用户输入不在String Array中的值时出错

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"};

我已经制作了上面的数组，包括字母拼写游戏中字母表中的所有字母。

如果用户输入的字母值超出这些值，我不确定如何显示错误信息，例如。一个号码？任何帮助将不胜感激。

Answer 1

您可以将其转换为List并使用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"))

如果您想要不区分大小写，那么您可以使用toLowerCase()。

Answer 2

你可以使用一个字符数组。这里给出了示例。然后，您可以使用indexof方法查看用户输入的值是否有效，如https://www.javatpoint.com/java-string-tochararray

中所述

Answer 3

您可以使用： -

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

How can I check if a single character appears in a string?

Answer 4

如果检查集合中元素是否存在的目的，则使用集合更合理，因为访问时间是不变的

所以相反

   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

然后检查此集合中是否存在字符串

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

Answer 5

如果您想坚持使用数组而不是其他数据结构，请使用此方法。

1. 如果数组中存在用户输入，则创建一个返回true的方法，否则返回false

   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. 在调用方法中，只需将用户输入传递给isValid

   即可

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