours() - currentTime.getHours())));
Console.log(JSON.stringify(currentTimeFixed)); //Return: "2021-09-15T13:52:09.891Z"
I wrote the following code blog where it makes service calls.. it will try to serializable the json in every post submission, it will format to local date it again.
protected async post(endPoint: string, data, panelName?: string, hasSuccessMessage: boolean = false): Promise<Observable<any>> {
const options = this.InitHeader(true);
const url: string = this._baseUrl + endPoint;
Date.prototype.toJSON = function () {
return moment(this).format("YYYY-MM-DDThh:mm:00.000Z");;
};
return await this._http.post(url, data, options).pipe(map(response => {
return this.Map<any>(response, null);
}));
}
All boils down to if your server backend is timezone-agnostic or not. If it is not, then you need to assume that timezone of server is the same as client, or transfer information about client's timezone and include that also into calculations.
a PostgreSQL backend based example:
select '2009-09-28T08:00:00Z'::timestamp -> '2009-09-28 08:00:00' (wrong for 10am)
select '2009-09-28T08:00:00Z'::timestamptz -> '2009-09-28 10:00:00+02'
select '2009-09-28T08:00:00Z'::timestamptz::timestamp -> '2009-09-28 10:00:00'
The last one is probably what you want to use in database, if you are not willing properly implement timezone logic.
Instead of toJSON
, you can use format function which always gives the correct date and time + GMT
This is the most robust display option. It takes a string of tokens and replaces them with their corresponding values.
I tried this in angular 8 :
create Model :
export class Model { YourDate: string | Date; }
in your component
model : Model; model.YourDate = new Date();
send Date to your API for saving
When loading your data from API you will make this :
model.YourDate = new Date(model.YourDate+"Z");
you will get your date correctly with your time zone.
In this case I think you need transform the date to UNIX timestamp
timestamp = testDate.getTime();
strJson = JSON.stringify(timestamp);
After that you can re use it to create a date object and format it. Example with javascript and toLocaleDateString
( https://developer.mozilla.org/fr/docs/Web/JavaScript/Reference/Objets_globaux/Date/toLocaleDateString )
newDateObject = new Date(JSON.parse(strJson));
newDateObject = newDateObject.toLocalDateStrin([
"fr-FR",
]);
If you use stringify to use AJAX, now it's not useful. You just need to send timestamp and get it in your script:
$newDateObject = new \DateTime();
$newDateObject->setTimestamp(round($timestamp/1000));
Be aware that getTime()
will return a time in milliseconds and the PHP function setTimestamp
take time in seconds. It's why you need to divide by 1000 and round
.
In Angular place the following in index.js
script section:
setTimeout(function (){
Date.prototype.toJSON = function(){
return new Date(this).toLocaleDateString("en-US") + " "+new Date(this).toLocaleTimeString();
}},1000);
let startDate = new Date('2023-01-01');
let date= new Date(Date.UTC(startDate.getFullYear(), startDate.getMonth(), startDate.getDate(), startDate.getHours(), startDate.getMinutes()))
let jsonString = JSON.stringify(date);
console.log(jsonString);
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2009-09-28T10:00:00Z
does not represent the same moment in time asMon Sep 28 10:00:00 UTC+0200 2009
. TheZ
in an ISO 8601 date means UTC, and 10 o'clock in UTC is a different moment in time to 10 o'clock in +0200. It would be one thing to want the date to be serialized with the right time zone, but you're asking us to help you serialise it to a representation that is unequivocally, objectively wrong. – Mark Amery Commented Feb 22, 2015 at 20:45