Question

下面你会找到方法规范和我写的匹配方法：

/**
 * This method prompts the user for a number, verifies that it is between min
 * and max, inclusive, before returning the number.  
 * 
 * If the number entered is not between min and max then the user is shown 
 * an error message and given another opportunity to enter a number.
 * If min is 1 and max is 5 the error message is:
 *      Expected a number from 1 to 5.  
 * 
 * If the user enters characters, words or anything other than a valid int then 
 * the user is shown the same message.  The entering of characters other
 * than a valid int is detected using Scanner's methods (hasNextInt) and
 * does not use exception handling.
 * 
 * Do not use constants in this method, only use the min and max passed
 * in parameters for all comparisons and messages.
 * Do not create an instance of Scanner in this method, pass the reference
 * to the Scanner in main, to this method.
 * The entire prompt should be passed in and printed out.
 *
 * @param in  The reference to the instance of Scanner created in main.
 * @param prompt  The text prompt that is shown once to the user.
 * @param min  The minimum value that the user must enter.
 * @param max  The maximum value that the user must enter.
 * @return The integer that the user entered that is between min and max, 
 *          inclusive.
 */

public static int promptUser(Scanner in, String prompt, int min, int max) {
    //initialize variables
    Integer userInput = 0;
    boolean userInteger = false;
    System.out.print(prompt);//prompts the user for input
    userInteger = in.hasNextInt();
    while (userInteger == false) {
            System.out.println("Expected a number from " + min + " to " + max +".");     
        in.nextLine();
        userInteger = in.hasNextInt();
    }

    while (userInteger == true) {
        userInput = in.nextInt();
        while (userInput > max || userInput < min) {
            System.out.println("Expected a number from " + min + " to " + max +".");
            in.nextLine();
            userInteger = in.hasNextInt();

            while (userInteger == false) {
                System.out.println("Expected a number from " + min + " to " + max +".");
                in.nextLine();
                userInteger = in.hasNextInt();

            }
            userInput = in.nextInt();                   
        }
        userInteger = false;
   }
   //userInteger = false; 
   return userInput; //FIXME
}

不幸的是，当我尝试使用以下值进行测试时： 4 4 五 4 哟哟哟

当我输入你的时，我打印出两个错误，而不是1。我知道我打印两次相同的print语句，它是while（userInteger = false）循环下的print语句。关于如何解决这个问题的任何想法？

Answer 1

当你输入＆＃34;哟哟＆＃34;在一个请求中，Java将其解释为单个字符串＆＃34; yo yo＆＃34;而不是2个不同的字符串＆＃34; yo＆＃34;和＆＃34;哟＆＃34;。

如果您想要2个不同的错误消息，则需要输入单个＆＃34; yo＆＃34;每一次。

Answer 2

当你有＆＃34;哟哟＆＃34;作为输入。 Scanner.nextInt（）接收第一个＆＃34; yo＆＃34;作为输入并打印您的错误消息，第二个＆＃34;哟＆＃34;留在输入缓冲区。当它第二次调用.nextInt（）时，它会接收第二个＆＃34; yo＆＃34;它位于缓冲区中，并打印出另一个错误输出。

您可能会重新考虑使用Scanner.nextLine（）然后解析输入而不是简单地使用Scanner.nextInt（）。

希望这有帮助。

循环打印2个错误而不是1个扫雷

2 个答案: