java - 为什么不显示异常错误

Question

      List<String> AccountList = new ArrayList<String>(); 
      AccountList.add("45678690");
      AccountList.add("7878787");
      Scanner AccountInput = new Scanner(System.in);
      int x = 1;
 do{
     try{ 

      System.out.println("Hi whats your pin code?");
      String Value = AccountInput.nextLine();

      for  (int counter = 0; counter < AccountList.size(); counter++){  
          if (AccountList.contains(Value)){ //If Input = ArrayList number then display "hi"
              System.out.println("Hi");

              x = 2;
          }

          }

} catch (Exception e){
    System.out.println("You cant do that");

}
}while (x==1); 
      }
}  

当我运行此命令时，错误“您不能这样做”不会出现，但它被视为无效并要求用户重新输入数字直到有效，我怎样才能显示错误行？

Answer 1

您需要抛出异常才能捕获异常:

for  (int counter = 0; counter < AccountList.size(); counter++){  
    if (AccountList.contains(Value)) { //If Input = ArrayList number then display "hi"
        System.out.println("Hi");
        x = 2;
    } else {
        throw new Exception();
    }
}

编辑:正如评论中所述，for 行不是必需的，因为了解该值是否包含在您的 AccountList 中的算法是由 contains 方法处理的。