當用戶輸入的值不在字符串數組中時出錯

Question

String[] alpha = {"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"}; 

我已經制作了包含字母猜測遊戲字母表中所有字母的數組。

我不確定如果用戶在字母表的這些值之外輸入某些內容，我將如何獲得錯誤消息。一個號碼？任何幫助將不勝感激。

Answer 1

如果你想區分大小寫的比較，那麼你可以使用toLowerCase()你可以將它轉換成List和使用contains，例如：

String[] alpha = {"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"}; 
List<String> list = Arrays.asList(alpha); 
System.out.println(list.contains("a")) 

。

Answer 2

你可以使用比那個字符數組。示例在這裏給出https://www.javatpoint.com/java-string-tochararray。然後，你可以使用的indexOf方法，看看如果用戶輸入值有效，這裏How can I check if a single character appears in a string?

Answer 3

您可以使用此解釋： -

if (! ArrayUtils.contains(alpha, "[i-dont-exist]")) { 
    try{ 
     throw new Exception("Not Found !"); 
    }catch(Exception e){} 
} 

Documentation Here

Answer 4

如果目的是檢查的存在裏面收藏了更合理的使用一組由於存取時間元素是constanct

這樣反而

Set<String> set = new HashSet<>();//if not java 8 make it HashSet<String> 
    set.put("a") // do this for all the strings you would like to check 

然後檢查是否字符串存在於這個設置

if(set.contains(str)) //str is the string you want to make sure it exists in the collections 

Answer 5

使用這種方法，如果你願意堅持使用數組，而不是其他的數據結構。

1. 創建如果用戶輸入是存在於陣列中返回true的方法，否則返回假

   public static boolean isValid(String input) { 
       String[] alpha = {"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", 
        "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", 
        "w", "x", "y", "z"}; 
   
       for(String s: alpha) { 
        if(input.equals(s)) { //use input.equalsIgnoreCase(s) for case insensitive comparison 
         return true; //user input is valid 
        } 
       } 
   
       return false; //user input is not valid 
   } 
   

2. 在調用方法，只需將用戶輸入傳遞到isValid

   //write some codes here to receive user input..... 
   
   if(!isValid(input)) 
       System.out.println("Oops, you have entered invalid input!");