23/12/2013 is mapping with MM/dd/yyyy format why, why not ParseException

Question

Can anybody help?

public void dateCalender() throws ParseException{
        System.out.println(new SimpleDateFormat("MM/dd/yyyy", Locale.ENGLISH).parse("120/12/2013").toString()); //OUTPUT (Unexpected): Mon Dec 12 00:00:00 IST 2022
        System.out.println(new SimpleDateFormat("MM/dd/yyyy", Locale.ENGLISH).parse("23/12/2013").toString()); //OUTPUT (Unexpected): Wed Nov 12 00:00:00 IST 2014
        System.out.println(new SimpleDateFormat("MM/dd/yyyy", Locale.ENGLISH).parse("Jan/12/2013").toString()); //OUTPUT (Expected): Unparseable date: "Jan/12/2013"
    }

Antoniossss · Accepted Answer · 2013-11-23T09:48:39.813

4

You have to invoke setLenient with false - otherwise SimpleDateFormat will try to "figure out" what month that is. So, first create SimpleDateFormat and invoke sdf.setLenient(false). Now when parsing, you will get exception.

edited Nov 23 '13 at 09:48

answered Nov 23 '13 at 08:55

Antoniossss

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1

It is `setLenient` (the spelling). – ADTC Nov 23 '13 at 09:46
1

You are right! OMG, my whole life is a lie! Thanks - edited!. – Antoniossss Nov 23 '13 at 09:48

Masudul · Answer 2 · 2013-11-23T09:09:50.103

According to docs , parsing is lenient, So you didn't get exception for invalid input. It converted to another value.

By default, parsing is lenient: If the input is not in the form used by this object's format method but can still be parsed as a date, then the parse succeeds. Clients may insist on strict adherence to the format by calling setLenient(false).

If you want to get java.text.ParseException: Unparseable date:, than apply setLenient(false); at SimpleDateFormat

    String dateStr="120/12/2013";
    SimpleDateFormat sdf=new SimpleDateFormat("MM/dd/yyyy", Locale.ENGLISH);
    sdf.setLenient(false);
    try {
        Date d=sdf.parse(dateStr);
    } catch (ParseException ex) {
        ex.printStackTrace();
    }

score 1 · Answer 3 · answered Nov 23 '13 at 08:55

1

From the documentation:

public void setLenient(boolean lenient)

You need to set the format to be non-lenient.

answered Nov 23 '13 at 08:55

devnull

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score 1 · Answer 4 · answered Nov 23 '13 at 09:00

On the first two:

MM/dd/yyyy 120/12/2013

The compiler will take this as '12/12/2013 + 118 months' and essentially solve for the correct date. In your example, it comes out as December 12, 2022 (12/12/2013 + 9 years).

MM/dd/yyyy 23/12/2013

The exact same thing happens. You get '12/12/2013 + 9 months', or 11/12/2014.

The third one isn't technically in the MM/dd/yyyy format. As from the other answers, you can do something like this:

SystemDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy", Locale.ENGLISH);
sdf.setLenient(true);
sdf.parse("Jan/12/2013");
System.out.println(sdf.toString());

score 1 · Answer 5 · answered Dec 31 '22 at 12:41

java.time

The question and the accepted answer use the java.util date-time API and their formatting API, SimpleDateFormat which was the right thing to do in 2013. In March 2014, the modern Date-Time API supplanted the legacy date-time API and since then it is strongly recommended to switch to java.time, the modern date-time API.

Solution using java.time, the modern Date-Time API:

The java.time is smart to catch most such problems without any explicit settings. A custom DateTimeFormatter is by default set to ResolverStyle#SMART which does not allow a month outside the range 1-12 or a day-of-month value beyond 31. However, it also means that 02/31/2022 can be parsed to 2022-02-28 with a DateTimeFormatter.ofPattern("MM/dd/uuuu"). It is only to disallow such a parsing that you need to set ResolverStyle.#STRICT.

Demo:

import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeParseException;
import java.util.Locale;

class Main {
    public static void main(String[] args) {
        DateTimeFormatter parser = DateTimeFormatter.ofPattern("MM/dd/uuuu", Locale.ENGLISH);
        String arr[] = {
                "120/12/2013", // 120 will be discarded against MM
                "23/12/2013", // Month can not be outside the range 1-12
                "Jan/12/2013", // Jan will discarded against MM. Requires MMM
                "12/31/2022", // Will be parsed normally
                "02/31/2022" // Will be parsed to 2022-02-28
        };
        for (String s : arr) {
            try {
                System.out.println("===========================");
                System.out.println("Trying to parse " + s + " using the format MM/dd/uuuu");
                System.out.println("Parsed to " + LocalDate.parse(s, parser));
            } catch (DateTimeParseException e) {
                System.out.println(e.getMessage());
                // throw e;
            }
        }
    }
}

Output:

===========================
Trying to parse 120/12/2013 using the format MM/dd/uuuu
Text '120/12/2013' could not be parsed at index 2
===========================
Trying to parse 23/12/2013 using the format MM/dd/uuuu
Text '23/12/2013' could not be parsed: Invalid value for MonthOfYear (valid values 1 - 12): 23
===========================
Trying to parse Jan/12/2013 using the format MM/dd/uuuu
Text 'Jan/12/2013' could not be parsed at index 0
===========================
Trying to parse 12/31/2022 using the format MM/dd/uuuu
Parsed to 2022-12-31
===========================
Trying to parse 02/31/2022 using the format MM/dd/uuuu
Parsed to 2022-02-28

ONLINE DEMO

Demo of ResolverStyle.#STRICT:

class Main {
    public static void main(String[] args) {
        DateTimeFormatter parser = DateTimeFormatter.ofPattern("MM/dd/uuuu", Locale.ENGLISH)
                .withResolverStyle(ResolverStyle.STRICT);
        String arr[] = {
                "12/31/2022", // Will be parsed normally
                "02/31/2022" // Will fail against ResolverStyle.STRICT check
        };
        for (String s : arr) {
            try {
                System.out.println("===========================");
                System.out.println("Trying to parse " + s + " using the format MM/dd/uuuu");
                System.out.println("Parsed to " + LocalDate.parse(s, parser));
            } catch (DateTimeParseException e) {
                System.out.println(e.getMessage());
                // throw e;
            }
        }
    }
}

Output:

===========================
Trying to parse 12/31/2022 using the format MM/dd/uuuu
Parsed to 2022-12-31
===========================
Trying to parse 02/31/2022 using the format MM/dd/uuuu
Text '02/31/2022' could not be parsed: Invalid date 'FEBRUARY 31'

ONLINE DEMO

Note: Here, you can use y instead of u but I prefer u to y.

Learn more about the modern Date-Time API from Trail: Date Time.

score -1 · Answer 6 · answered Nov 23 '13 at 09:09

-1

You are using the format MM/dd/yyyy .So its taking Two positions

U try this way:

System.out.println(new SimpleDateFormat("MMM/dd/yyyy", Locale.ENGLISH).parse("Jan/12/2013").toString());

Here not throws unparsable exception but am not shoore about that is working or not

I think this will be wrong way every time using only MM/dd/yyyy format as "1/12/2013"

answered Nov 23 '13 at 09:09

user2996174

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Why don't you delete that answer? – Antoniossss Nov 23 '13 at 09:14

23/12/2013 is mapping with MM/dd/yyyy format why, why not ParseException

6 Answers6

java.time

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