I'm Working in Windows Application. I'm Developing Simple Notification System based c#.net with datetime concept. this project working in my system. I need validate datetime as dd/MM/yyyy while validate in client system error occur as string not recognized as a valid datetime. because client system have system formatted : dd MMMM yyyy
Let me know how to accept all kind of datetime format in my project..
Used Code:
public bool DateValidation(string stringDateValue)
{
bool A = false;
try
{
CultureInfo CultureInfoDateCulture = new CultureInfo("fr-FR");
DateTime d = DateTime.ParseExact(stringDateValue, "dd/MM/yyyy", CultureInfoDateCulture);
A = true;
}
catch
{
A = false;
}
return A;
}
If this is on a windows client you don't need to use a specific culture. When the application starts the Thread.CurrentThread.CurrentCulture gets set with a Culture matching the client regional settings. DateTime.TryParse parses a string and returns a boolean to indicate success or failure while also creating a DateTime struct bases on the string. Instead of keeping the stringDateValue (with its undetermined culture) you better adapt your code so you can use that DateTime type.
public bool DateValidation(string stringDateValue)
{
DateTime dtm;
bool success = DateTime.TryParse(stringDateValue, out dtm);
// you might want to keep the `dtm` value,
// either by returning it instead of the bool,
// our make it an out parameter on your method.
return success;
}
If you really need to be sure all dates in your client application use the same datetime format you could force the application to use a specific culture by setting the CurrentCulture at the start of your app:
Thread.CurrentThread.CurrentCulture = new new CultureInfo("fr-FR");
and instruct your users to use the dateformat prescribed in that culture (for french that is dd/MM/yyyy for the Shortdate pettern and dddd d MMMM yyyy for the long date pettern.
Related
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 writing test script for signup page and i need to put email address on each time. Can someone help me how to increment by value 1 in the email address as i execute the script for example, test#test.com and next time value should be test1#test.com. I an try with time stamp but not successfully work.
public class GetCurrentTimeStamp
{
public static void main( String[] args )
{
java.util.Date date= new java.util.Date();
System.out.println(new Timestamp(date.getTime()));
}
}
If you are trying to provide always unique email id, then you can use date with seconds as it keep changing also you can use
System.currentTimeMillis()
which gives number always unique. so you can append/concatenate it to email, i hope you know it.
You can use below code to get date
DateFormat dateFormat = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");
Date date = new Date();
System.out.println(dateFormat.format(date)); //2016/04/19 16:05:48
depends of simple date format provide 'yyyy/MM/dd HH:mm:ss' output will be displayed.
Thank You,
Murali
Use java.util.Date class instead of Timestamp and format it like so.
String timeStamp = new SimpleDateFormat("yyyy.MM.dd.HH.mm.ss").format(new Date());
String email= "test"+ timestamp + "#test.com";
Use below code:-
int num = 1; // Put this stament outside the for loop or put it as global variable.
Now use below code :-
num++;
String email= "test"+ num + "#test.com";
Hope it will help you :)
This question is similar to How to parse ZonedDateTime with default zone? but addinitional condition.
I have a string param that represent a date in UK format: "3/6/09". It doesn't contain time, only date. But may contain it and even time zone.
And I want to parse it to ZonedDateTime.
public static ZonedDateTime parse(String value) {
DateTimeFormatter formatter = DateTimeFormatter.ofLocalizedDateTime(SHORT).withLocale(Locale.UK).withZone(ZoneId.systemDefault());
TemporalAccessor temporalAccessor = formatter.parseBest(value, ZonedDateTime::from, LocalDateTime::from, LocalDate::from);
if (temporalAccessor instanceof ZonedDateTime) {
return ((ZonedDateTime) temporalAccessor);
}
if (temporalAccessor instanceof LocalDateTime) {
return ((LocalDateTime) temporalAccessor).atZone(ZoneId.systemDefault());
}
return ((LocalDate) temporalAccessor).atStartOfDay(ZoneId.systemDefault());
}
But, it fails with exception:
java.time.format.DateTimeParseException: Text '3/6/2009' could not be parsed at index 6
It's a bug for me, or isn't?
In my opinion is not a bug. Your approach is flawed.
First of all you are returning a ZonedDateTime so it is expected that the String contains full date, time and zone information. The string "3/6/09" should be parsed to a LocalDate.
Second, you are delegating a runtime detection of format to the library. Again, you should be parsing/formatting an expected format. Your application should know wether is expecting a full date & time or a partial (only date or only time).
Anyway you will have more luck detecting the format and then using different parsing methods.
Only local date:
DateTimeFormatter
.ofLocalizedDate(FormatStyle.SHORT)
.parse(value, LocalDate::from)`
Zoned date and time:
DateTimeFormatter
.ofLocalizedDateTime(FormatStyle.SHORT, FormatStyle.SHORT)
.parse(value, ZonedDateTime::from)`
The format used can be seen using the getLocalizedDateTimePattern() method:
String fmt = DateTimeFormatterBuilder.getLocalizedDateTimePattern(
FormatStyle.SHORT, FormatStyle.SHORT, IsoChronology.INSTANCE, Locale.UK);
The result is "dd/MM/yy HH:mm".
As such, the format is expecting both a date and a time with a space separator, so that is what must be provided.
In addition, the format/parse expects there to be two digits for the day-of-month and two digits for the month-of-year. Thus, you would need to pass in "03/06/09 00:00" in order to get the result you expect, in which case you can parse directly to a LocalDateTime.
Alternatively, use ofLocalizedDate():
DateTimeFormatter formatter =
DateTimeFormatter.ofLocalizedDate(FormatStyle.SHORT).withLocale(Locale.UK);
LocalDate date = LocalDate.parse("03/06/99", formatter);
Note that the input must still have two digits for the day and month.
Alternatively, parse using a specific pattern that can handle the missing leading zeroes:
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("d/M/yy");
LocalDate date = LocalDate.parse("3/6/99", formatter);
LocalDate date = LocalDate.parse("03/06/99", formatter);
// handles both "3/6/99" and "03/06/99"
Update: Lenient parsing also handles this case:
DateTimeFormatter formatter = new DateTimeFormatterBuilder()
.parseLenient().appendPattern("dd/MM/yy").toFormatter();
LocalDate date = LocalDate.parse("3/6/99", formatter);
LocalDate date = LocalDate.parse("03/06/99", formatter);
// handles both "3/6/99" and "03/06/99"
I need t use the CQ5 xtype 'datefield' since I need only the date and not time tombe entered by the author.
But the issue is that 'datefield' saves the date in JCR as a String and not as a timestap [ as it does when 'datetime' is used.]
Is there a work around to save the date as a timestamp ?
I don't think it is possible to save the date as timestamp using datefield, without meddling with the default script. But as a workaround, you can use datetime and set the property hideTime to true, to hide the time part, so that the author will not be able to author it.
The json for the config is shown below.
{ "fieldLabel":"Date",
"xtype":"datetime",
"hideTime":true,
"name":"./date",
"defaultValue":"now",
"jcr:primaryType":"cq:Widget"
}
You can add defaultValue to 'now', if you want current date to be initialized as the default, if not explicitly filled in by the author, else it can be ignored.
NOTE: The defaultValue: 'now' doesnt work for me in IE (i am using IE 11 and emulating the previous versions through dev tools), but it works fine in Chrome and Mozilla.
A rough workaround for your jsp:
<%#page import="java.text.SimpleDateFormat,java.util.Date"%>
<%
SimpleDateFormat displayDateFormat = new SimpleDateFormat("dd MMM yyyy");
String dateField = properties.get("nameofdatefield", "");
SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yy");
Date formattedDate = sdf.parse(dateField);
String formattedDateStr = displayDateFormat.format(formattedDate);
out.println('Example of formated string'+formattedDateStr);
%>
From the above, you can also convert it to a Date Object, depending on what you wish to manipulate.
Let me know if the above example helps