如何暂停程序执行，等待输入，输入完后继续

怎么暂停程序执行，等待输入，输入完后继续

 	private void button1_Click(object sender, EventArgs e)

        {

            Thread threadstart = new Thread(new ThreadStart(Mystart));

            threadstart.IsBackground = true;

            threadstart.Start();

        }

        private void richTextBox1_TextChanged(object sender, EventArgs e)

        {

            if (richTextBox1.Text.Length == 4)

            {

                detailCollectedEvent.Set();

            }

        }

        AutoResetEvent detailCollectedEvent = new AutoResetEvent(false);

        private void Mystart()

        {

            Con();

            detailCollectedEvent.WaitOne();

            SendPost(richTextBox1.Text);

        }

为啥我在richTextBox中输入完文本，程序不继续执行.求指点
------解决方案--------------------
我这里测试可以。

按照你的程序修改的，新建一个窗体，添加一个RichTextBox一个Button，编写如下代码：

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Windows.Forms;

using System.Threading;


namespace WindowsFormsApplication1

{

    public partial class Form1 : Form

    {

        public Form1()

        {

            InitializeComponent();

        }


        private void button1_Click(object sender, EventArgs e)

        {

            Thread threadstart = new Thread(new ThreadStart(Mystart));

            threadstart.IsBackground = true;

            threadstart.Start();

        }

        private void richTextBox1_TextChanged(object sender, EventArgs e)

        {

            if (richTextBox1.Text.Length == 4)

            {

                detailCollectedEvent.Set();

            }

        }

        AutoResetEvent detailCollectedEvent = new AutoResetEvent(false);

        private void Mystart()

        {

            Step1();

            detailCollectedEvent.WaitOne();

            Step2();

        }


        private void Step2()

        {

            Invoke(new Action(() => this.Text = "2"));

        }


        private void Step1()

        {

            Invoke(new Action(() => this.Text = "1"));

        }

    }

}

看看问题是不是出在con sendpost里面。