爲什麼此循環的第二次迭代會跳過第一次掃描？

Question

我想寫一個代碼，要求用戶輸入他們的名字，將它存儲在列表中，然後要求他們輸入他們的年齡，將它存儲在另一個列表中。然後詢問用戶是否想再次嘗試[是/否]。 當測試代碼時，我輸入「Y」，並期望循環要求我輸入另一個名稱。相反，它跳過名稱輸入並跳轉到年齡輸入。我無法弄清楚爲什麼。 代碼如下

import java.util.ArrayList; 
import java.util.Scanner; 

public class PatternDemoPlus { 
    public static void main(String[] args){ 
     Scanner scan = new Scanner(System.in); 
     ArrayList<String> names = new ArrayList<String>(); 
     ArrayList<Integer> ages = new ArrayList<Integer>(); 
     String repeat = "Y"; 
     while(repeat.equalsIgnoreCase("Y")){ 
      System.out.print("Enter the name: "); 
      names.add(scan.nextLine()); 
      System.out.print("Enter the age: "); 
      ages.add(scan.nextInt()); 

      System.out.print("Would you like to try again? [Y/N]"); 
      repeat = scan.next(); 
      //Notice here that if I use "repeat = scan.nextLine(); instead, the code does not allow me to input anything and it would get stuck at "Would you like to try again? [Y/N] 
      System.out.println(repeat); 

      //Why is it that after the first iteration, it skips names.add and jumps right to ages.add? 
     } 
    } 
} 

我希望得到您的幫助。謝謝。

Answer 1

使用next（）將只返回空格前的內容。 nextLine（）在返回當前行後自動向下移動掃描器。

嘗試更改您的代碼，如下所示。

public class PatternDemoPlus { 
    public static void main(String[] args){ 
     Scanner scan = new Scanner(System.in); 
     ArrayList<String> names = new ArrayList<String>(); 
     ArrayList<Integer> ages = new ArrayList<Integer>(); 
     String repeat = "Y"; 
     while(repeat.equalsIgnoreCase("Y")){ 
      System.out.print("Enter the name: "); 
      String s =scan.nextLine(); 
      names.add(s); 
      System.out.print("Enter the age: "); 
      ages.add(scan.nextInt()); 
      scan.nextLine(); 
      System.out.print("Would you like to try again? [Y/N]"); 
      repeat = scan.nextLine(); 

      System.out.println(repeat); 


     } 
    } 
}