Java有效日期閏年

Question

這是我更新的代碼，它仍然無效。 它返回2月29日的所有案例的日期，如果它是閏年，它應該只返回一天，如果它不是閏年，則應返回1。

public int checkDay (int day)
{
    // For months with 30 days.
    if ((month == 4 || month == 6 || month == 9 || month == 11) && day <= 30)
        return day;
    // For months with 31 days.
    if ((month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12) && (day <= 31))
        return day;
    // For leap years.
    // If February 29th...
    if (month == 2 && day == 29)
    {
        // Check if year is a leap year.
        if ((year%4 == 0 && year%100!=0) || year%400 == 0) 
        {
            // If year is a leap year return day as 29th.
            return day; 
        }
        // If not a leap year, return day as 1st.
        else return 1;
    }
    // If Date if February 1st through 28th return day, as it is valid.
    if (month == 2 && (day >= 1 && day <= 28))
        return day;
// Return day as 1st for all other cases.
return 1;
}

Answer 1

試試GregorianCalendar http://docs.oracle.com/javase/6/docs/api/java/util/GregorianCalendar.html

GregorianCalendar gc = new GregorianCalendar();
if (gc.isLeapYear(year) ) 

Answer 2

嘗試將代碼更改為

if (year%4==0&&(year%100!=0&&year%400==0)) 

Answer 3

if ((year%4==0 && year%100!=0) || year%400==0)

這解決了你的問題，你的邏輯是假的:)

嘗試此代碼：如果返回的布爾值為false，則可以將日期設置為1，因為日期無效。

public bool checkDay (int day, int month, int year){
    bool valid = false;
    if(day >=1){
        // For months with 30 days.
        if ((month == 4 || month == 6 || month == 9 || month == 11) && day <= 30){
            valid = true;
        }
        // For months with 31 days.
        if ((month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12) && day <= 31){
            valid = true;
        }
        // For February.
        if (month == 2)
        {
            if(day <= 28){
                valid = true;
            } else if(day == 29){
                if ((year%4 == 0 && year%100!=0) || year%400 == 0){
                    valid = true;
                } //else invalid
            }
        }
    } //else date is not valid
    return valid;
}

更好的做法是在每個方法中只有一個return語句。 這使得更容易理解代碼，並通過它來調試它並找到可能的錯誤。 如果您有任何問題，請隨時提出。