How to convert this date 2017-07-09T17:50:21.000-0500 | C#

Question

when i run the below code, string dt = "2017-07-09T17:50:21.000-0500"; DateTime date = Convert.ToDateTime(dt);

it gives me output as

7/10/2017 4:20:21 AM

where as i want my output to be

2017-07-09 17:50

update

the code @alexander-petrov gave worked string dt = "2017-07-09T17:50:21.000-0500"; string date = DateTimeOffset.Parse(dt).DateTime.ToString("yyyy-MM-dd HH:mm");

gives output

2017-07-09 17:50

but on inserting the same to database it is adding +5 hrs to the time and inserting as

2017-07-09 22:50

"It is showing wrong time" - what time do you _expect_ it to be? It's a [standard format](https://en.wikipedia.org/wiki/ISO_8601) - you need to show what you're doing to parse it, and why you think it's different to what it should be. — James Thorpe, Jun 09 '17 at 12:48
2017-07-09T17:50:21.000-0500 is in string? How are you converting it into DateTime? Can you share that code? — Yogesh Patel, Jun 09 '17 at 12:49
Probably a problem with timezones. Please post your conversion code. — Myrtle, Jun 09 '17 at 12:49

Jony Adamit · Answer 1 · 2017-06-09T13:08:15.693

1

This is a Round-Trip format of a DateTime specified with a DateTimeKind.Local kind.
You need to decide if your program needs to be aware of time zones or not.

You could try parsing it while supplying the System.Globalization.DateTimeStyles.RoundtripKind or System.Globalization.DateTimeStyles.AdjustToUniversal parameter to the Parse method.

edited Jun 09 '17 at 13:08

answered Jun 09 '17 at 12:59

Jony Adamit

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score 0 · Answer 2 · answered Jun 09 '17 at 12:56

0

I couldn't get your date to work, as I think there is a colon missing in the last part. Adding that colon back allows me to convert the XSD date time into a SQL DATETIME using this script:

DECLARE @stringDate VARCHAR(30);
SELECT @stringDate = '2017-07-09T17:50:21.000-05:00';
DECLARE @xmlDate XML;
SELECT @xmlDate = CAST('' AS XML);
SELECT @xmlDate.value('xs:dateTime(sql:variable("@stringDate"))', 'datetime');

Results:

2017-07-09 22:50:21.000

answered Jun 09 '17 at 12:56

Richard Hansell

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The OP doesn't mention SQL in his question or tags. – Jony Adamit Jun 09 '17 at 13:00
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_"converting this to c# date time object"_ via calling SQL which transforms the date into XML? Innovative. – stuartd Jun 09 '17 at 13:01
I could have sworn this was a SQL question, my bad ;D – Richard Hansell Jun 09 '17 at 13:05

score 0 · Answer 3 · answered Jun 09 '17 at 13:10

0

Try:

string date = "2017-07-09T17:50:21.000-0500";
DateTime d = DateTime.ParseExact(date, "yyyy'-'MM'-'dd'T'HH':'mm':'ss'.'fffzzzz", null);

answered Jun 09 '17 at 13:10

Pablo notPicasso

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Alexander Petrov · Accepted Answer · 2017-06-09T18:18:39.750

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If you want take offset into account then use DateTimeOffset type.

string dt = "2017-07-09T17:50:21.000-0500";

DateTimeOffset date = DateTimeOffset.Parse(dt);

// format on my machine
// 09.07.2017 17:50:21 - 05:00
Console.WriteLine(date);

// without offset
// 09.07.2017 17:50:21
Console.WriteLine(date.DateTime);

edited Jun 09 '17 at 18:18

answered Jun 09 '17 at 16:43

Alexander Petrov

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on converting the same to DateTime result is changing, i need to save // // 09.07.2017 17:50 in database – stacker Jun 09 '17 at 18:10
@stacker - Don't convert to `DateTime`. Always use `DateTimeOffset`. [Must read](https://stackoverflow.com/q/4331189/5045688). – Alexander Petrov Jun 09 '17 at 18:14
the code you gave worked `string dt = "2017-07-09T17:50:21.000-0500"; string date = DateTimeOffset.Parse(dt).DateTime.ToString("yyyy-MM-dd HH:mm");` gives output >> 2017-07-09 17:50 but on inserting the same to database it is adding 5hrs to the time and inserting as >> 2017-07-09 22:50 – stacker Jun 10 '17 at 03:20
@stacker - ask another question about inserting into database. – Alexander Petrov Jun 10 '17 at 10:14
ok..added [here](https://stackoverflow.com/questions/44486866/inserting-iso-date-in-sql-server) – stacker Jun 11 '17 at 17:49

How to convert this date 2017-07-09T17:50:21.000-0500 | C#

4 Answers4