密碼檢查程序的Java程序

Question

我想創建一個程序，將檢查用戶的密碼。我希望程序在用戶正確的時候結束，但如果沒有，我只希望它只問4次。

問題：即使您正確獲取密碼，程序也會繼續詢問密碼。如果你弄錯了，它會問錯。我怎樣才能解決這個問題？

import java.util.Scanner; 
public class HowToAdd { 
    public static void main (String [] args){ 
     Scanner input = new Scanner (System.in); 
     int trys = 0; 
     String password=null; 
     do{ 
      System.out.println("Guess the password"); 
      password = input.next(); 

      if(password.equalsIgnoreCase("Password")){ 
       System.out.println("Great job"); 
      }else{ 
       System.out.println("Try again"); 
       input.next(); 
       trys++; 
      } 
     }while(trys<2); 
     System.out.println("Try again later!"); 

     input.close(); 
    } 
} 

Answer 1

你只需要添加一休：

if(password.equalsIgnoreCase("Password")){ 
System.out.println("Great job"); 
break; 
} 

Answer 2

我不僅添加break修復不會離開環路的問題，當輸入正確的密碼，但我加了幾個其他的事情來幫助你看到如下：

public static void main (String [] args){ 
    Scanner input = new Scanner (System.in); 
    int trys = 0; 
    String password=null; 
    System.out.println("Guess the password:"); 
    while(trys<2) 
    { 
     password = input.next(); 

     if(password.equalsIgnoreCase("Password")){ 
      System.out.println("Great job"); 
      break; 
     }else{ 
      trys++; 
      if(trys != 2) 
      { 
       System.out.println("Try again:"); 
      } 
     } 
    } 
    if(trys == 2) 
    { 
     System.out.println("Try again later!"); 
    } 

    input.close(); 
} 

試試這個它會跳出循環，如果它使用break;說法是正確的。此外，它只會顯示猜測密碼的第一次嘗試，然後再試一次。如果他們猜對了，它也不會再說再試，因爲它會檢查他們是否猜錯了兩次。

Answer 3

您可以用break關鍵字修復它，就像這樣：

 if(password.equalsIgnoreCase("Password")){ 
      System.out.println("Great job"); 
      break; 
     } 

的break在這裏解釋：

Branching Statements

Answer 4

需要實現能夠在不使用break報表做些什麼。 試着看看我的代碼流。

Scanner scn = new Scanner(System.in); 
    int tries=0; 
    String pw = ""; 

    System.out.println("Guess the password"); 
    pw = scn.nextLine(); 

    while(!pw.equalsIgnoreCase("Password") && tries < 3){ 
     tries++; 
     System.out.println("Try again"); 
     pw = scn.nextLine();    
    } 

    if(tries >= 3) 
     System.out.println("Try again later!"); 
    else 
     System.out.println("Great job!"); 

TEST RUN 1：

Guess the password 
1st try 
Try again 
2nd try 
Try again 
3rd try 
Try again 
last try 
Try again later! 

TEST RUN 2：

Guess the password 
password 
Great job! 

Answer 5

這裏是代碼另一種變型：

public static void main(String[] args) { 
    Scanner input = new Scanner(System.in); 
    int trys = 0; 
    String password = null; 
    // create variable for a number of tries 
    int numberOfCheks = 4; 
    do { 
     System.out.println("Guess the password"); 
     password = input.nextLine(); 

     if (password.equalsIgnoreCase("Password")) { 
      System.out.println("Great job"); 
      break; 
     } else if (++trys == numberOfCheks) { 
      //Break if you've used all your tries 
      System.out.println("Try again later!"); 
      break; 
     } else { 
      System.out.println("Try again"); 
      // input.next(); - Useless. You don't handle it. Removed 
     } 
    //There is already tries number check. So 'true' can be used as the condition. 
    } while (true); 
    input.close(); 
}