Question

我目前正在制作Tic Tac Toe游戏。当一名球员获胜并被问及是否想再次参赛时，我会调用Playing();方法重新开始比赛。我遇到的问题是它会在第一个Playing();停止Scanner而不停止。

如何让循环停在第一个Scanner（要求用户输入他们的[row]？

当玩家同意重新启动时

Output：

Enter [Row] : Please enter a number : Enter [Row] :

Playing()方法

public static void Playing() {

    String line;

    // Check if the user's input is a number. If not, retry!

    while (true) {

        try {
            System.out.print("\nEnter [Row] : "); //Tell user to input Row
            line = input.nextLine();
            row = Integer.parseInt(line) - 1;

            System.out.print("Enter [Col] : "); //Tell user to input Col
            line = input.nextLine();
            col = Integer.parseInt(line) - 1;
            break;
        } catch (NumberFormatException e) {

            System.out.print("Please enter a number : ");
            break;

        }

    }

    //Check if their input for [row][col] is valid
    if (row < 0 || row > 2 || col < 0 || col > 2) {
        System.out.print("Oops! Invalid input. Try again");
        Playing();

    } else if (Board[row][col] != '_') {
        System.out.print("This position is already taken. Try again");
        Playing();

    } else {
        Board[row][col] = player;
        moveCount++;
        DisplayBoard();
        GameOver(); //Check if anyone won. If not, continue the game

    }
}

Replay()方法

    public static void Replay() {
    /*
     * Check if the user wants to play again
     */
    System.out.print("Would you like to play again?(Y or N) : ");
    while (true) {
        String retry = input.next();

        if ("y".equalsIgnoreCase(retry)) {
            for (int r = 0; r < 3; r++) {
                for (int c = 0; c < 3; c++) {
                    Board[r][c] = '_';
                }
            }
            System.out.print("-----------------------------------------------------\n");
            DisplayBoard();
            Playing();
            break;
        } else if ("n".equalsIgnoreCase(retry)) {
            System.out.println("Thank you for Playing!");
            System.exit(1);

            // If the user enters an invalid input, this will ask them to try again
        } else {
            System.out.print("Invalid input. Would you like to play again?(Y or N) : ");

        }
    }
}

Answer 1

首先，使用

的原因是什么

<?php

$servername = "";
$user = "";
$pw = "";
$dbname = "";

$conn = new mysqli($servername, $user, $pw, $dbname);

if ($conn->connect_error) {
die("Could not connect to the server.");
}

$someone = $_GET["user"];

$sql = "SELECT * FROM some_table WHERE some_row='$someone'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "Welcome, " . $row["some_row"]. "<span id='date'></span>
<script>
var date = new Date();
var days = ['Sun', 'Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat'];
document.getElementById('date').innerHTML = days[date.getDay()];
</script>
";
}
} else {
echo "Error";
}

?>

其中一个人可以使用Scanner提供的nextInt（）方法实际获取整数

line = input.nextLine();
row = Integer.parseInt(line) - 1;

为了回答你的问题，我认为这是导致跳过你的输入的罪魁祸首

line = input.nextInt();
row = line - 1;

如果您输入一些关键字，请在此行中输入＆＃34; Hello＆＃34;然后点击输入＆＃34; Hello \ n＆＃34;下一个方法只需要＆＃34; Hello＆＃34;并且\ n将跳过你的nextLine（）方法。所以我建议你尝试在这一行之后添加另一个nextLine（）

String retry = input.next();

注意：这只是一个问题的猜测，我还没有通过在我的终端系统上运行来调试你的代码。

重新启动我的游戏会跳过每个Scanner类和循环？

1 个答案: