在用户输入错误（超出范围）值后如何提示用户输入正确值？

Question

System.out.println("Enter the first test score:");

double test1 = input.nextInt();

if (!(test1 >= 0  && test1 <= 100))
System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");

System.out.println("Enter the second test score:");

double test2 = input.nextInt();

if (!(test2 >= 0  && test2 <= 100))
System.out.println("This is out of the acceptable range, please enter a number between 0 and 100 .");

如果我提示用户输入 test1 的值及其在特定范围内的 nor，我如何再次提示他输入正确的值？现在如果我运行并输入一个错误的值“这超出了可接受的范围，请输入一个介于 0 和 100 之间的数字”消息会出现，但它会直接进入 test2。我试着这样做

System.out.println("Enter the first test score:");

double test1 = input.nextInt();

if (!(test1 >= 0  && test1 <= 100))
System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");
double test1 = input.nextInt();

但我收到一条错误消息。 我是否必须使用循环，如果是这样怎么办？

Answer 1

那是因为你声明了两次test1。

double test1 = input.nextInt();

if (!(test1 >= 0  && test1 <= 100))
System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");
-->  double test1 = input.nextInt(); //Look at this line

只需尝试以下操作： 双 test1 = input.nextInt();

if (!(test1 >= 0  && test1 <= 100))
System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");
test1 = input.nextInt(); 

对于更简洁的代码，请使用以下方法：

public double getValue(){
        double test1=input.nextInt();
        if (!(test1 >= 0  && test1 <= 100)){
            return getValue();
        }else{
            return test1;
        }
    }

除非根据!(test1 >= 0 && test1 <= 100) 条件输入正确，否则上述代码将继续接受用户输入。

Answer 2

首先，您要声明 test1 两次，其次 - 您可以使用循环

    double test1 = input.nextInt();
    while(!(test1 >= 0 && test1 <= 100))
    {
        System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");
        /*double */test1 = input.nextInt();
    }

Answer 3

试试类似的东西

double test1 = input.nextInt();
while(!(test1 >= 0 && test1 <= 100)) {
     System.out.println("Enter a valid number"); 
     test1.nextInt();
}

这个想法是提示用户直到他输入一个有效的输入。如果您使用简单的 if 语句或固定的 for 循环，您最终只会要求固定数量的正确值。

您应该记住的另一件事是：

if (!(test1 >= 0  && test1 <= 100))
System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");
double test1 = input.nextInt(); //There is no need to declare test1 again. Remove double

将被编译为

if (!(test1 >= 0  && test1 <= 100))
     System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");

test1 = input.nextInt(); //This is not part of the if!!!!

所以你必须使用括号，如

if (!(test1 >= 0  && test1 <= 100)) {
    System.out.println("This is out of the acceptable range, please enter a  number between 0 and 100 .");
    test1 = input.nextInt(); 
}