计算尝试次数并要求用户再次播放循环

Question

我正在尝试制作一款有趣的新游戏。游戏是猜一个数字。我使用Microsoft Visual Studio制作申请表。该程序将使用户10尝试获得正确的猜测。它还会告诉数字太高或太低。该程序具有循环功能来监控尝试。如果用户猜了10次，程序将停止。问题是显示屏显示未更改的尝试。例如，用户猜测了五次，但节目的显示仍显示10次尝试。如果用户使用10次尝试，则停止程序也是一样的。该程序应在10次尝试后停止，但不起作用。我无法找到问题的来源。如果有人可以提供帮助，将不胜感激。

public class Main : Form

    int randomNumber;
    int count;

    private void Form1_Load(object sender, EventArgs e)
    {
        count = 10;
        randomNumber = generateRandomNumber();
    }

    private void button1_Click(object sender, EventArgs e)
    {  
        int guess = 0;
        guess = Convert.ToInt32(textBox1.Text);

        for (int i = 1; i <= count; i++)
        {
            label4.Text = "Guess Attempt: " + i;

            if (guess == randomNumber)
            {
                label4.Text = "Your guess is correct! Game Won";
                label2.Visible = false;
                label3.Visible = false;
                button1.Visible = false;
                if (MessageBox.Show("Do you want to play again?", "Start a new game", MessageBoxButtons.YesNo, MessageBoxIcon.Information) == DialogResult.Yes)
                {
                    Application.Restart();
                }
                else
                {
                    Application.Exit();
                } 
                break;
            }
            else 
                if (guess < randomNumber)
                {
                    label2.Text = "Too low!";

                }
                else if (guess > randomNumber)
                {
                    label2.Text = "Too high!";

                }

                else if (i == count)
                {
                    label2.Text = "Sorry, You missed it! Game Lost";
                    label3.Text = "The answer is " + randomNumber;
                    if (MessageBox.Show("Do you want to play again?", "Start a new game", MessageBoxButtons.YesNo, MessageBoxIcon.Information) == DialogResult.Yes)
                {
                    Application.Restart();
                }
                else
                {
                    Application.Exit();
                } 
                }
        }
    }

    private int generateRandomNumber()
    {
        Random number = new Random();
        int randomNumber = number.Next(1, 100);

        return randomNumber;
    }

Answer 1

您需要一个变量来存储与count和randomNumber处于同一级别的尝试次数，并在每次单击按钮时将其递增。

public class Main : Form

    int randomNumber;
    int count;
    int attempt;

    private void Form1_Load(object sender, EventArgs e)
    {
        attempt = 0;
        count = 10;
        randomNumber = generateRandomNumber();
    }

    private void button1_Click(object sender, EventArgs e)
    {  
        int guess = 0;
        guess = Convert.ToInt32(textBox1.Text);

        attempt++;
        label4.Text = "Guess Attempt: " + attempt;

        if (guess == randomNumber)
        {
            label4.Text = "Your guess is correct! Game Won";
            label2.Visible = false;
            label3.Visible = false;
            button1.Visible = false;
            if (MessageBox.Show("Do you want to play again?", "Start a new  game", MessageBoxButtons.YesNo, MessageBoxIcon.Information) == DialogResult.Yes)
            {
                Application.Restart();
            }
            else
            {
                Application.Exit();
            } 
            break;
        }
        else
            if (guess < randomNumber)
            {
                label2.Text = "Too low!";

            }
            else if (guess > randomNumber)
            {
                label2.Text = "Too high!";
            }
            else if (attempt == count)
            {
                label2.Text = "Sorry, You missed it! Game Lost";
                label3.Text = "The answer is " + randomNumber;
                if (MessageBox.Show("Do you want to play again?", "Start a new game", MessageBoxButtons.YesNo, MessageBoxIcon.Information) == DialogResult.Yes)
                {
                    Application.Restart();
                }
                else
                {
                    Application.Exit();
                } 
            }
        }
    }

    private int generateRandomNumber()
    {
        Random number = new Random();
        int randomNumber = number.Next(1, 100);

        return randomNumber;
    }