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如何多次发送文件

tcp-ip c#

-2

Error 404 · 技术社区 · 8 年前

有两个C#项目:一个用于客户端,另一个用于服务器。第一步是运行服务器,然后选择目标文件夹,然后运行客户端项目,选择一些文本。txt发送到服务器的目标文件夹。 只有客户端才能将文件发送到服务器

演示:

1.choosing file target                       2.client sends
   +------------+                                
   | tar folder |          <----------------       text.txt 
   +------------+

这很好,但我希望它能发送文本。每隔几分钟自动发送一次txt,所以我添加了 while(true) 循环和 Thread.Sleep(10000) 但总有例外。我的意思是,它不会让我再发送一次。

在客户端类中,我尝试放置一个while循环:

 public Form1()
        {
            InitializeComponent();
        }


        private void button1_Click(object sender, EventArgs e)//browse btn
        {
            string n;
            byte[] b1;
            TcpClient client;
            OpenFileDialog op;
            Stream s;
            StreamWriter sw;
            op = new OpenFileDialog();

            if (op.ShowDialog() == DialogResult.OK)
            {
                string t = textBox1.Text;
                t = op.FileName;
                FileInfo fi = new FileInfo(textBox1.Text = op.FileName);
                n = fi.Name + "." + fi.Length;

                client = new TcpClient("127.0.0.1", 8100);//"127.0.0.1", 5055

                sw = new StreamWriter(client.GetStream());
                s = client.GetStream();
                b1 = File.ReadAllBytes(op.FileName);

                //  s = client.GetStream();
                while (true)
                {

                    s.Write(b1, 0, b1.Length);
                    sw.WriteLine(n);
                     sw.Flush();
                    //sw.Close();
                    s.Close();//added by me
 Thread.Sleep(18000);

                                   }
                sw.Flush();
                client.Close();

                // label2.Text = "File Transferred....";
            }
        }

服务器类别:

public Form1()
        {
            InitializeComponent();
        }

        string rd;
        byte[] b1;
        string v;
        int m=20;//number of byts
        TcpListener list;
        TcpClient client;
        int port = 8100;//5050
        int port1 = 8100;//5055
        IPAddress localAddr = IPAddress.Parse("127.0.0.1");
        private void button1_Click(object sender, EventArgs e)//browse button
        {

            if (folderBrowserDialog1.ShowDialog() == DialogResult.OK)
            {

                    textBox1.Text = folderBrowserDialog1.SelectedPath;
                    list = new TcpListener(localAddr, port1);
                    list.Start();

                    Thread incoming_connection = new Thread(ic);
                    incoming_connection.Start();

            }
        }

        private void ic()
        {

            client = list.AcceptTcpClient();
            Stream s = client.GetStream();
            b1 = new byte[m];
            s.Read(b1,0, b1.Length);
            MessageBox.Show("pathh "+textBox1.Text);
            File.WriteAllBytes(textBox1.Text+"\\flag.txt", b1);// the left side us the name of the written file
            //list.Stop();
            //client.Close();
          //  label1.Text = "File Received......";
        }

        private void Form2_Load(object sender, EventArgs e)
        {
            list = new TcpListener(localAddr, port);
           // TcpListener list = new TcpListener(port);
            list.Start();
            TcpClient client = list.AcceptTcpClient();
            MessageBox.Show("Client trying to connect");
            StreamReader sr = new StreamReader(client.GetStream());
            rd = sr.ReadLine();
            v = rd.Substring(rd.LastIndexOf('.') + 1);
            m = int.Parse(v);
           // list.Stop();
           // client.Close();
        }

1

Anderson Pimentel 8 年前

尝试更改为:

while (true)
{
    using (var client = new TcpClient("127.0.0.1", 8100))//"127.0.0.1", 5055
    using (var sw = new StreamWriter(client.GetStream()))
    using (var s = client.GetStream())
    {
        b1 = File.ReadAllBytes(op.FileName);
        //  s = client.GetStream();
        s.Write(b1, 0, b1.Length);
        sw.WriteLine(n);
        sw.Flush();
    }
    Thread.Sleep(18000);
}

using 语句有助于处理对象。

或者,如果要保持连接打开,请尝试:

using (var client = new TcpClient("127.0.0.1", 8100))//"127.0.0.1", 5055
while (true)
{
    using (var sw = new StreamWriter(client.GetStream()))
    using (var s = client.GetStream())
    {
        b1 = File.ReadAllBytes(op.FileName);
        //  s = client.GetStream();
        s.Write(b1, 0, b1.Length);
        sw.WriteLine(n);
        sw.Flush();
    }
    Thread.Sleep(18000);
}

2

1

Kickass 8 年前

好的,这是你的问题:

在第一个循环中,它工作正常,因为所有内容都已正确初始化,但在它发送第一个文件后,您调用s.Close();

while (true)
{
s.Write(b1, 0, b1.Length);
sw.WriteLine(n);
sw.Flush();
//sw.Close();
s.Close();//added by me
Thread.Sleep(18000);
}

现在发生的事情是在第一个循环之后关闭流,然后在第二个循环上它已经被“释放”。

固定移动s.Close();退出循环并进入断开按钮事件或应用程序关闭事件。

此外,我建议不要使用线程。睡眠(间隔); 这是因为睡眠会冻结线程和主线程。NET用于您的接口,因此线程非常糟糕。

创建一个新线程,如前一篇文章中所述:

using System.Threading;

Stream s;
byte[] b1;
TcpClient client;
OpenFileDialog op;
StreamWriter sw;
Thread thread = new Thread(sendFile);
string fn;
string n;
threadExit = false;
Int32 interval = 18000;

public Form1()
{
  InitializeComponent();
}
private void Form1_OnLoad(object sender,EventArgs e) {
  op = new OpenFileDialog();
  client = new TcpClient("127.0.0.1", 8100);//"127.0.0.1", 5055
  sw = new StreamWriter(client.GetStream());
  s = client.GetStream();
  //It would be your responsibility to move this code to the proper place
  //because now it tries to open the connection when your app loads
}
private void sendFile() {
  while (!threadExit)
  {
    b1 = File.ReadAllBytes(fn);
    s.Write(b1, 0, b1.Length);
    sw.WriteLine(n);
    sw.Flush();
    Thread.Sleep(interval);
  }
}

private void button1_Click(object sender, EventArgs e)
{
  if (op.ShowDialog() == DialogResult.OK)
  {
    fn = op.FileName;
    FileInfo fi = new FileInfo(textBox1.Text = op.FileName);
    n = fi.Name + "." + fi.Length;
    thread.Start();
  }
  //It would be a good idea to add an indicator to signal that another thread is trying to send the file at a set interval

}
private void Form1_Close(object sender, EventArgs e) {
  s.Close();
  client.Close();
  threadExit = true;
  Application.Exit();
}

请记住为该线程使用try-catch块,否则如果发生异常,将无法很好地处理。还要记住,在该线程运行时关闭应用程序将使你的应用程序保持打开状态,直到你手动终止进程,因此在应用程序关闭事件中,请记住终止该线程。

关闭时也会发送一个信号,让线程退出,但干净地退出可能需要18秒,另一种解决方案是使用线程。中止();但这是一个肮脏的出口,很可能引发异常。

再次请记住阅读线程技术,这是一个快速而糟糕的版本,它在生产环境中应该是怎样的。