嘗試將字符串解析爲int時出錯

Question

在此Java程序中，用戶應猜測1到100之間的數字，然後如果按S它會顯示嘗試的摘要。問題是我正在輸入字符串並將其轉換爲數字，以便將其與範圍進行比較，但是我還需要能夠將該字符串用作菜單輸入。 更新如何在用戶猜測正確後讓程序回到菜單選項。用戶贏得打完，我想爲顯示其可以通過使用S-

這裏以其他方式訪問的總結報告中的問題，是我的代碼

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

您應該首先檢查S/Q值，然後將字符串解析爲一個整數。如果您發現NumberFormatException（拋出Integer.parseInt()），則可以確定輸入是否爲有效值。我會做這樣的事情：

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."); 
    } 
} 

有了這個算法（我敢肯定，這可以優化），你對下列案件：

• 用戶輸入的S/
• 用戶輸入q/Q
• 用戶進入非有效值（非數字）
• 用戶進入非有效數（小於0或大於100）
• 用戶ENT一個有效的號碼

Answer 2

要檢查一個字符串是否是一個整數，只是嘗試將其解析爲一個整數，如果引發異常，則它不是整數。

參見：

http://bytes.com/topic/java/answers/541928-check-if-input-integer

String input = .... 
try { 
    int x = Integer.parseInt(input); 
    System.out.println(x); 
} 
catch(NumberFormatException nFE) { 
    System.out.println("Not an Integer"); 
} 

Answer 3

的Integer.parseInt（命令）會給你NumberFormatException如果該字符串是無效的。如果用戶輸入無法解析爲int值的'S'或'E'，那麼在代碼中可能會出現這種情況。

我修改了你的代碼。檢查驗證碼：

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; 
      if(!command.equals("S") && !command.equals("E")) { 
      // Only then parse the command to string 

      int number = Integer.parseInt(command); 
      if (number >= 0 && number <= 100) { 
      if(randomInt == number){ 

Answer 4

你試圖進入String轉換爲int你檢查它的轉義序列（S或Q）前。

嘗試重新安排您的if聲明以檢查S和Q，然後嘗試將該值轉換爲int。

我會還建議您纏繞Integer.parseInt調用（它的後續，依賴代碼）在try-catch塊，這樣你就可以提供錯誤的語句給用戶，如果他們中的任何類型，它是不是一個int