Unparseable Date Exception SimpleDateFormat for Hmm format

Question

While validating the given string text is of Hmm format getting this error

java.text.ParseException: Unparseable date: "1637"

but this is working for text "747".

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;

    public class Validation{
    
         public void validateDateFormat(String format,String text){
              SimpleDateFormat sdfrmt = new SimpleDateFormat(format);
              sdfrmt.setLenient(false);
              Date testDate = null;
              try {
                    testDate = sdfrmt.parse(text);
    
              } catch (ParseException e) {
                    System.out.println("dateFormat Exception :");
                    e.printStackTrace();
            }
         }
         
         public static void main(String []args){
            Validation val = new Validation();
            val.validateDateFormat("Hmm","747"); //working 
            val.validateDateFormat("Hmm","1637");//not working
         }
    }

This is for validating the given columns from the uploaded file.So wrote this to be dynamic based on the format written in the config for each column.

Well, you shouldn't use `Date` and `SimpleDateFormat`, for they are obsolete and troublesome. You know why? [Here it is](https://stackoverflow.com/questions/1969442/whats-wrong-with-java-date-time-api). Use classes from the `java.time` package instead. — MC Emperor, Jun 21 '20 at 12:31
I recommend you don’t use `SimpleDateFormat` and `Date`. Those classes are poorly designed and long outdated, the former in particular notoriously troublesome. Instead use `LocalTime` and `DateTimeFormatter`, both from [java.time, the modern Java date and time API](https://docs.oracle.com/javase/tutorial/datetime/). — Ole V.V., Jun 21 '20 at 18:28

score 1 · Answer 1 · answered Jun 21 '20 at 12:44

1

Well, it should work.

But you should use the newer Java Date and Time API (JSR 310) available in the java.time package.

If you then replace Date with LocalTime, SimpleDateFormat with DateTimeFormatter (using the ofPattern factory method) and ParseException with DateTimeParseException, it'll work.

answered Jun 21 '20 at 12:44

MC Emperor

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Great link! Thanks. – Ole V.V. Jun 21 '20 at 18:28

score 1 · Answer 2 · answered Jun 21 '20 at 12:53

This is happening because the SimpleDateFormat is unable to parse the given date using the format you have given.

Let's understand - your format is Hmm and you have given the date as 747 then, of course during parsing, the first letter 7 of the date is mapped to the first letter "H" of the format i.e. hours and 47 is mapped to mm i.e. minutes and hence it is able to convert correctly but for the next date 1637 it is failing because it does not know which letter to assign to H.

Here are some options you can try to make it more generic, choose format such as HHmm and always give the date of length 4, for e.g. for 747 enter the input as 0747 and it should work.

Or choose a format that is more clear for the parser to map for e.g. you can choose the format as H:mm and give the inputs as 7:47 or 16:37 since there is a delimiter : between hours and minutes, the parser will be able to parse all types of time irrespective of the length of the given input 3 or 4.

Unparseable Date Exception SimpleDateFormat for Hmm format

2 Answers2