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我打算用 Java 创建一个程序来确定输入的年份是否是闰年和有效日期。取那个日期，我希望将其转换为完整的书面名称（2013 年 4 月 2 日 = 2013 年 4 月 2 日），然后确定该日期在一年中的编号（2013 年 4 月 2 日 = 第 92 天）。

有很多程序可以做一个或另一个，但我正在学习方法/获得关于如何将它们组合在一起的想法，如果可能的话。

为了检查闰年，这就是我使用的：

public class LeapYear {
    public static void main (String[] args) {
    int theYear;
    System.out.print("Enter the year: ");
    theYear = Console.in.readInt();
    if (theYear < 100) {
        if (theYear > 40) {
        theYear = theYear + 1900;
        }
        else {
        theYear = theYear + 2000;
        }
    }
    if (theYear % 4 == 0) {
        if (theYear % 100 != 0) {
        System.out.println(theYear + " is a leap year.");
        }
        else if (theYear % 400 == 0) {
        System.out.println(theYear + " is a leap year.");
        }
        else {
        System.out.println(theYear + " is not a leap year.");
        }
    }
    else {
        System.out.println(theYear + " is not a leap year.");
    }
    }

}

我意识到我需要对其进行一些更改以读取一年中的月份和日期，但对于这种情况，我只是检查年份。我怎样才能输入相同的日期并将其转换为完整的书面姓名？我是否必须创建一个 if 语句，例如：

if (theMonth == 4){
     System.out.println("April");
         if (theDay == 2){
          System.out.print(" 2nd, " + theYear + ".");
         }
    }

这似乎是很多硬编码的工作。我正在尝试限制所需的硬编码数量，以便获得类似：

Output:
 Valid entry (4/2/2013).
 It is April 2nd, 2013.
 It is not a leap year.
 It is day 92.

如果出现错误，例如无效日期，我希望程序重新提示用户，直到收到有效条目，而不是必须运行程序（在编写“退出”时结束程序）。

我想我可能只是为主要方法（获取日期）创建不同的类，检查它是否是闰年、转换方法，也许还有验证方法。

4

3 回答 3

1 
public void testFormatDate() throws ParseException {
        final String[] suffixes =
                //    0     1     2     3     4     5     6     7     8     9
                { "th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th",
                        //    10    11    12    13    14    15    16    17    18    19
                        "th", "th", "th", "th", "th", "th", "th", "th", "th", "th",
                        //    20    21    22    23    24    25    26    27    28    29
                        "th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th",
                        //    30    31
                        "th", "st" };
        SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy");
        SimpleDateFormat odf = new SimpleDateFormat("MMMM"); // Gives month name.
        Calendar dayEntered = new GregorianCalendar();
        dayEntered.setTime(sdf.parse("04/02/2013"));
        System.err.println("You chose date: " + odf.format(dayEntered.getTime()) + " " + dayEntered.get(Calendar.DAY_OF_MONTH) + suffixes[dayEntered.get(Calendar.DAY_OF_MONTH)]
        + " " + dayEntered.get(Calendar.YEAR));
        System.err.println("This is " + (((GregorianCalendar)dayEntered).isLeapYear(dayEntered.get(Calendar.YEAR)) ? "" : "not ") + "a leap year.");

        System.err.println("This is day: " + dayEntered.get(Calendar.DAY_OF_YEAR));
    }
于 2013-04-02T23:34:43.317 回答
1

这是一种方法。问题是没有办法SimpleDateFormatter打印当天的序数值。我无耻地从这里getDayOfMonthSuffix偷走了方法。

public static void main(String[] args) {
    final String input = "4/2/2013";
    final SimpleDateFormat parser = new SimpleDateFormat("MM/dd/yyyy");
    final SimpleDateFormat formatter1 = new SimpleDateFormat("MMMM");
    final GregorianCalendar cal = (GregorianCalendar) GregorianCalendar.getInstance();
    final Date date;
    try {
        date = parser.parse(input);
        cal.setTime(date);
    } catch (ParseException ex) {
        System.out.println("Invalid input \"" + input + "\".");
        return;
    }
    if (cal.isLeapYear(cal.get(Calendar.YEAR))) {
        System.out.println("The year is a leap year");
    } else {
        System.out.println("The year is not a leap year");
    }
    System.out.println("The day of the year is " + cal.get(GregorianCalendar.DAY_OF_YEAR));
    final int dayOfMonth = cal.get(GregorianCalendar.DAY_OF_MONTH);
    System.out.println("The date is " +
            formatter1.format(date) +
            " " +
            dayOfMonth +
            getDayOfMonthSuffix(dayOfMonth) +
            ", " +
            cal.get(GregorianCalendar.YEAR));
}

static String getDayOfMonthSuffix(final int n) {
    if (n < 1 || n > 31) {
        throw new IllegalArgumentException("illegal day of month: " + n);
    }
    if (n >= 11 && n <= 13) {
        return "th";
    }
    switch (n % 10) {
        case 1:
            return "st";
        case 2:
            return "nd";
        case 3:
            return "rd";
        default:
            return "th";
    }
}
于 2013-04-02T23:38:26.073 回答
0

正如您所说，由于您需要更多的框架 - 来吧：

package com.yours

import java.io.Console;
import java.util.Calendar;

public class DoStuffWithADate {

    private Calendar parsedDate;

    public static void main(String[] args) {
        DoStuffWithADate soundsNaughty = new DoStuffWithADate();
        System.out.println("Enter a date my friend.  You should use the format: (MM/dd/yyyy)");
        Console theConsole = System.console();
        String enteredDate = theConsole.readLine();

        if (soundsNaughty.isValidDate(enteredDate)) {
            soundsNaughty.writeTheDateInNewFormat();
            soundsNaughty.writeIfItsALeapYear();
            soundsNaughty.writeTheDayOfYearItIs();
        }
    }

    private boolean isValidDate(String enteredDate) {
        //logic goes here.
        parsedDate = null;// if it's valid set the parsed Calendar object up.
        return true;
    }

    private void writeTheDateInNewFormat() {
        System.out.println("The new date format is: ");
    }

    private void writeIfItsALeapYear() {
        System.out.println("The year is a leap year.");
    }

    private void writeTheDayOfYearItIs() {
        System.out.println("The day of year is: ");
    }


}
于 2013-04-03T00:05:53.983 回答