如何在 Java 中检查用户输入的类型？

Question

我想知道如何对用户输入进行类型检查。这里我有一个简单的测试来检查用户输入是否在 1 到 10 之间。我想设置它以便用户也可以输入字母，主要是我可以使用输入 'q' 来退出程序。

扫描仪有可以打字检查的部分吗？我的想法是有一个if语句：如果用户输入type int continue，如果不是type int，检查是否是q退出程序，否则输出这是一个错误。以下是我到目前为止的内容，由于类型不匹配，因此在输入数字时会抛出一个表达式。

public static void main(String[] args) {
    //Create new scanner named Input
    Scanner Input = new Scanner(System.in);
    //Initialize Number as 0 to reset while loop
    int Number = 0;

    do
    {
        //Ask user to input a number between 1 and 10
        System.out.println("At anytime please press 'q' to quit the program.");
        System.out.println();
        System.out.print("Please enter a number between 1 and 10:");
        Number = Input.nextInt();

        if (Number == 'Q'|| Number == 'q')
        {
            System.out.println("You are now exiting the program");
        }
        else if (Number >= 1 && Number <= 10)
        {
            System.out.println("Your number is between 1 and 10");
        }   
        else
        {
            System.out.println("Error: The number you have entered is not between"
                    + " 1 and 10, try again");
        }
    }
    //Continue the loop while Number is not equal to Q
   while (Number != 'Q' & Number != 'q');
}

}

感谢大家的回复。我有点新，所以 try 语句对我来说是新的，但看起来它会起作用（似乎有点自我解释它的作用）。我会更多地研究它的使用并正确地实现它。

Answer 1

我会使用nextLine() 和parseInt() 来查看它是否是int：

    int Number;
    String test = "";
    boolean isNumber = false;
    ......

    test = Input.nextLine();
    try
    {
        Number = Integer.parseInt(test);
        isNumber = true;
    }
    catch(NumberFormatException e)
    {
         isNumber = false;
    }
    if(isNumber)
    {
        if (Number >= 1 && Number <= 10)
        {
            System.out.println("Your number is between 1 and 10");
        }   
        else
        {
            System.out.println("Error: The number you have entered is not between"
                + " 1 and 10, try again");
        }
    }
    else
    {
        if (test.equalsIgnoreCase("q"))
        {
            System.out.println("You are now exiting the program");
        }
    }

.........
while (!test.equalsIgnoreCase("q"));

Answer 2

试试这个

    Scanner Input = new Scanner(System.in);
    //Initialize Number as 0 to reset while loop
    String Number = "0";

    do {
        try {
            //Ask user to input a number between 1 and 10
            System.out.println("At anytime please press 'q' to quit the program.");
            System.out.println();
            System.out.print("Please enter a number between 1 and 10:");
            Number = Input.next();

            if (Number.equalsIgnoreCase("q")) {
                System.out.println("You are now exiting the program");
            } else if (Integer.valueOf(Number) >= 1 && Integer.valueOf(Number) <= 10) {
                System.out.println("Your number is " + Number);
            } else {
                System.out.println("Error: The (" + Number + ") is not between 1 and 10, try again");
            }
        } catch (NumberFormatException ne) {
            System.out.println("Error: The (" + Number + ") is not between 1 and 10, try again");
        } catch (Exception e) {
            System.out.println(e.getMessage());
        }
    } //Continue the loop while Number is not equal to Q
    while (!Number.equalsIgnoreCase("q"));

Answer 3

将您的输入作为字符串读入，以检查 Q/q 的输入。然后，您可以通过以下方式将输入字符串从输入字符串解析为整数，捕获潜在的 NumberFormatExceptions。

String input = "5"
try {
int number = Integer.parseInt(input);
} catch (NumberFormatException e) {
System.out.println("Not a number!") 
}

或类似的东西。

Answer 4

您可以通过这种方式检查它是否为int：

String inputArg = Input.next();

private boolean isInt(String inputArg){
    boolean isInt = true;
    try {
         Integer.parseInt(inputArg);
    } catch(NumberFormatException e) {
         isInt = false;
    }
    return isInt;
}

其他方式是使用regular expressions

Answer 5

找到这段代码检查一下，它对我有用。

          Scanner input = new Scanner (System.in);
          if (input.hasNextInt()) {
              int a = input.nextInt();
              System.out.println("This input is of type Integer."+a);
          }   
                   
          else if (input.hasNextLong())
            System.out.println("This input is of type Long.");
       
          else if (input.hasNextFloat())
          System.out.println("This input is of type Float.");
          
          else if (input.hasNextDouble()) 
          System.out.println("This input is of type Double."); 
           
          else if (input.hasNextBoolean())
           System.out.println("This input is of type Boolean.");  
          
          else if (input.hasNextLine())
              System.out.println("This input is of type string."); 

Answer 6

您可以将输入作为字符串获取，然后对所述字符串使用正则表达式匹配。

String txt = Input.nextLine();
if(txt.matches("[0-9]+") // We have a number
{
   // Safe to parse to number
}
else if(txt.matches("[A-Za-z]+") // We have alphabeticals
{
   // Check for various characters and act accordingly (here 'q')
}