相关疑难解决方法(0)

我正在尝试创建一种方法,在用户下载时不断传输zip文件(这样就不会浪费流媒体)

我添加了一个thread.sleep来模拟延迟

public override void ExecuteResult(ControllerContext context) {
    HttpResponseBase response = context.HttpContext.Response;

    response.Clear();
    response.ClearContent();
    response.ClearHeaders();
    response.Cookies.Clear();
    response.ContentType = ContentType;
    response.ContentEncoding = Encoding.Default;
    response.AddHeader("Content-Type", ContentType);
    context.HttpContext.Response.AddHeader("Content-Disposition", 
                            String.Format("attachment; filename={0}", 
                            this.DownloadName));
    int ind = 0;
    using (ZipOutputStream zipOStream = 
                new ZipOutputStream(context.HttpContext.Response.OutputStream))
    {
        foreach (var file in FilesToZip)
        {
            ZipEntry entry = new ZipEntry(FilesToZipNames[ind++]);
            zipOStream.PutNextEntry(entry);
            Thread.Sleep(1000);
            zipOStream.Write(file, 0, file.Length);
            zipOStream.Flush();
        }
        zipOStream.Finish();
    }    
    response.OutputStream.Flush();
}

看起来拉链将不会开始流式传输,直到所有文件都是压缩文件.有没有办法连续流？也许有一个不同的图书馆？

我在C#MVC应用程序中有一个函数,它创建一个临时目录和一个临时文件,然后使用FileStream打开文件,将FileStream返回给调用函数,然后需要删除临时文件.但是,我不知道如何删除临时目录和文件,因为它总是错误地说"进程无法访问该文件,因为它正被另一个进程使用".这是我尝试过的,但FileStream仍在使用finally块中的临时文件.如何返回FileStream并删除临时文件？

public FileStream DownloadProjectsZipFileStream()
{
    Directory.CreateDirectory(_tempDirectory);
    // temporary file is created here
    _zipFile.Save(_tempDirectory + _tempFileName);

    try
    {
        FileStream stream = new FileStream(_tempDirectory + _tempFileName, FileMode.Open);
        return stream;
    }
    finally
    {
        File.Delete(_tempDirectory + _tempFileName);
        Directory.Delete(_tempDirectory);
    }
}

FileStream返回的函数如下所示:

public ActionResult DownloadProjects ()
{
    ProjectDownloader projectDownloader = new ProjectDownloader();

    FileStream stream = projectDownloader.DownloadProjectsZipFileStream();
    return File(stream, "application/zip", "Projects.zip");
}

更新:我忘了提到zip文件是380 MB.使用MemoryStream时,我遇到了系统内存不足的问题.