字符串到整數輸入驗證轉換

我在寫一個類來驗證來自另一個類的輸入。似乎一切似乎工作，但我的驗證將不會接受確切的高範圍價值。我不知道爲什麼它不接受它 - 例如，如果highRange是10，用戶輸入10，！（10> 10）。如果你們可以關心我的代碼，我將不勝感激！字符串到整數輸入驗證轉換

import javax.swing.JOptionPane; 

public class InputVerification { 
private String input; 
private int lowRange; 
private int highRange; 
private int invalidNum; 

public InputVerification(String input, int lowRange, int highRange, int invalidNum) { 
    this.input = input; 
    this.lowRange = lowRange; 
    this.highRange = highRange; 
    this.invalidNum = invalidNum; 
} 

public int intVerify() { 

    String userInput = null; 
    int intInput = 0; 
    do { 
     do { 
      try { 

       /** 
       * handles any text entered instead of numbers enter -2 if 
       * you don't need an invalid number 
       */ 

       userInput = JOptionPane.showInputDialog(this.input); 

       intInput = Integer.parseInt(userInput); 

       if ((intInput > highRange || intInput < lowRange) && !(userInput.matches("[a-zA-Z]+"))) { 

        JOptionPane.showMessageDialog(null, 
          "Error! Please pick a number between " + lowRange + "-" + highRange + "."); 

       } 

      } catch (NumberFormatException e) { 

        JOptionPane.showMessageDialog(null, 
          "Error! Please pick a number between " + lowRange + "-" + highRange + "."); 

      } 

     } while (!userInput.matches("^[0-9]")); 

     if ((intInput > highRange || intInput < lowRange)) { 

      /** 
      * 
      * sends an error message if the number is higher than 100 or 
      * lower than 1 as long as the input was not text 
      * 
      */ 

      JOptionPane.showMessageDialog(null, 
        "Error! Please pick a number between " + lowRange + "-" + highRange + "."); 

     } 

     if (invalidNum != -2 && intInput == invalidNum) { 
      JOptionPane.showMessageDialog(null, "Error! Please pick a number between " + lowRange + "-" + highRange 
        + " that is not " + invalidNum + "."); 
     } 

    } while ((intInput > highRange || intInput < lowRange || intInput == invalidNum) 
      && !(userInput.matches("[a-zA-Z]+"))); 

    return intInput; 

    } 
}

來源

2017-01-29 Greg Smith

您應該定義一個布爾函數來驗證整數並從GUI中使用它。 – Aubin

歡迎來到堆棧溢出！它看起來像你需要學習使用調試器。請幫助一些[互補調試技術]（https://ericlippert.com/2014/03/05/how-to-debug-small-programs/）。如果您之後仍然有問題，請隨時返回更多詳情。 –

@JoeC謝謝，我會檢查出來！ –

import javax.swing.JOptionPane; 

public class InputVerification { 

    public static Integer parseInt(String value, int min, int max) { 
     try { 
     final int iValue = Integer.parseInt(value); 
     if((min <= iValue) && (iValue <= max)) { 
      return iValue; 
     } 
     } 
     catch(final Throwable t) {/**/} 
     return null; 
    } 

    public static int getInteger(String message, int lowRange, int highRange) { 
     Integer intValue = null; 
     do { 
     final String userInput = JOptionPane.showInputDialog(message); 
     intValue = parseInt(userInput, lowRange, highRange); 
     if(intValue == null) { 
      JOptionPane.showMessageDialog(null, 
       "Error! Please pick a number in [" + 
       lowRange + ".." + highRange + "]"); 
     } 
     } while(intValue == null); 
     return intValue.intValue(); 
    } 

    public static void main(String[] args) { 
     getInteger("Hello!", 0, 10); 
    } 
}

來源

2017-01-29 08:48:44 Aubin