Regular expression for validating date format ddmmyyyy

We might think that something as conceptually trivial as a date validation should be an easy job for a regular expression. The main issue is that regular expressions don’t deal directly with numbers.

to match 3 followed by 0 or 1, or to match 1 or 2 followed by any digit, or to match an optional 0 followed by 1 to 9. [0-9]$"; Pattern pattern = Pattern.compile(regex); for(String date : dates) List dates = new Array List(); //With leading zeros dates.add("01/01/11"); dates.add("01/01/2011"); //Missing leading zeros dates.add("1/1/11"); dates.add("01/1/2011"); dates.add("1/11/2011"); dates.add("1/11/11"); dates.add("11/1/11"); String regex = "^[0-3][0-9]/[0-3][0-9]/(?

Recipe 4.7 shows how to validate date and time formats according to the ISO 8601 standard.

Recipe 6.7 explains how you can create a regular expression to match a number in a given range of numbers.

^(((((0[1-9])|(1\d)|(2[0-8]))\/((0[1-9])|(1[0-2])))|((31\/((0[13578])|(1[02])))|((29|30)\/((0[1,3-9])|(1[0-2])))))\/((20[0-9][0-9])|(19[0-9][0-9])))|((29\/02\/(19|20)(([02468][048])|([13579][26]))))$ Hi Use this following Regular Expression Details, This will support leap year also.

This is easily done with a pair of word boundaries.

The other issue is that regular expressions don’t deal directly with numbers. That’s because the characters for the digits 0 through 9 occupy consecutive positions in the ASCII and Unicode character tables.

You can’t tell a regular expression to “match a number between 1 and 31”, for instance. See Chapter 6 for more details on matching all kinds of numbers with regular expressions.

Because of this, you have to choose how simple or how accurate you want your regular expression to be.

List dates = new Array List(); dates.add("1/1/11"); dates.add("01/01/11"); dates.add("01/01/2011"); dates.add("01/1/2011"); dates.add("1/11/2011"); dates.add("1/11/11"); dates.add("11/1/11"); String regex = "^[0-3]?

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