I am storing a date string in SharedPreferences as 2022-9-14 19:34, but this is not recognized as DateTime when using
DateTime dt1 = DateTime.parse(_inicio);
print("fecha······· ${(dt1)}");
I am getting the following error:
Invalid date format
2022-9-14 19:34
What I need is to show the date string as 14-09-2022 19:34
EDIT
The date string is saved as follows:
var hoy = DateTime.now();
var ano = hoy.year.toString();
var mes = hoy.month.toString();
var dia = hoy.day.toString();
var turno = "";
var fechaHora = ano+"-"+mes+"-"+dia+" "+_timeOfDay.format(context).toString();
And _timeOfDay is the picked time from TimePicker.
2022-09-14 19:34 is one of the valid case. So you need to add zeros where there is a single digit (month in this case)
But a better solution would be to store DateTime.now().toString() directly in shared-preferences. You could also use a package like intl to format the date in your choice of format.
Since you're using time as well from timepicker, you could use DateTime constructor will all 5 values like this:
new DateTime(year, month, date, hour, minute).toString();
Related
I have a date which is a string and looks like this:
String day = "30-11-2022 12:27";
I am trying to convert the above string to DateTime object and convert the 24hr time to 12hr. I am using the following code:
DateFormat("dd-MM-yyyy hh:mm a").parse(day);
It was working before but today the parsing is causing format exception error. The error message is shown below:
[ERROR:flutter/runtime/dart_vm_initializer.cc(41)] Unhandled Exception: FormatException: Trying to read from 30-11-2022 12:27 at position 17
Why am I getting error while parsing now? How to fix it?
There must be a simpler solution, but it works this way:
final str = "30-11-2022 12:27";
final date = DateFormat("dd-MM-yyyy hh:mm").parse(str);
final formated = DateFormat("dd-MM-yyyy h:mm a").format(date);
The format for a 24-hr time is 'HH:mm'. The key is to call add_jm(). Just do this:
final day = '30-11-2022 12:27';
final dateTime = DateFormat('dd-MM-yyyy HH:mm').parse(day);
final formatted = DateFormat('dd-MM-yyy').add_jm().format(dateTime);
print(formatted);
The output is:
30-11-2022 12:27 PM
can try this. hope your problem solved
DateTime dateTime = DateFormat('yyyy/MM/dd h:m').parse(date);
final DateFormat formatter = DateFormat('yyyy-MM-dd h:m a');
final String formatted = formatter.format(dateTime);
It really doesn't matter for datetime that the input date is in am format or not, because you can parse it to what ever format you want. For both option when you parse your string, it will save it in regular form. So just try this:
String day = "30-11-2022 12:27";
DateTime date = normalFormat.parse(day);
and when ever you want show it as AM, just format date;
Why am I getting error while parsing now? How to fix it?
You specified a DateFormat format string that uses a, which indicates that you want to parse an AM/PM marker, but the string you try to parse is "30-11-2022 12:27", which lacks it.
If you're trying to convert a 24-hour time to a 12-hour time, those are fundamentally two different formats, so you will two different DateFormat objects. Note that the format pattern for hours is different for 12-hour times (h) than for 24-hour ones (H):
String day = "30-11-2022 12:27";
var datetime = DateFormat("dd-MM-yyyy HH:mm").parse(day);
// Prints: 30-11-2022 12:27 PM
print(DateFormat('dd-MM-yyyy hh:mm a').format(datetime));
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?
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
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.)
import 'package:intl/intl.dart';
DateTime brazilianDate = new DateFormat("dd/MM/yyyy").parse("11/11/2011");
you can just use : DateTime.parse("your date string");
for any extra formating, you can use "Intl" package.
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;
}
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
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;
}
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
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?
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
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.)
import 'package:intl/intl.dart';
DateTime brazilianDate = new DateFormat("dd/MM/yyyy").parse("11/11/2011");
you can just use : DateTime.parse("your date string");
for any extra formating, you can use "Intl" package.
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;
}
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
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;
}
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
I am getting date and time from store. In data base its look like this
Need to know how can I show this as DD/MM/YY
I am trying to do like this
String timeString = snapshot.data[index]['lastupdate'].toString();
DateTime date = DateTime.parse(timeString);
print(DateFormat('yyyy-MM-dd').format(date));
Its showing error of Invalid date format
Try like this your lastupdate date is not convertime inDate thats why its showing error
DateTime date = DateTime.parse(snapshot.data[index]['lastupdate'].toDate().toString());
print(DateFormat('dd-MMM-yyy').format(date));
Use function to get date from Timestamp like this:
readDate(Timestamp dateTime) {
DateTime date = DateTime.parse(dateTime.toDate().toString());
// add DateFormat What you want. Look at the below comment example
//String formatedDate = DateFormat('dd-MMM-yyy').format(date);
String formatedDate = DateFormat.yMMMMd().format(date);
return formatedDate;
}
Use the function when you want it.
For example:
Text(readDOB(streamSnapshot.data!["dob"]))
For this you should install intl package. read
I have a date-time in epoch format: 1633247247
I want to convert it into timestamps like this: Sunday, 3 October 2021 or just October 3 2021
I am writing this code
final DateTime timeStamp = DateTime.fromMillisecondsSinceEpoch(1633247247);
But it is returning 1970-01-19 18:25:11.247
Edit I ran this code
final DateTime timeStamp = DateTime.fromMillisecondsSinceEpoch(1633247247 * 1000);
Got the output in datetime. I am now trying to convert into string
String formattedDate = DateFormat('yyyy-MM-dd – kk:mm').format(date);
It gives this error The instance member 'date' can't be accessed in an initializer.
It represent to milliseconds, you need to multiple it with 1000 like below :
final DateTime timeStamp = DateTime.fromMillisecondsSinceEpoch(1633247247 * 1000);
It seems like that the time you've is 'seconds since epoch' so just multiplying by 1000 should give you the correct time.
final DateTime timeStamp = DateTime.fromMillisecondsSinceEpoch(1633247247 * 1000);
Download and Import
intl Pacakge
import 'package:intl/intl.dart';
Initialize a variable
String datetime;
Get Current date and conversion
Future<void> Datetimes() async{
setState(() {
DateTime now = DateTime.now();
String formattedDate = DateFormat('EEEE d MMM hh:mm:ss').format(now);
datetime=formattedDate;
});
}
Output
Saturday 3 oct 2:07:29
Refer this for DateandTime class for more formatting parameters
DateFormat-class
You need to use DateTime Formats. You can find a solution here:
https://stackoverflow.com/a/51579740/11604909