如何從Java中的do while循環退出？

Question

一個簡單的程序，將號碼輸出為電話號碼格式，我似乎無法退出循環我不確定自己在做什么錯，我認為！PhoneNumber.equals（“ 999”）; 當用戶輸入999但不起作用時，它將退出循環。 誰能幫我這是我的代碼

import javax.swing.*;

public class PhoneNumberFormat 
{

    public static void main(String[] args) 
    {
      String PhoneNumber;
      int numLength= 10;

      do
      {

          PhoneNumber = JOptionPane.showInputDialog(null, 
                 "Enter your 10 digit phone number or enter 999 to quit");
          while(PhoneNumber.length() != numLength)
          {
              PhoneNumber = JOptionPane.showInputDialog(null,
                  "Error: You Entered " + PhoneNumber.length() + " digits\nPlease"
                      + " Enter a 10 digit Phone number");    
          }

            StringBuilder str = new StringBuilder (PhoneNumber);
            str.insert(0, '(');
            str.insert(4, ')');
            str.insert(5,' ');
            str.insert(9, '-');

        JOptionPane.showMessageDialog(null, "Your telephone number is " +str.toString());

      }while(!PhoneNumber.equals("999"));

    }
}

Answer 1

如果希望在999時存在，則需要添加一個if條件來對其進行監視。

public static void main(String[] args) {
    String PhoneNumber;
    int numLength = 10;

    do {
        PhoneNumber = JOptionPane.showInputDialog(null,
                "Enter your 10 digit phone number or enter 999 to quit");

        // add this condition to exit the loop, as well protect against NPE
        if (PhoneNumber == null || PhoneNumber.equals("999")) {
            break;
        }

        while (PhoneNumber.length() != numLength) {
            PhoneNumber = JOptionPane.showInputDialog(null,
                    "Error: You Entered " + PhoneNumber.length()
                            + " digits\nPlease"
                            + " Enter a 10 digit Phone number");

            //protect against NPE
            if(PhoneNumber == null) 
               PhoneNumber = "";
        }

        StringBuilder str = new StringBuilder(PhoneNumber);
        str.insert(0, '(');
        str.insert(4, ')');
        str.insert(5, ' ');
        str.insert(9, '-');

        JOptionPane.showMessageDialog(null, "Your telephone number is "
                + str.toString());

    } while (!PhoneNumber.equals("999"));

}

Answer 2

您將數字強制為10位，那么您期望什么？ 它永遠不會等於三位數的“ 999”。

也許您打算這樣做：

while(!PhoneNumber.startsWith("999"));

Answer 3

如果要使用999作為輸入選項退出，請首先允許用戶輸入999作為有效輸入。

  int numLength= 10;
  do
  {

      PhoneNumber = JOptionPane.showInputDialog(null, 
             "Enter your 10 digit phone number or enter 999 to quit");
      while(PhoneNumber.length() != numLength)
      {
          PhoneNumber = JOptionPane.showInputDialog(null,
              "Error: You Entered " + PhoneNumber.length() + " digits\nPlease"
                  + " Enter a 10 digit Phone number");    
      }

在這里，如果輸入的長度不是10，則拒絕考慮輸入。

Answer 4

具有修復潛在NPE並解決問題的代碼應如下所示：

import javax.swing.*;

public class PhoneNumberFormat 
{

    public static void main(String[] args) 
    {
      String PhoneNumber;
      int numLength= 10;

      do
      {

          PhoneNumber = JOptionPane.showInputDialog(null, 
                 "Enter your 10 digit phone number or enter 999 to quit");
          while(PhoneNumber!=null && PhoneNumber.length() != numLength)
          {
              PhoneNumber = JOptionPane.showInputDialog(null,
                  "Error: You Entered " + PhoneNumber.length() + " digits\nPlease"
                      + " Enter a 10 digit Phone number");    
          }

            StringBuilder str = new StringBuilder (PhoneNumber);
            str.insert(0, '(');
            str.insert(4, ')');
            str.insert(5,' ');
            str.insert(9, '-');

        JOptionPane.showMessageDialog(null, "Your telephone number is " +str.toString());

      } while(!PhoneNumber.substring(0,3).equals("999"));

    }
}

修復程序已上線

while(PhoneNumber!=null && PhoneNumber.length() != numLength)

和

while(!PhoneNumber.substring(0,3).equals("999"));