The problem is that replace(“st”, “”) also removes the st ending of August, resulting in the input string seen in the error message. To handle this, you need to make sure the st suffix is right after a digit, so…
String dateStr = “04/05/2010”; SimpleDateFormat curFormater = new SimpleDateFormat(“dd/MM/yyyy”); try { dateObj = curFormater.parse(dateStr); } catch (ParseException e) { // TODO Auto-generated catch block e.printStackTrace(); } SimpleDateFormat postFormater = new SimpleDateFormat(“MMMM dd, yyyy”); String newDateStr = postFormater.format(dateObj); Toast.makeText(getApplicationContext(), newDateStr, 1).show();…
Just follow the documentation and use the formatting elements that correspond to the string you have: String strDate = “12/19/2015 8:00:00 AM”; DateFormat df = new SimpleDateFormat(“MM/dd/yyyy hh:mm:ss a”); Date date = df.parse(strDate); solved How to parse this date in…
Your date format MM/dd/yyyy’T’HH:mm:ss which your using is incorrect 1/31/2018 8:58:12 AM +00:00 and return different value from expected. SimpleDateFormat sdf = new SimpleDateFormat(“MM/dd/yyyy’T’HH:mm:ss”); return 1/31/2018T8:58:12 And you are trying to pass below 1/31/2018 8:58:12 AM +00:00 1 solved Exception…
To use date formatting, you need to start with a DateTime object, not a string. So in your scenario you’ll have to parse your string as a DateTime, then format it out again to a different string format. There’s no…
This is due to the format you used as “yyyy-MM-DD”. The parser will parse in the sequence way: Your input value is “2018-03-08” yyyy – will bring to the year 2018 MM – will bring to the month MARCH But…
At First Post Your Code . Please follow How do you format date and time in Android Code for SimpleDateFormat . Just check the logic . SimpleDateFormat timeStampFormat = new SimpleDateFormat(“dd-yyyy-MM HHmm”); Date GetDate = new Date(); String DateStr =…
String year1= String.valueOf (year); String month1= String.valueOf(month); String day1= String.valueOf(day); must be changed to String year1= String.valueOf (num1); String month1= String.valueOf(num2); String day1= String.valueOf(num3); because these 3 variables only hold the values entered by the user. Also, not sure why…
This should work as you specified in your question. DateFormat dateFormat = new SimpleDateFormat(“yy-MMM-dd HH:mm:ss a”); System.out.println(dateFormat.format(new Date())); The date format you have in your question is something like: DateFormat dateFormat = new SimpleDateFormat(“yyyyMMddHmmss”); 2 solved Convert String to Date…
I would use java.time for a task like this. You can define two java.time.format.DateTimeFormatter instances (one for parsing the input string to java.time.YearMonth and another for formatting the obtained YearMonth to a string of the desired format). Define a method…
You should be using SimpleDateFormat(“dd-MM-yyyy”). DD is for day in year, like there are 365 days in a year. 3 solved SimpleDateFormat returning incorrect day on given date [closed]
Using SimpleDateFormatter you can convert from date string to date and date to date String public final class DateUtil { private static final String DEFAULT_DATE_FORMAT = “dd-MMM-yyyy”; private DateUtil() { } public static final String formatDate(final Date date, final String…
You are experiencing SimpleDateFormat behaving as designed. It’s a behaviour that comes as a negative surprise to most. There are two solutions: the recommended one and the discouraged one. Recommended solution: LocalTime DateTimeFormatter timeFormat = DateTimeFormatter.ofPattern(“hh:mm a”, Locale.ROOT); try {…
The problem is in your lines DateFormat dfq = new SimpleDateFormat(“EEE MMM dd HH:mm:ss z yyyy”, Locale.FRENCH); Date date1 = dfq.parse(dateq); You got a ParseException because in ” Tue Feb 08 12:30:27 +0000 2011 ” you have leading and trailing…
Use these two method for convert second to 24hours format public static String convertSecToHoursMinute(long Sec) { long hours = Sec / 3600; long minutes = (Sec % 3600) / 60; long seconds = (Sec % 3600) % 60; String amPm…