How do I convert a date/time string to a DateTime object in Dart?

Question

Say I have a string

"1974-03-20 00:00:00.000"

It is created using DateTime.now(), how do I convert the string back to a DateTime object?

Answer 1

DateTime has a parse method

var parsedDate = DateTime.parse('1974-03-20 00:00:00.000');

https://api.dartlang.org/stable/dart-core/DateTime/parse.html

Answer 2

There seem to be a lot of questions about parsing timestamp strings into DateTime. I will try to give a more general answer so that future questions can be directed here.

• Your timestamp is in an ISO format. Examples: 1999-04-23, 1999-04-23 13:45:56Z, 19990423T134556.789. In this case, you can use DateTime.parse or DateTime.tryParse. (See the DateTime.parse documentation for the precise set of allowed inputs.)

• Your timestamp is in a standard HTTP format. Examples: Fri, 23 Apr 1999 13:45:56 GMT, Friday, 23-Apr-99 13:45:56 GMT, Fri Apr 23 13:45:56 1999. In this case, you can use dart:io's HttpDate.parse function.

• Your timestamp is in some local format. Examples: 23/4/1999, 4/23/99, April 23, 1999. You can use package:intl's DateFormat class and provide a pattern specifying how to parse the string:

  import 'package:intl/intl.dart';
  
  ...
  
  var dmyString = '23/4/1999';
  var dateTime1 = DateFormat('d/M/y').parse(dmyString);
  
  var mdyString = '04/23/99'; 
  var dateTime2 = DateFormat('MM/dd/yy').parse(mdyString);
  
  var mdyFullString = 'April 23, 1999';
  var dateTime3 = DateFormat('MMMM d, y', 'en_US').parse(mdyFullString));
  

  See the DateFormat documentation for more information about the pattern syntax.

  DateFormat limitations:

  • DateFormat cannot parse dates that lack explicit field separators. For such cases, you can resort to using regular expressions (see below).
  • Prior to version 0.17.0 of package:intl, yy did not follow the -80/+20 rule that the documentation describes for inferring the century, so if you use a 2-digit year, you might need to adjust the century afterward.
  • As of writing, DateFormat does not support time zones. If you need to deal with time zones, you will need to handle them separately.

• Last resort: If your timestamps are in a fixed, known, numeric format, you always can use regular expressions to parse them manually:

  var dmyString = '23/4/1999';
  
  var re = RegExp(
    r'^'
    r'(?<day>[0-9]{1,2})'
    r'/'
    r'(?<month>[0-9]{1,2})'
    r'/'
    r'(?<year>[0-9]{4,})'
    r'$',
  );
  
  var match = re.firstMatch(dmyString);
  if (match == null) {
    throw FormatException('Unrecognized date format');
  }
  
  var dateTime4 = DateTime(
    int.parse(match.namedGroup('year')!),
    int.parse(match.namedGroup('month')!),
    int.parse(match.namedGroup('day')!),
  );
  

  See https://stackoverflow.com/a/63402975/ for another example.

  (I mention using regular expressions for completeness. There are many more points for failure with this approach, so I do not recommend it unless there's no other choice. DateFormat usually should be sufficient.)

Answer 3

import 'package:intl/intl.dart';

DateTime brazilianDate = new DateFormat("dd/MM/yyyy").parse("11/11/2011");

Answer 4

you can just use : DateTime.parse("your date string");

for any extra formating, you can use "Intl" package.

Answer 5

a string can be parsed to DateTime object using Dart default function DateTime.parse("string");

final parsedDate = DateTime.parse("1974-03-20 00:00:00.000"); 

Example on Dart Pad

Answer 6

void main() {
  
  var dateValid = "30/08/2020";
  
  print(convertDateTimePtBR(dateValid));
  
}

DateTime convertDateTimePtBR(String validade)
{
   DateTime parsedDate = DateTime.parse('0001-11-30 00:00:00.000');
  
  List<String> validadeSplit = validade.split('/');
  
  if(validadeSplit.length > 1)
  {
      String day = validadeSplit[0].toString();
      String month = validadeSplit[1].toString(); 
      String year = validadeSplit[2].toString();
    
     parsedDate = DateTime.parse('$year-$month-$day 00:00:00.000');
  }
 
  return parsedDate;
}

Answer 7

String dateFormatter(date) {

date = date.split('-'); DateFormat dateFormat = DateFormat("yMMMd"); String format = dateFormat.format(DateTime(int.parse(date[0]), int.parse(date[1]), int.parse(date[2]))); return format; }

Answer 8

I solved this by creating, on the C# server side, this attribute:

using Newtonsoft.Json.Converters;

public class DartDateTimeConverter : IsoDateTimeConverter
{
    public DartDateTimeConverter() 
    {
        DateTimeFormat = "yyyy'-'MM'-'dd'T'HH':'mm':'ss.FFFFFFK";
    }
}

and I use it like this:

[JsonConverter(converterType: typeof(DartDateTimeConverter))]
public DateTimeOffset CreatedOn { get; set; }

Internally, the precision is stored, but the Dart app consuming it gets an ISO8601 format with the right precision.

HTH