Java 8 LocalDateTime is parsing invalid datetime

Question

I wanted to validate datetime in client side so I wrote the following code. But instead of getting an exception I am getting a proper datetime object for 31st of February date string, which is clearly an invalid date.

public class Test {

    public static void main(String[] args) {
        String dateFormat = "HH:mm:ss MM/dd/yyyy";
        String dateString = "11:30:59 02/31/2015";
        DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern(dateFormat, Locale.US);
        try {
            LocalDateTime date = LocalDateTime.parse(dateString, dateTimeFormatter);
            System.out.println(date);
        } catch (Exception e) {
            // Throw invalid date message
        }
    }
}

Output : 2015-02-28T11:30:59

Does anyone know why LocalDateTime is parsing this datetime instead of throwing an exception.

@Satya: What makes you think that? Something clearly knows that date doesn't exist, as it's converting it to the 28th... — Jon Skeet
– Jon Skeet, Commented Sep 28, 2015 at 12:56
@JonSkeet My apologies for misleading you earlier, the problem was not the missing milliseconds. It was in fact the missing Era. It still requires a DateTimeFormatterBuilder but with a default for the Era, not for milliseconds. If you want to edit & undelete your answer, feel free. — RealSkeptic
– RealSkeptic, Commented Sep 28, 2015 at 14:29

Jon Skeet · Accepted Answer · 2015-09-28 15:26:26Z

54

You just need a strict ResolverStyle.

Parsing a text string occurs in two phases. Phase 1 is a basic text parse according to the fields added to the builder. Phase 2 resolves the parsed field-value pairs into date and/or time objects. This style is used to control how phase 2, resolving, happens.

Sample code - where withResolverStyle(ResolverStyle.STRICT) is the important change, along with the use of uuuu rather than yyyy (where uuuu is "year" and "yyyy" is "year of era", and therefore ambiguous):

import java.time.*;
import java.time.format.*;
import java.util.*;

public class Test {

    public static void main(String[] args) {
        String dateFormat = "HH:mm:ss MM/dd/uuuu";
        String dateString = "11:30:59 02/31/2015";
        DateTimeFormatter dateTimeFormatter = DateTimeFormatter
            .ofPattern(dateFormat, Locale.US)
            .withResolverStyle(ResolverStyle.STRICT);
        try {
            LocalDateTime date = LocalDateTime.parse(dateString, dateTimeFormatter);
            System.out.println(date);
        } catch (DateTimeParseException e) {
            // Throw invalid date message
            System.out.println("Exception was thrown");
        }
    }
}

edited Sep 28, 2015 at 15:26

answered Sep 28, 2015 at 12:53

Jon Skeet

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1 Comment

Palamino Over a year ago

Change the yyyy to uuuu in format string then there is no need to mess with ERA

Palamino · Accepted Answer · 2015-09-28 14:47:25Z

11

The Java 8 DateTimeFormatter uses yyyy to mean YEAR_OF_ERA, and uuuu to mean YEAR. You need to modify your pattern string as follows:

String dateFormat = "HH:mm:ss MM/dd/uuuu";

The DateTimeFormatter defaults to using the SMART resolver style, but you want it to use the STRICT resolver style. Modify your dateTimeFormatter initialization code as follows:

DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern(dateFormat, Locale.US)
                                                       .withResolverStyle(ResolverStyle.STRICT);

answered Sep 28, 2015 at 14:47

Palamino

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Comments

Edwin Buck · Accepted Answer · 2015-09-28 14:56:38Z

3

It is not rounding down. February has never had 31 days, and it is impossible to use a validating date / time object to represent a day that doesn't exist.

As a result, it takes the invalid input and gives you the best approximation to the correct date (the last date of February that year).

SimpleDateFormat inherits from DateFormat which has a setLenient(boolean value) method on it. I would expect that if you called setLenient(true) prior to parsing, it would probably complain more, as detailed in the javadocs.

answered Sep 28, 2015 at 14:56

Edwin Buck

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Comments

SatyaTNV · Accepted Answer · 2015-09-28 15:19:54Z

2

try {
    SimpleDateFormat df = new java.text.SimpleDateFormat("HH:mm:ss MM/dd/yyyy");
    df.setLenient(false);
    System.out.println(df.parse("11:30:59 02/29/2015"));
} catch (java.text.ParseException e) {
  System.out.println(e);
}

I found one solution to recognize date as a valid date with DateFormat.setLenient(boolean). If you try to parse any invalid date it will throws parse exception.

Edit:

Java 8, but this will raise exception if a month is not between 1 and 12, if a day is more than 32. Exactly not working. But for month its working.

try {
TemporalAccessor ta = DateTimeFormatter.ofPattern("HH:mm:ss MM/dd/yyyy").parse("11:30:59 02/32/2015");
} catch (Exception e) {
System.out.println(e);
}

Output:

java.time.format.DateTimeParseException: Text '11:30:59 02/32/2015' could not be
 parsed: Invalid value for DayOfMonth (valid values 1 - 28/31): 32

edited Sep 28, 2015 at 15:19

answered Sep 28, 2015 at 13:12

SatyaTNV

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1 Comment

Zeeshan Over a year ago

SimpleDateFormat doesn't work with Java 8 LocalDateTime. I am looking for a solution with Java8 LocalDateTime

jste89 · Accepted Answer · 2015-09-28 14:30:22Z

0

LocalDateTime.parse will only throw an error if the String passed in contains invalid characters, a number of days exceeding 31 or a month exceeding 12.

For example, if you modified your code as such:

String dateString = "11:30:59 0zz2/31/2015";

an exception would be thrown caused by the invalid 'zz' characters within your given date. As to why it's 'rounding-down' the date so to speak, that I don't know.

Source: https://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html#parse-java.lang.CharSequence-

edited Sep 28, 2015 at 14:30

answered Sep 28, 2015 at 12:59

jste89

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