Regular expression for validating date format ddmmyyyy dating antiques joinery

Posted by / 09-Sep-2020 00:52

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]/(?

The final two solutions are a little more complex, so we’re presenting these in both condensed and free-spacing form. Unfortunately, all it does is match valid numbers in a date time input, and 31st Feb will be marked as valid, but as so many have said, regex really isn't the best tool to do this test. \d)$ Your regexp does not work for years that "are multiples of 4 and 100, but not of 400". – Much To Learn (^(((0[1-9]|1[0-9]|2[0-8])[\/](0[1-9]|1[012]))|((29|30|31)[\/](0[13578]|1[02]))|((29|30)[\/](0[4,6,9]|11)))[\/](19|[2-9][0-9])\d\d$)|(^29[\/]02[\/](19|[2-9][0-9])(00|04|08|12|16|20|24|28|32|36|40|44|48|52|56|60|64|68|72|76|80|84|88|92|96)$) For those who look at these and get completely confused, here is an excerpt from my script. ([0-9](([2468][048]|[02468][48])|[13579][26])|([13579][26]|[02468][048]|0[0-9]|1[0-6])00))$ ^([0-9][-/]? (0[1-9]|1[0-9]|2[0-8]))|([0-9](([2468][048]|[02468][48])|[13579][26])|([13579][26]|[02468][048]|0[0-9]|1[0-6])00)[-/]?

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The only difference between the two forms is readability. The final two solutions allow all of the date formats, just like the first two examples.

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