OK, you asked for a regex. Here it is. I think it’s immediately obvious why it’s not a good idea to validate a date with a regular expression: First, the verbose, commented version to at least make understanding this beast possible: ^ # start of string (?: # either match… (?: (?: # 31st day … Read more
The length of a year is (more or less) 365.242196 days. So we have to subtract, more or less, a quarter of a day to make it fit : 365.242196 – 0.25 = 364.992196 (by adding 1 day in 4 years) : but oops, now it’s too small!! lets add a hundreth of a day … Read more
It’s safer to use Date objects for datetime stuff, e.g. isLeap = new Date(year, 1, 29).getMonth() == 1 Since people keep asking about how exactly this works, it has to do with how JS calculates the date value from year-month-day (details here). Basically, it first calculates the first of the month and then adds N … Read more
Most efficient leap year test: if ((year & 3) == 0 && ((year % 25) != 0 || (year & 15) == 0)) { /* leap year */ } This code is valid in C, C++, C#, Java, and many other C-like languages. The code utilizes a single TRUE/FALSE expression that consists of three separate … Read more
The correct implementation is: public static boolean isLeapYear(int year) { Calendar cal = Calendar.getInstance(); cal.set(Calendar.YEAR, year); return cal.getActualMaximum(Calendar.DAY_OF_YEAR) > 365; } But if you are going to reinvent this wheel then: public static boolean isLeapYear(int year) { if (year % 4 != 0) { return false; } else if (year % 400 == 0) { … Read more