I wanted to validate date in client side so I wrote the following code. But instead of getting an exception I am getting a proper date object for 31st of February date strin
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);
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-
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");
}
}
}
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
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.