Java code for validating date
): If the year is divisible by 4, it's a leap year UNLESS it's divisible by 100, in which case it's not a leap year UNLESS it's divisible by 400, in which case it is a leap year.
If the year is not divisible by 4, it's not a leap year. An if statement that determines the number of days in the month entered and stores that value in variable days In Month.
// updated via the `Scheduled Thread Pool Executor` private int year; private int month; private int day; private static final Pattern DATE_PATTERN = Pattern.compile("\s*(\d)/(\d)/(\d))\s*"); public static boolean is Valid Date(String date) As already said, solution reading the input char by char would be faster, but I'd go the opposite direction.
Most probably, the speed is more than good enough, and the problem is called "stringly typed programming".
* * @param p Date String date to be tested * @return a Testable Lastly, this method is difficult to test. While you may not use Java 8, you can certainly make a What happens if there is a non-numeric character in the year string?
Does it work when now falls in a DST overlap period? You could either accept a three digit year or reject a four digit year. -800 would be acceptable under this check but 2016 would not.
You catch anything that is a fails or if the year is non-positive.
If the year is between 0 and 100, it assumes that it is abbreviated and asks for more digits.
* * @param p Date String date to be tested * @return true if date (@ am) = yesterday (@am) */ public static boolean is Valid Future Date(String p Date String) /** * This method tests whether the date string passed in represents * a real date in mm/dd/YYYY format.
Scanner; public class Dates Here is my code that I have so far completed.
It contains some errors that I have no idea how to solve.
For date in European (English) format, we can rearrange the regular expression as follows. So the date 17/08/2009 is valid but 08/17/2009 will not pass this validation.
The logic is same but the elements are re-arranged.