2016-10-01 71 views
-1

嗨,大家好,請讓我知道我錯了,而試圖喲運行該程序並獲得java.lang.NumberFormatException錯誤Java輸入輸出測試通過GUI

import javax.swing.*; 
public class InputOutputTest { 
public static void main(String[] args) { 
    //takes input through GUI 
    String input = JOptionPane.showInputDialog("Enternumber"); 
    int number = Integer.parseInt(input); 
    int square = number * number; 
    //Display square on console 
    System.out.println("square:" + square); 
    //Display square on GUI 
    JOptionPane.showMessageDialog(null, "square:"+ square); 
    System.exit(0); 
} 
} 
+1

在'InputDialog'中輸入數字。它不得包含任何字母或符號字符。 – asissuthar

回答

3

你應該在輸入對話框中只輸入數字。 parseInt正在導致異常。添加如下錯誤處理

 String input = JOptionPane.showInputDialog("Enter number"); 
     try { 
      int number = Integer.parseInt(input); 
      int square = number * number; 
      System.out.println("square:" + square); 
      JOptionPane.showMessageDialog(null, "square:" + square); 
     } catch (NumberFormatException exception) { 
      JOptionPane.showMessageDialog(null, "Only Numbers are accepted"); 
     }