TL;博士

Question

我正在做一个项目，包括：'编写一个提示日期（月，日，年）的程序，并报告该日期的星期几。有可能有助于知道1601年1月1日是星期一。'。这是一本“构建Java程序 - 回归基础知识方法，第2版”的练习，这本书是我自学Java自学的书。任何反馈都受到高度赞赏，但我确实要求你解释为什么你会以某种方式做某事。谢谢！

所以，我的问题是，对于接近1600年的日期，它给出了正确的一天（我相信），最近几天也是如此，他们有三天的偏移（至少我是选中）。为什么会发生这种情况，我该如何解决？谢谢！

我的代码：

// finds the day of the week of the given date
public static String dayFinder(int month, int day, int year) {
    // handle invalid input
    if (month > 12 || month < 1 || day > 31 || day < 1) {
        throw new IllegalArgumentException("Month must be between "
                + "1 and 12 and Day must be between 1 and 31.");
    }

    // convert to "absolute" day, covering day and month
    int absoluteDay = monthToDay(month, day, year);

    // convert year to days and add to "absolute" day
    absoluteDay += yearToDay(year);

    if (absoluteDay % 7 == 1) {
        return "Monday";
    } else if (absoluteDay % 7 == 2) {
        return "Tuesday";
    } else if (absoluteDay % 7 == 3) {
        return "Wednesday";
    } else if (absoluteDay % 7 == 4) {
        return "Thursday";
    } else if (absoluteDay % 7 == 5) {
        return "Friday";
    } else if (absoluteDay % 7 == 6) {
        return "Saturday";
    } else { // absoluteDay % 7 == 0
        return "Sunday";
    }
}

// calculates the number of days present in a given
// date since the beginning of the year
public static int monthToDay(int month, int day, int year) {

    // convert to "absolute" day
    int absoluteDay = 0, daysTo31 = 0;

    // iterate through months
    for (int i = 0, loopMonth = month; i < month; i++) {
        if (loopMonth == 1 || loopMonth == 3 || loopMonth == 5 
                || loopMonth == 7 || loopMonth == 8 || loopMonth == 10
                || loopMonth == 12) {
            absoluteDay += 31;
            daysTo31 = 0;
        } else if (loopMonth == 2) {
            if (year % 4 != 0) {
                absoluteDay += 28;
                daysTo31 = 3;
            } else { // leap year
                absoluteDay += 29;
                daysTo31 = 2;
            }
        } else { // month = 4, 6, 9 or 10
            absoluteDay += 30;
            daysTo31 = 1;
        }
        loopMonth--;
    }

    // adjust to specific day
    absoluteDay -= (31 - day - daysTo31);       
    return absoluteDay;
}

// calculates the number of days between 
// (the beginning of) the given year and
// (the beginning of) the reference year 1601
public static int yearToDay(int year) {

    // convert to "absolute" day
    int absoluteDay = 0;
    year -= 1601; 

    // iterate through years
    for (int i = 0, loopYear = year; i < year; i++) {
        if (loopYear % 4 != 0) {
            absoluteDay += 365;
        } else { // leap year
            absoluteDay += 366;
        }
        loopYear--;
    }

    return absoluteDay;
}

// 1604年（MDCIV）是从周四开始的闰年

Answer 1

你的问题可能与闰年有关。

由于维基百科：

Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100; the centurial years that are exactly divisible by 400 are still leap years. For example, the year 1900 is not a leap year; the year 2000 is a leap year. [link]

这就是为什么你有三天太多（1700,1800,1900）。

Answer 2

Java.util日期/日历功能充满了疯狂。我认为Java7已经改进了一点（我听说过，但没有调查过），但我推荐第三方库Joda-Time。做你想和Joda做的事情是：

 DateTime anyDateTime = new DateTime(
     1800, //year
     4,    //month
     19,   //day
     0,    //hour
     0);   // minutes
System.out.println("DOW = " + anyDateTime.dayOfWeek().getAsText());

打印“DOW = Saturday” 我意识到你的目标是学习Java，但并不总是需要重新发明轮子。

Answer 3

TL;博士

int dayOfWeekNumber = 
    LocalDate.of( 2016 , 12 , 22 )  // A date-only object, no time-of-day nor time zone.
             .getDayOfWeek()        // `DayOfWeek` enum object.
             .getValue() ;          // 1-7 for Monday-Sunday.

java.time

问题实际上是关于算法的。但是，仅供参考，这个功能是使用java.time类构建到Java中的。

`LocalDate`

LocalDate类表示没有时间且没有时区的仅限日期的值。

LocalDate ld = LocalDate.of( 2016 , 12 , 22 );

`DayOfWeek`

DayOfWeek枚举为一周中的每一天定义七个对象。

DayOfWeek dow = ld.getDayOfWeek() ;

通常最好使用DayOfWeek对象。但如果你绝对需要一个整数，你可以要求一个。星期一至星期日每ISO 8601编号为1-7。

int dowNumber = dow.getValue();

关于java.time

java.time框架内置于Java 8及更高版本中。这些类取代了麻烦的旧日期时间类，例如java.util.Date，.Calendar和＆amp; java.text.SimpleDateFormat

现在位于Joda-Time的maintenance mode项目建议迁移到java.time。

要了解详情，请参阅Oracle Tutorial。并搜索Stack Overflow以获取许多示例和解释。

大部分java.time功能都被反向移植到Java 6＆amp; ThreeTen-Backport中的7，并进一步适应Android中的ThreeTenABP（见How to use…）。

ThreeTen-Extra项目使用其他类扩展java.time。该项目是未来可能添加到java.time的试验场。您可以在此处找到一些有用的课程，例如Interval，YearWeek，YearQuarter和more。

Answer 4

import java.util.*;

此代码适用于任何日期。

也许会帮助某人。

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner console = new Scanner(System.in);
    System.out.println("Give me a special date !!!");
    System.out.print("Month: ");
    int month = console.nextInt();
    System.out.print("Day: ");
    int day = console.nextInt();
    System.out.print("Year:");
    int year = console.nextInt();
    String date = dayFinder(month, day, year);
    System.out.println("It is the date " + month + "/" + date + "/" + year);

    console.close();
}

public static String dayFinder(int month, int day, int year) {
    // handle invalid input
    if (month > 12 || month < 1 || day > 31 || day < 1) {
        throw new IllegalArgumentException("Month must be between "
                + "1 and 12 and day must be between 1 and 31");
    } else {
        // convert to "absolute" day, covering day and month
        int absoluteDay = monthToDay(month, day, year);
        
        // convert year to days and add to "absolute" day
        absoluteDay += yearToDay(year);
        
        if (absoluteDay % 7 == 2) {
            return "Monday";
        } else if (absoluteDay % 7 == 3) {
            return "Tuesday";
        } else if (absoluteDay % 7 == 4) {
            return "Wednesday";
        } else if (absoluteDay % 7 == 5) {
            return "Thursday";
        } else if (absoluteDay % 7 == 6) {
            return "Friday";
        } else if (absoluteDay % 7 == 0) {
            return "Saturday";
        } else { // absoluteDay % 7 == 1
            return "Sunday";
        }
        
    }
}

// calculates the number of days between
// (the beginning of) the given year
// (the beginning of) the reference year 1601;
public static int yearToDay(int years) {
    
    // covert to "absolute" day;
    int absoluteDay = 0;
    int leapYears = 0;
    
    // iterate through years;
    for (int i = 0; i < years; i++) {
        if (((i % 4) == 0) && ((i % 100) != 0)) {
            leapYears +=1;
        } else if (i % 400 == 0) {
            leapYears ++;
        }
    }
    absoluteDay = (leapYears * 366) + (((years - 1) - leapYears) * 365);
    return absoluteDay;
} 

// Calculates the numbers of days present in a given
// date since the beginning of the year;
public static int monthToDay (int month, int day, int year) {
    // convert to absolute day;
    int absoluteDay = 0;
    
    // iterate through months
    for (int i = 1; i < month; i++) {
        if ((i == 4) || (i == 6) || (i == 9) || (i == 11)) { // 30 day
            absoluteDay += 30;
        } else if (i == 2) {
            if ((year % 4 == 0) && (year % 100 != 0)) {
                absoluteDay += 29;
            } else if (year % 400 == 0) {
                absoluteDay += 29;
            } else {
                absoluteDay += 28;
            }
        } else {
            absoluteDay += 31; // 1,3,5,7,8,10,12 months
        }
    }
    return absoluteDay + day;
}

}

Java中的工作日查找器

4 个答案:

TL;博士

java.time

`LocalDate`

`DayOfWeek`

关于java.time