试图循环到程序的开头

Question

如果输入错误，我不知道如何让程序重新启动。我是 java 新手，不知道如何有效地使用 if 语句或循环。import java.util.Scanner; public class TuitionRates {public static void main(String[] args) {    //variables    String name = "";    String studentType = "";    int creditNumber = 0;    double totalTF = 0.00;    double studentACTFee = 4.60;    double parkingFee = 2.00;    double tuition = 180.40;    double capitalFee1 = 0.00;    double capitalFee2 = 21.00;    double perCreditR = 187.00;    double perCreditV = 187.00;    double perCreditD = 187.00;    double perCreditM = 208.00;    double perCreditB = 268.00;    double perCreditO = 387.25;    Scanner input = new Scanner(System.in);    //Asking user for NameSystem.out.println("Welcome to the NOVA Tuition and Fees Calculator.");System.out.println("Please enter your name: ");name= input.nextLine();//Ask user to choose option    System.out.println("Please enter the type of student that your are from     the choices below:\n"            +             "R for Virginia Resident\r\n" +             "M for Military Contract Out-of-State\r\n" +             "V for Military Veterans and Dependents\r\n" +             "D for Dual Enrolled\r\n" +             "B for Business Contract Students\r\n" +             "O for Out of State Students\r\n");    studentType = input.nextLine();if (studentType.equalsIgnoreCase("R"))   {   System.out.println("Please enter the number of credits that you are    taking:");   creditNumber = input.nextInt();如果积分数超过 18，我希望我的程序重新启动 if 语句。因此，如果我输入 19，它会说重新输入积分并重新开始 if 语句。

Answer 1

您可以尝试将您的if语句包装在一个do-while循环中：

if (studentType.equalsIgnoreCase("R"))

{

    do{

       System.out.println("Please enter the number of credits that you are taking:");

       creditNumber = input.nextInt();

        if (creditNumber <= 18)

        {

            System.out.println("Tuition and fees report for " + name);          

            System.out.println("Tuition: "+ tuition);           

            System.out.println( "Capital Fee:  \t"+ capitalFee1);           

            System.out.println( "Student Activities Fee: \t "+   studentACTFee);            

            System.out.println( "Parking Infrastructure Fee: \t " +  parkingFee);           

            System.out.println("Tuition & Fees Per Credit: " +  perCreditR);            

            System.out.println("X Number of Credits: " + creditNumber);

            totalTF = creditNumber * perCreditR;            

            System.out.println("Total Tuition and Fees: \t" +  totalTF);

            System.out.println("Bodly NOVA");

        }

        else {System.out.println("Please re-enter credit Number ");}

    }while(creditNumber > 18);

}

Answer 2

那么我可以建议你重新考虑设计。但这不是你的问题——对吧？所以，如果你真的想要，你所要求的，这里就是一个休息（“穷人例外”）：

exitpoint:{

     //your code ...

     break exitpoint;

     //more code

     break exitpoint;

     //....

}

或有一些循环：

exitpoint:

while( ){

    // code....

    for(;;){

        //...

        break exitpoint;

    }

}

处理错误（如错误的用户输入）的更好方法是异常。但这也不是问题——是吗？

Answer 3

我会做这样的事情：

if (studentType.equalsIgnoreCase("R"))

{

System.out.println("Please enter the number of credits that you are 

taking:");

creditNumber = input.nextInt();

while(creditNumber > 18)

{

    System.out.println("Please re-enter the number of credits that you are 

    taking:");

    creditNumber = input.nextInt();

}

System.out.println("Tuition and fees report for " + name);          

System.out.println("Tuition: "+ tuition);           

System.out.println( "Capital Fee:  \t"+ capitalFee1);           

System.out.println( "Student Activities Fee: \t "+   studentACTFee);            

System.out.println( "Parking Infrastructure Fee: \t " +  parkingFee);           

System.out.println("Tuition & Fees Per Credit: " +  perCreditR);            

System.out.println("X Number of Credits: " + creditNumber);

totalTF = creditNumber * perCreditR;            

System.out.println("Total Tuition and Fees: \t" +  totalTF);

System.out.println("Bodly NOVA");

}

这使用了一个 while() 语句来检查初始 creditNumber 是否大于 18 并不断地重新提示用户进行新输入。然后，一旦它们提供小于或等于 18 的值，它就会执行您希望它执行的所有其他操作。请注意，我没有对此进行测试，但它应该可以工作。