Java：接受输入为字符串或整数

Question

我正在尝试制作一个简单的偶数或奇数程序。 我希望它继续运行直到用户输入“ q”。 但是我很难接受'q'作为字符串。

import java.util.Scanner;

class EvenOrOdd {

    public static void main(String[] args) {
        Scanner myScanner = new Scanner(System.in);

        System.out.println("Welcome to my program that checks if a number is even or odd.");

        while (true) {
            System.out.println();
            System.out.print("Please type number in a number ['q' to quit]: ");

            int number;
            String quit;
            try {
                number = myScanner.nextInt();
            } finally {
                quit = myScanner.nextLine();
            }

            if (quit.equals("q")) {
                break;
            } else if (number % 2 == 0) {
                System.out.println(number + " is Even.");
            } else {
                System.out.println(number + " is Odd.");
            }
        }
    }
}

当我输入数字时，该程序运行正常，但是当我输入“ q”时，控制台将引发错误：

Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at EvenOrOdd.main(EvenOrOdd.java:19)

我知道这对你们中的许多人来说可能很容易，但是我刚拿起一本Java书籍，正在尝试完成任务。 任何帮助将不胜感激！

Answer 1

从Scanner中获取一个字符串，检查是否为'q'，如果不是，则将其转换为int，然后检查偶数或奇数。

public static void main(String[] args) {
    Scanner myScanner = new Scanner(System.in);

    System.out.println("Welcome to my program that checks if a number is even or odd.");

    while (true) {
        System.out.println();
        System.out.print("Please type number in a number ['q' to quit]: ");

        String inText = myScanner.next();

        if (inText.equals("q")){
            break;
        }
        int number = Integer.valueOf(inText);
        if (number % 2 == 0) {
            System.out.println(number + " is Even.");
        } else {
            System.out.println(number + " is Odd.");
        }
    }
}

Answer 2

你可以做这样的事情，我发现在这种情况下，布尔值对循环更好，而不是while（true）和break：

public class EvenOrOdd {
public static void main(String[] args) {
    Scanner myScanner = new Scanner(System.in);

    System.out
            .println("Welcome to my program that checks if a number is even or odd.");
    boolean enterLoop = true;
    while (enterLoop) {
        System.out.println();
        System.out.print("Please type number in a number ['q' to quit]: ");

        String scannerinput = myScanner.nextLine();
        if (scannerinput.equals("q")) {
            enterLoop = false;
        } else {
            checkNumber(scannerinput);
        }

    }
}

private static void checkNumber(String scannerinput) {
    try {
        int number = Integer.parseInt(scannerinput);
        if (number % 2 == 0) {
            System.out.println(number + " is Even.");
        } else {
            System.out.println(number + " is Odd.");
        }
    } catch (Exception e) {
        System.out.println("No Number!");
    }
}

}

Answer 3

import java.util.Scanner;

class EvenOrOdd {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.println("Welcome to my program that checks if a number is even or odd.");

        while (true) {
            System.out.print("\nPlease type number in a number ['q' to quit]: ");
            String input = scanner.next();
            if (input.equals("q")) {
                break;
            } else {
                int number = Integer.parseInt(input);
                System.out.print(number + " is ");
                System.out.print(number%2 == 0 ? "Even." : "Odd.");
            }
        }
    }
}

会的。 :)

Answer 4

除了使用myScanner.nextInt() ，还可以使用myScanner.next()获取String 。 然后，如果不是“ q”，则使用Integer.valueOf(inputString)获取int并检查其偶数/奇数。

while (true) {
    String input = myScanner.next();
    if ("q".equals(input)) {
        break;
    } else {
        int number = Integer.valueOf(input);
        if (number % 2 == 0) {
            System.out.println(number + " is Even.");
        } else {
            System.out.println(number + " is Odd.");
        }
    }
}

Answer 5

这是因为程序需要输入int类型，基本上是程序输出：请在数字中键入数字['q'退出”，然后它将到达myScanner.nextInt(); 行并且将等待输入，并且由于“ q”不是整数，因此将引发异常。

一种快速的解决方案是使用myScanner.nextLine() ，然后将字符串转换为整数，除非它等于'q'。 像这样：

import java.util.Scanner;

public class EvenOrOdd {

    public static void main(String[] args) {
        Scanner myScanner = new Scanner(System.in);

        System.out.println("Welcome to my program that checks if a number is even or odd.");

        while (true) {
            System.out.println();
            System.out.print("Please type number in a number ['q' to quit]: ");

            String string = myScanner.nextLine();
            int number = 0;

            if (string.equals("q")) {
                myScanner.close(); // Close the scanner.
                break;
            } else if ((number = toInteger(string)) == -1){  // Is the string a number, less than Integer.MAX_VALUE and greater than Integer.MIN_VALUE?
                System.out.printf("%s is not a valid integer!%n",string);
            } else if (number % 2 == 0) {
                System.out.println(number + " is Even.");
            } else {
                System.out.println(number + " is Odd.");
            }
        }
    }

    private static int toInteger(String str){
        try{
            return Math.abs(Integer.parseInt(str));
        }catch(NumberFormatException e){
            return -1;
        }
    }
}

顺便说一句，请始终关闭扫描仪，否则可能发生资源泄漏。

Answer 6

试试这种方法。 查看代码注释以获取更多详细信息。

import java.util.Scanner;

class EvenOrOdd {

public static void main(String[] args) {
    Scanner myScanner = new Scanner(System.in);

    System.out.println("Welcome to my program that checks if a number is even or odd.");

    String input=null;
    int number;
    boolean flag=true;  // loop flag
    do {
        System.out.println();
        System.out.print("Please type number in a number ['q' to quit]: ");

        // Take user input as String
        input=myScanner.nextLine();


        try 
        {
            // convert the string value to integer value
            number = Integer.parseInt(input);

            if (number % 2 == 0)
            {
                System.out.println(number + " is Even.");
            }
            else
            {
                System.out.println(number + " is Odd.");
            }
        } 
        catch (NumberFormatException nfe) 
        {
           // control goes here if input is not integer value
            if(input.equals("q"))   // exist option
                flag=false;
            else    // invalid input
                System.out.println("Invalid input, Please enter integer value or (q) to exist");


        }


    } while (flag);

}
}

Answer 7

您应该更新方法流程，实际上是在尝试通过和对字符串进行整数验证，因此首先将输入作为String并检查它是否是退出字符q，然后如果不尝试将字符串解析为int并使用Integer.parseInt(input)并测试它是奇还是偶。 如果假设该过程既不是q也不是数字（任何其他字符）而失败，那么将提示用户输入一条消息，告诉他使用有效数字或“ q”退出：

import java.util.Scanner;

class EvenOrOdd {

public static void main(String[] args) {
  Scanner myScanner = new Scanner(System.in);

  System.out.println("Welcome to my program that checks if a number is even or odd.");

  while (true) {
    System.out.println();
    System.out.print("Please type number in a number ['q' to quit]: ");

    int number;
    String input = myScanner.next();
    if (input.equals("q")) {
      break;
    } else {
      try {
        number = Integer.parseInt(input);
        if (number % 2 == 0) {
           System.out.println(number + " is Even.");
        } else {
           System.out.println(number + " is Odd.");
        }
        } catch (NumberFormatException nfe) {
           System.out.println("Enter valid number or \"q\" to quit!");
        }
      }
    }
  }
}