Can any one suggest me the following:
i> how to convert 3/12/2010 10:15 to the format Mar 12, 2010 10:15 in c#.
ii> how to remove time part out of it(i.e after the conversion i want Mar 12, 2010
Thanks in advance.
nimesh,
The String.Format function should be able to accommodate your needs (updated to reflect my comment):
string inputString = "03/12/2012";
DateTime dt = DateTime.MinValue;
try {
dt = DateTime.Parse(inputString);
}
catch (Exception ex) {
// handle the exception however you like.
return;
}
string formattedDate = String.Format("{0:MMM d, yyyy}", dt);
To avoid the above, check out TryParse:
http://msdn.microsoft.com/en-us/library/system.datetime.tryparse.aspx
You can format it in lots of ways:
http://www.csharp-examples.net/string-format-datetime/
Use DateTime.Parse(...) to parse the input string into a DateTime object, and then use Format to make the string as posted by mikey.
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
When parsing date and time from an API call which is formatted like this 2018 - 04 - 18T21:51:00 I get and error that states
and I don't understand what am I doing wrong.
/// I set in which format should my data be, and parse the data from the API call
Text('Date and Time:
${DateFormat("yyyy-MM-dd hh:mm:ss").format(DateTime.parse(apicall.timeanddate))}'),
After that I would format the Date and Time according to Locale language preference of the device.
Any help is appreciated.
The format you are passing is invalid, so you need to transform it:
String getDateTime(String dateFromServer){
dateFromServer = dateFromServer.replaceAll(" ","");
dateFromServer = dateFromServer.replaceAll("T"," ");
return dateFromServer;
}
I replaced all spaces and T that were causing problems you can now use this with:
DateFormat("yyyy-MM-dd hh:mm:ss").format(getDateTime(apicall.timeanddate));
Hope this would solve your problem.
def cleantz( time : String ) : String = {
var sign_builder= new StringBuilder ++= time
println(sign_builder)
var clean_sign = ""
if (sign_builder.charAt(23).toString == "-"){
clean_sign= sign_builder.replace(23,24,"-").toString()
}else{
clean_sign = sign_builder.replace(23,24,"+").toString()
}
var time_builder= new StringBuilder ++= clean_sign
if (time_builder.charAt(26).toString == ":"){
val cleanz = time_builder.deleteCharAt(26)
cleanz.toString()
}else{
time_builder.toString()
}
}
val start = ISO8601Format.parse(cleantz(01/01/2017 6:54 PM))
I get this error:
java.lang.StringIndexOutOfBoundsException: String index out of range: 23
java.time
For the sake of completeness I should like to contribute the modern answer. It’s quite simple and straightforward.
I am sorry that I can neither write Scala code nor test it on my computer. I have to trust you to translate from Java.
private static DateTimeFormatter inputFormatter
= DateTimeFormatter.ofPattern("MM/dd/yyyy h:mm a", Locale.US);
public static String cleantz(String time) {
return LocalDateTime.parse(time, inputFormatter)
.atOffset(ZoneOffset.ofHours(1))
.toString();
}
Now cleantz("01/01/2017 6:54 PM") returns 2017-01-01T18:54+01:00, which is in ISO 8601 format. I would immediately suppose that you’re set. If for some reason you want or need the seconds too, replace .toString(); with:
.format(DateTimeFormatter.ISO_OFFSET_DATE_TIME);
Now the result is 2017-01-01T18:54:00+01:00. In both cases the milliseconds would have been printed if there were any.
Since AM and PM are hardly used in other languages than English, I suggest you give an English-speaking locale to DateTimeFormatter.ofPattern() (in my example I used Locale.US). Failing to provide a locale will cause the code to fail on many computers with non-English language settings.
Why java.time?
SimpleDateFormat and friends are long outdated and notoriously troublesome. I cannot count the questions asked on Stack Overflow because SimpleDateFormat behaved differently from what every sane programmer would have expected, or offered no help to debug the simple errors we all make from time to time.
Joda-Time was good for a long time. Today the Joda-Time homepage says:
Note that Joda-Time is considered to be a largely “finished” project.
No major enhancements are planned. If using Java SE 8, please migrate
to java.time (JSR-310).
java.time is the modern Java date & time API built using the experience from Joda-Time and under the same lead developer, Stephen Colebourne. It is built into Java 8 and later, and a backport exists for Java 6 and 7, so you can use the same classes there too.
Assuming that your input string is 01/01/2017 6:54 PM: it has 18 characters. When you call charAt(23), it tries to get the character at position 23, which doesn't exist: the string has positions from zero (the first 0) to 17 (the M). If you try to get a position greater than that, it throws a StringIndexOutOfBoundsException.
But you don't need to do all this string manipulation. If you have a string that represents a date in some format, and want to convert it to another format, all you need is:
parse the original string to a date
format this date to another format
So you need 2 different Joda formatter's (one for each step). But there's one additional detail.
The input has a date (01/01/2017) and a time (6:54 PM), and the output has a date (2017-01-01), a time (18:54:00.000) and the UTC offset (+0100). So you'll have an additional step:
parse the original string to a date
add the +0100 offset to the parsed date
format this date to another format
With Joda-Time, this can be achieved with the following code:
import org.joda.time.DateTimeZone
import org.joda.time.LocalDateTime
import org.joda.time.format.DateTimeFormat
import org.joda.time.format.ISODateTimeFormat
val fmt = DateTimeFormat.forPattern("dd/MM/yyyy h:mm a")
// parse the date
val localDate = LocalDateTime.parse("01/01/2017 6:54 PM", fmt)
// add the +01:00 offset
val dt = localDate.toDateTime(DateTimeZone.forOffsetHours(1))
// format to ISO8601
print(ISODateTimeFormat.dateTime().print(dt))
The output will be:
2017-01-01T18:54:00.000+01:00
Note that the offset is printed as +01:00. If you want exactly +0100 (without the :), you'll need to create another formatter:
val formatter = DateTimeFormat.forPattern("yyyy-MM-dd'T'HH:mm:ss.SSSZ")
println(formatter.print(dt))
The output will be:
2017-01-01T18:54:00.000+0100
This is the code I used to achieve the same result. The error occurred because I was trying to parse the wrong date format.
val inputForm = new SimpleDateFormat("MM/dd/yyyy h:mm a")
val outputForm = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSZ")
val dateFormat1 = start_iso
val dateFormat2 = stop_iso
val start = outputForm.format(inputForm.parse(start_iso))
val stop = outputForm.format(inputForm.parse(stop_iso))
println(start)
println(stop)