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值，我该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到100之间的数字”msg会显示但它会直接进入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();

但我收到错误信息。我是否必须使用循环，如果是这样的话？

Answer 1

那是因为你宣布test1两次。

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

试试以下内容： 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 .");
test1 = input.nextInt();

对于更干净的代码，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，否则以上代码将继续接受用户输入。

Answer 2

首先，你宣布test1两次，第二次 - 你可以使用循环

    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

尝试类似

的内容

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

我们的想法是提示用户 ，直到 输入有效输入。如果你使用简单的if语句或一个固定的for-loop，你最终只会要求一个固定数量的正确值。

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!!!!

所以你必须使用像

这样的括号

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(); 
}

如何在输入错误（超出范围）值后提示用户输入正确的值？

3 个答案: