Convert DateTime to a specific format

Question

What is the best and fastest way to convert a DateTime to this format?

2015-03-26T18:02:58.145798Z

Currently I receive a date from a server and I'm able to parse it and convert the date in to DateTime and the ToString() output is something like this:

26/03/2015 18:02:58

For converting the date I'm using this line of code:

var parsedDate = DateTime.Parse("2015-03-26T18:02:58.145798Z", CultureInfo.InvariantCulture, DateTimeStyles.RoundtripKind);

What is the best way to convert parsedDate back to the original format?

EDIT: I want to convert the DateTime to this format 2015-03-26T18:02:58.145798Z as string

What is it that you want? Do you want to convert it back to String? — Praveen Paulose, Apr 23 '15 at 08:19
the ToString("O") almost do the job, but the problem is that is missing the Z at the end of the string, any idea? — frenk91, Apr 23 '15 at 08:29
I'm writing a universal app for Windows Phone 8.1 and Windows 8.1 under Windows Runtime — frenk91, Apr 23 '15 at 08:32
don't add the "final code" in your question. If you feel that the answers does not explain it enough, add your own answer with the final code. Separate the question from the answer. — default, Apr 23 '15 at 08:47
I rolled back your revision, since the code you added is already explained in the added answers. — default, Apr 23 '15 at 08:48

Dirk · Accepted Answer · 2015-04-23T08:47:04.623

7

If you have a DateTime object you can convert it to a string with that particular format by using O as format specifier:

parsedDate.ToString("O")

or

parsedDate.ToUniversalTime().ToString("O") // if parsedDate is not UTC

returns "2015-03-26T18:02:58.1457980Z".

If the DateTimeKind of your DateTime object is not Utc then you won't get the Z extension at the end of the string according to ISO8601. In the example you provided the Z is present because DateTime.Parse will recognize it and return a DateTime in Utc. Should the Z be missing in the original string you parse you can still assume it's UTC by using ToUniversalTime() on the date time object.

edited Apr 23 '15 at 08:47

answered Apr 23 '15 at 08:27

Dirk

10,668
2
35
49

the uotput using ToString("O") is this 2015-04-23T08:03:39.0000000 it miss the Z at the end – frenk91 Apr 23 '15 at 08:31
1

@Default @frenk91 Because you use `DateTime.Now` which has `DateTimeKind.Local`. The parsed `DateTime` in the example has `DateTimeKind.Utc` in which case there will be a `Z` at the end. – Dirk Apr 23 '15 at 08:31
@Dirk you need to correct parsedDate.UtcNow, it will not work. Add DateTime.UtcNow as an alternative and explain why the Z is only on the end for UTC – Jaycee Apr 23 '15 at 08:38
@Jaycee there is no `parsedDate.UtcNow`, If he parses the string the way he says he will get a UTC time. – Dirk Apr 23 '15 at 08:42
@Dirk what I am saying is you are using the original DateTime variable and calling UtcNow on it, it is a static property! – Jaycee Apr 23 '15 at 08:44
1

@Default fixed it now. I checked for existance of the method but did not notice that is was static. My bad! – pyrocumulus Apr 23 '15 at 08:48

Tamás Deme · Answer 2 · 2015-04-23T08:42:20.243

4

The answer is almost what @Dirk said:

parsedDate.ToString("O") is the line, but you have to convert the DateTime to UTC: that's how you get the "Z" at the end.

See https://msdn.microsoft.com/en-us/library/az4se3k1%28v=vs.110%29.aspx for more info.

Edit:

To convert a DateTime to UTC, use the ToUniversalTime() method.

edited Apr 23 '15 at 08:42

answered Apr 23 '15 at 08:33

Tamás Deme

2,194
2
13
31

score 0 · Answer 3 · answered May 18 '21 at 06:20

The FASTEST way, known to me, is this:

        ///<summary>Format the date time value as a parsable ISO format: "2008-01-11T16:07:12Z".</summary>
    public static string ISO( this DateTime dt ) {
        var ca = new char[] {
            (char) ( dt.Year / 1000 % 10 + '0' ),
            (char) ( dt.Year / 100 % 10 + '0' ),
            (char) ( dt.Year / 10 % 10 + '0' ),
            (char) ( dt.Year % 10 + '0' ),
            '-',
            (char) ( dt.Month / 10 + '0' ),
            (char) ( dt.Month % 10 + '0' ),
            '-',
            (char) ( dt.Day / 10 + '0' ),
            (char) ( dt.Day % 10 + '0' ),
            'T',
            (char) ( dt.Hour / 10 + '0' ),
            (char) ( dt.Hour % 10 + '0' ),
            ':',
            (char) ( dt.Minute / 10 + '0' ),
            (char) ( dt.Minute % 10 + '0' ),
            ':',
            (char) ( dt.Second / 10 + '0' ),
            (char) ( dt.Second % 10 + '0' ),
            'Z',
        };
        return new string( ca );
    }

Convert DateTime to a specific format

3 Answers3