java - 尝试将字符串解析为 int 时出错

Question

在这个 Java 程序中，用户应该猜测 1 到 100 之间的数字，然后如果您按 S，它会显示尝试的摘要。问题是我正在获取输入字符串并将其转换为数字，以便我可以将其与范围进行比较，但随后我还需要能够使用该字符串作为菜单输入。更新在用户猜测正确后，如何使程序返回到菜单选项。因此，在用户获胜后，我希望问题能够显示摘要报告，该报告可以通过使用 S

进行访问

这是我的代码

public class GuessingGame {
  public static void main(String[] args) {


    // Display list of commands
                System.out.println("*************************");
                System.out.println("The Guessing Game-inator");
                System.out.println("*************************");  
                System.out.println("Your opponent has guessed a number!");
                System.out.println("Enter a NUMBER at the prompt to guess.");
                System.out.println("Enter [S] at the prompt to display the summary report.");
                System.out.println("Enter [Q] at the prompt to Quit.");
                System.out.print("> ");


    // Read and execute commands
    while (true) {

      // Prompt user to enter a command
      SimpleIO.prompt("Enter command (NUMBER, S, or Q): ");
      String command = SimpleIO.readLine().trim();

      // Determine whether command is "E", "S", "Q", or
      // illegal; execute command if legal.
      int tries = 0;
      int round = 0;
      int randomInt = 0;
      int number = Integer.parseInt(command);
      if (number >= 0 && number <= 100) {
        if(randomInt == number){

                System.out.println("Congratulations! You have guessed correctly." +
                                " Summary below");
                round++;
        }
        else if(randomInt < number)
        {
                System.out.println("your guess is TOO HIGH. Guess again or enter Q to Quit");
                tries++;
        }      
        else if(randomInt > number){
                System.out.println("your guess is TOO LOW. Guess again or enter Q to Quit");
                tries++;
        }

      } else if (command.equalsIgnoreCase("s")) {
         // System.out.println("Round        Guesses");
         // System.out.println("-------------------------");
        //  System.out.println(round + "" + tries);



      } else if (command.equalsIgnoreCase("q")) {
        // Command is "q". Terminate program.
        return;

      } else {
        // Command is illegal. Display error message.
        System.out.println("Command was not recognized; " +
                           "please enter only E, S, or q.");
      }

      System.out.println();
    }
  }
}

Answer 1

您应该首先检查 S/Q 值，然后将字符串解析为整数。如果您捕获 NumberFormatException(由 Integer.parseInt() 引发)，您可以确定输入是否为有效值。我会做类似的事情:

if ("s".equalsIgnoreCase(command)) {
    // Print summary
} else if ("q".equalsIgnoreCase(command)) {
    // Command is "q". Terminate program.
    return;
} else {
    try {
        Integer number = Integer.parseInt(command);
        if(number < 0 || number > 100){
            System.out.println("Please provide a value between 0 and 100");
        } else if(randomInt == number){
            System.out.println("Congratulations! You have guessed correctly." +
                        " Summary below");
            round++;
        } else if(randomInt < number) {
            System.out.println("your guess is TOO HIGH. Guess again or enter Q to Quit");
                 tries++;
        } else if(randomInt > number) {
            System.out.println("your guess is TOO LOW. Guess again or enter Q to Quit");
            tries++;
        }
    } catch (NumberFormatException nfe) {
        // Command is illegal. Display error message.
        System.out.println("Command was not recognized; " +
                       "please enter only a number, S, or q.");
    }
}

使用这个算法(我确信它可以优化)，您可以处理以下情况:

• 用户输入 s/S
• 用户输入 q/Q
• 用户输入了无效值(不是数字)
• 用户输入了无效数字(小于 0 或大于 100)
• 用户输入有效号码