Question

import java.util.HashMap;
import java.util.Map;
import java.util.Iterator;
public class Details
{
    public static void main(String [] args)
    {
        HashMap<Integer, String> hmap = new HashMap<Integer, String>();
        //Adding elements to HashMap
        hmap.put(1, "January");
        hmap.put(2, "January");
        hmap.put(3, "January");
        hmap.put(4, "January");
        hmap.put(5, "January");
        hmap.put(6, "January");
        hmap.put(7, "January");
        hmap.put(8, "January");
        hmap.put(9, "January");
        hmap.put(10, "January");
        //FOR LOOP
        System.out.println("For Loop:");
        for (Map.Entry me : hmap.entrySet()) {
          System.out.println("Key: "+ me.getKey() + " & Value: " + me.getValue());
        }

        //WHILE LOOP & ITERATOR
        System.out.println("While Loop:");
        Iterator iterator = hmap.entrySet().iterator();
        while (iterator.hasNext()) {
             Map.Entry me2 = (Map.Entry) iterator.next();
          System.out.println("Key: "+ me2.getKey() + " & Value: " + me2.getValue());
        } 
    }
}

我的输出：

For Loop:
Key: 1 & Value: January
Key: 2 & Value: January
Key: 3 & Value: January
Key: 4 & Value: January
Key: 5 & Value: January
Key: 6 & Value: January
Key: 7 & Value: January
Key: 8 & Value: January
Key: 9 & Value: January
Key: 10 & Value: January
While Loop:
Key: 1 & Value: January
Key: 2 & Value: January
Key: 3 & Value: January
Key: 4 & Value: January
Key: 5 & Value: January
Key: 6 & Value: January
Key: 7 & Value: January
Key: 8 & Value: January
Key: 9 & Value: January
Key: 10 & Value: January

我想要这个输出：

1: January 1, 1910
2: January 2, 1910
3: January 3, 1910
4: January 4, 1910
5: January 5, 1910.....

up to

41758: December 28, 2025
41759: December 29, 2025
41760: December 30, 2025

感谢您的帮助

Answer 1

你可以这样做。查看 java.time 包以获取更多有用的日期/时间类。

import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.time.temporal.ChronoUnit;


LocalDate start = LocalDate.of(1910,1,1);
LocalDate end = LocalDate.of(2025,12,31);
DateTimeFormatter fmt = DateTimeFormatter.ofPattern("MMMM d, yyyy");
int key = 1;
while (start.isBefore(end)) {
    System.out.println(key++ + " : " + start.format(fmt));
    start = start.plus(1, ChronoUnit.DAYS);
}

打印类似这样的东西。

1 : January 1, 1910
2 : January 2, 1910
3 : January 3, 1910
4 : January 4, 1910
5 : January 5, 1910
6 : January 6, 1910
7 : January 7, 1910
8 : January 8, 1910
9 : January 9, 1910
10 : January 10, 1910
11 : January 11, 1910
12 : January 12, 1910
...
...

另一方面，如果您不想使用任何导入的类来执行此操作，您可以按如下方式进行：

创建月份名称数组
创建一个每月最多天数的数组（非闰年）。
创建闰年方法（稍后解释）。

String[] monthNames = { "January", "February", "March",
        "April", "May", "June", "July", "August", "September",
        "October", "November", "December" };
int[] monthDays =
        { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int day = 1;
int month = 0; // months go from 0 to 11
int year = 1910;
int count = 0;

用于打印日期范围的简单 while 循环驱动程序。

while (!(year == 2025 && month == 11 && day == 31)) {
    count++;
    System.out.printf("%d : %s %d, %d%n", count,
            monthNames[month], day, year);

     if (day++ >= monthDays[month]) {
         // take care of February in a leap year
         if (month == 1 && day == 29 && isLeap(year)) {
             continue;
         }
         day = 1;        // reset day if past end of month.
         month++;
         if (month == 12) {
             month = 0;  // reset month and increment year
             year++;
         } 
     }
         
}

只有能被 4 整除的非世纪年和能被 400 整除的世纪年才是闰年。首先检查非世纪年份，因为它们更频繁地出现。

public static boolean isLeap(int year) {
    return year % 100 != 0 && year % 4 == 0 || year % 400 == 0;
}

Answer 2

这是一种生成请求数据的方法，正确地（希望）考虑闰年等。我会注意到运行没有与示例输出相同的最后一个数字。

如果需要将数据存储到某个数据数组中，请根据需要替换 output 方法。

    private static DateTimeFormatter FMTR = DateTimeFormatter.ofPattern("MMMM d, yyyy");
    
    public static LocalDate genNextDate(LocalDate inpDate)
    {
        return inpDate.plusDays(1);
    }
    
    
    public static void output(int cntr, LocalDate inpDate)
    {

        System.out.printf("%6d: %s%n", cntr, FMTR.format(inpDate));
    }
    
    
    public static void doIt()
    {
        final LocalDate endDate = LocalDate.of(2025, 12, 31);
        int cntr = 1;
        LocalDate date = LocalDate.of(1910, 1, 1);
        
        while (date.isBefore(endDate)) {
            output(cntr, date);
            date = genNextDate(date);
            cntr++;
        }
    }

   public static void main(String[] args)
    {
        doIt();
    }

示例输出

     1: January 1, 1910
     2: January 2, 1910
     3: January 3, 1910
     4: January 4, 1910
     5: January 5, 1910
     6: January 6, 1910
     7: January 7, 1910
     8: January 8, 1910
     9: January 9, 1910
     ...
 42364: December 26, 2025
 42365: December 27, 2025
 42366: December 28, 2025
 42367: December 29, 2025
 42368: December 30, 2025

打印日期范围从 1910 年 1 月 1 日到 2025 年 12 月 30 日

2 个答案: