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Now that we successfully matched our general date format, we need to constrain that further – to make sure the dates are actually correct: In order to match leap years correctly we must first identify when we have encountered a leap year, and then make sure that we accept February 29th as a valid date for those years.As the number of leap years in our restricted range is large enough we should use the appropriate divisibility rules to filter them: Since we haven't optimized this pattern in favor of a better readability, feel free to experiment with a length of it. [1-9] # 01-09 or 1-9 | # .[0-9] # 10-19 or 20-29 | # .3 # 30, 31 ) #end of group #1 / # follow by a "/" ( # start of group #2 0? [1-9] # 01-09 or 1-9 | # .1 # 10,11,12 ) # end of group #2 / # follow by a "/" ( # start of group #3 (19|20)\d\d # 19[0-9][0-9] or 20[0-9][0-9] ) # end of group #3 The above regular expression is used to validate the date format in “dd/mm/yyyy”, you can easy customize to suit your need. Mostly programming in Java, Spring Framework, Hibernate / JPA. For an introduction to regular expressions, refer to our Guide To Java Regular Expressions API.The primary purpose of this article was not to learn an efficient way of testing a string for its membership in a set of all possible dates.Consider using provided by Java8 if a reliable and fast approach to validating a date is needed.
“333/2/2008”, “29/02/200a” – day is invalid, year is invalid 7.In the code snippet below if the user does not enter a letter the code will keep asking for a valid letter.It loops until the length of the inputted letters equals to the length of secret word. “29/a/2008”, “a/02/2008” – month is invalid, day is invalid 6.
“30/2/2008”, “31/02/2008” – leap year in February has 29 days only 5.We're going to define a valid date in relation to the international Gregorian calendar.