所以我试图在java中创建一个简单的计算器,但不知道如何获取输入操作.该程序要求两个数字,然后要求运算符.当我运行它直到我进入操作类型时,一切正常.编译器只是跳过它,甚至不让我输入任何内容.有线索吗？

Scanner keyboard = new Scanner(System.in);

double x;
double y;
double sum =0;
double minus =0;
double times =0;
double divide =0;
double nCr =0;
String Op;

System.out.print("X: ");
x = keyboard.nextDouble();
System.out.print("Y: ");
y = keyboard.nextDouble();
System.out.print("Op: ");
Op = keyboard.nextLine();

解决方法:

尝试：

System.out.print("X: ");
x = Double.parseDouble(keyboard.nextLine());
System.out.print("Y: ");
y = Double.parseDouble(keyboard.nextLine());
System.out.print("Op: ");
op = keyboard.nextLine();

这将解析您输入的数字为Double并读取回车符.

如果您想要错误控制(如果用户输入字符串而不是数字)：

Boolean check = true;

do
{
    try
    {
        //The code to read numbers
    }
    catch(InputMismatchException e)
    {
        System.err.println("Please enter a number!");
        check = false;
    }
}
while(check == false)