C#中的随机数

Question

当我尝试运行应用程序时,我收到以下错误我确信它很简单,但我没有看到它.当我点击我标记为Play的按钮时,我正在尝试这样做.我想调用一个名为randomnumber的方法.然后我希望结果显示在lblPickFive_1中.我有2x2,Pick5和强力球.每个随机数将显示在我创建的自己的标签中.

现在我只是想通过生成一个随机数并将其显示在一个标签中然后我将移动到其余部分.如果我无法弄清楚剩下的问题,我肯定会发布更多问题.

错误1方法'RandomNumber'的重载没有'0'参数

using System;
using System.Windows.Forms;

namespace LotteryTickets
{
public partial class Form1 : Form
{
    /// <summary>
    /// no-args Constructor
    /// </summary>
    public Form1()
    {
        InitializeComponent();

    }


    #region "== Control Event Handlers =="
    private void Form1_Load(object sender, EventArgs e)
    {
        ClearWinningNumbers();
    }


    #endregion "== End Control Event Handlers =="


    #region "== Methods ==";
    /// <summary>
    /// Clears the text inside the winning number "balls"
    /// </summary>
    private void ClearWinningNumbers()
    {
        this.lblPickFive_1.Text = "";
        this.lblPickFive_2.Text = "";
        this.lblPickFive_3.Text = "";
        this.lblPickFive_4.Text = "";
        this.lblPickFive_5.Text = "";

        this.lblTwoByTwo_1.Text = "";
        this.lblTwoByTwo_2.Text = "";

        this.lblPowerball_1.Text = "";
        this.lblPowerball_2.Text = "";
        this.lblPowerball_3.Text = "";
        this.lblPowerball_4.Text = "";
        this.lblPowerball_5.Text = "";

        this.lblPowerball_PB.Text = "";
    }
    #endregion "== End Methods ==";

    private void cblTwoByTwo_2_SelectedIndexChanged(object sender, EventArgs e)
    {

    }

    private void cblTwoByTwo_1_SelectedIndexChanged(object sender, EventArgs e)
    {

    }

    private void btnPlay_Click(object sender, EventArgs e)
    {
        RandomNumber();

    }

    private void lblPickFive_1_Click(object sender, EventArgs e)
    {

    }
    private void RandomNumber(int min, int max)
    {
        int num = new Random().Next(min, max);
        lblPickFive_1.Text = num.ToString();
    }
    }
}

Answer 1

首先,每次你想要一个新的随机数时,你不应该新建一个随机数发生器.您应该将生成器设置为静态或成员变量,并为每个新数字引用它.

其次,你必须将min和max传递给你的RandomNumber方法.