如何通过不获取另一种情况的值来制作if语句在switch情况下？

Question

现在我有一个问题，我想以此来运行我的代码，以便检查案例2的februay（如果year为true），而不是运行execute if语句（如果它为false），它将变为默认值，但是现在转到case 3语句如何解决不是转到case 3而是转到默认值的问题？

//这是我的代码

public static void main(String[] args) {

    getDaysInMonth(1, 2018);

    System.out.println(getDaysInMonth(2, 2018));

}

public static boolean isLeapYear(int year) {

    if(year < 1 && year > 9999) {

        return false;

    } else {

        if(year % 4 == 0) {

            return true;

        } else if(year % 100 == 0) {

            return false;

        } else if(year % 400 == 0) {

            return true;

        } else {

            return false;

        }

    }         

}

public static int getDaysInMonth(int month, int year) {

    if((month < 1 || month > 12) && (year < 1 || year > 9999)) {

        return -1;

    }

    isLeapYear(year);

    int days;

    switch(month) {

        case 1:

            days = 31;

            break;

        case 2:

            if(year == 1) {

                days = 29;

                break;

            } 

        case 3:

            days = 31;

            break;

        case 4:

            days = 30;

            break;

        case 5:

            days = 31;

            break;

        case 6:

            days = 30;

            break;

        case 7:

            days = 31;

            break;

        case 8:

            days = 31;

            break;

        case 9:

            days = 30;

            break;

        case 10:

            days = 31;

            break;

        case 11:

            days = 30;

            break;

        case 12:

            days = 31;

            break;                

        default:

            days = 28;

            break;             

    }

      return days; 

}

Answer 1

发生这种情况是因为您break在IF范围内->如果年份不是1，则继续第3种情况。您可以使用if else子句：&nbsp; &nbsp; case 2:&nbsp; &nbsp; &nbsp; &nbsp; if(year == 1) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; days = 29;&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; days = 28;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; break;但是我要提醒您，year != 1这并不意味着它的2月获得28天，也许您想要这样：&nbsp; &nbsp; case 2:&nbsp; &nbsp; &nbsp; &nbsp; if(!isLeapYear(year)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; days = 29;&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; days = 28;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; break;

Answer 2

除了关于if语句和的switch问题外，首先使用leap年检查的方法还需要解决一些问题。这是下面原始代码的开头：public static boolean isLeapYear(int year) {&nbsp; &nbsp; if(year < 1 && year > 9999) {&nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; ...这种情况year < 1 && year > 9999永远不会发生，因为year不能同时小于1和大于9999，所以这是多余的。除此之外，用于确定年份是否为a年的算法如下（以纯英语显示）：看看该年份是否可以被4整除。如果不是，则该年份不能是a年（返回false）。在这里，一年可以平均除以4（在步骤1中确定）。如果不能被100整除，则为is年（返回true）。在这里，年份可以被4除以100，也可以除以100。如果该年份也可以被400除以，那么它就是leap年。否则不是。将以上所有考虑因素放入代码中，可以使其更具可读性：public static boolean isLeapYear(int year) {&nbsp; &nbsp; if (year % 4 != 0) {&nbsp; &nbsp; &nbsp; &nbsp; // year is not evenly divisible by 4 (it has a remainder, can't be a leap year).&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; }&nbsp; &nbsp; // year is evenly divisible by 4&nbsp; &nbsp; if (year % 100 != 0) {&nbsp; &nbsp; &nbsp; &nbsp; // divisible by 4 and not 100, it's a leap year&nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; }&nbsp; &nbsp; // divisible by 4 and also 100&nbsp; &nbsp; if (year % 400 != 0) {&nbsp; &nbsp; &nbsp; &nbsp; // divisible by 4, 100 and not by 400&nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; }&nbsp; &nbsp; // divisible by 4, 100 and 400&nbsp; &nbsp; return true;}考虑到您的getDaysInMonth方法，我们有以下情况，其中1 =一月，2 =二月，依此类推：30天：9月（9），4月（4），6月（6），11月（11）31天：除2月（2）以外的所有其他日子（a年28、29）public static int getDaysInMonth(int month, int year) {&nbsp; &nbsp; switch (month) {&nbsp; &nbsp; &nbsp; &nbsp; case 9:&nbsp; &nbsp; &nbsp; &nbsp; case 4:&nbsp; &nbsp; &nbsp; &nbsp; case 6:&nbsp; &nbsp; &nbsp; &nbsp; case 11:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 30;&nbsp; &nbsp; &nbsp; &nbsp; case 2:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (isLeapYear(year)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 29;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 28;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; default:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 31;&nbsp; &nbsp; }}关于参数的验证，如果您正在接收用户输入，则应在用于计算的方法之外对所有参数进行验证（传递已被验证的方法输入，而不是直接在其中进行验证）。