Question

我正在制作程序，通过FTP获取截图并在网络主机上上传该图像。我有问题，当程序上传图片时，那时PC冻结了第二次，我想第二次是图片上传时的时间。用户不应该感觉某些东西正在减慢他的PC。

如何消除那一秒冻结？

注意：这不是某种类型的病毒

/* Upload File */
public void upload(string remoteFile, string localFile)
{
    try
    {
        /* Create an FTP Request */
        ftpRequest = (FtpWebRequest)FtpWebRequest.Create(host + "/" + remoteFile);
        /* Log in to the FTP Server with the User Name and Password Provided */
        ftpRequest.Credentials = new NetworkCredential(user, pass);
        /* When in doubt, use these options */
        ftpRequest.UseBinary = true;
        ftpRequest.UsePassive = true;
        ftpRequest.KeepAlive = true;
        /* Specify the Type of FTP Request */
        ftpRequest.Method = WebRequestMethods.Ftp.UploadFile;
        /* Establish Return Communication with the FTP Server */
        ftpStream = ftpRequest.GetRequestStream();
        /* Open a File Stream to Read the File for Upload */
        FileStream localFileStream = new FileStream(localFile, FileMode.Open);
        /* Buffer for the Downloaded Data */
        byte[] byteBuffer = new byte[bufferSize];
        int bytesSent = localFileStream.Read(byteBuffer, 0, bufferSize);
        /* Upload the File by Sending the Buffered Data Until the Transfer is Complete */
        try
        {
            while (bytesSent != 0)
            {
                ftpStream.Write(byteBuffer, 0, bytesSent);
                bytesSent = localFileStream.Read(byteBuffer, 0, bufferSize);
            }
        }
        catch (Exception ex) { MessageBox.Show(ex.ToString()); }
        /* Resource Cleanup */
        localFileStream.Close();
        ftpStream.Close();
        ftpRequest = null;  
    }
    catch (Exception ex) { MessageBox.Show(ex.ToString()); }
    return;
}

这是我的上传功能

Answer 1

更改方法声明：

public async Task upload(string remoteFile, string localFile)

然后将调用更改为Write：

await ftpStream.WriteAsync(byteBuffer, 0, bytesSent);

该方法中的其余I / O可能足够快，不需要切换到async / await方法。但是你也可以考虑改掉它们。

请注意，您还必须更改upload()的呼叫网站。你需要＆＃34;冒泡＆＃34; async / await模式一直到初始UI事件处理程序，可以是async void而不是async Task（这样你就可以匹配所需的事件处理程序方法签名）。

Answer 2

只要您调用upload()方法，请使用Task以异步方式运行它：

var task = new Task(() =>
{
    upload(remotePath, localPath);
});
task.Start();

如果您需要向UI线程报告某些内容，而在任务正在运行....进度，例如，只需使用Dispatcher：

var task = new Task(() =>
{
    upload(remotePath, localPath);
    Application.Current.Dispatcher.BeginInvoke(DispatcherPriority.Normal,
    new Action(() =>
    {
        DoSomethingToUI();
    }));
});
task.Start();

它适用于我，并且不需要“冒泡”异步回复给任何人。

在图像上传PC冻结第二

2 个答案: