c# - C#根据输入参数从url下载一个zip文件

Question

我有一个要求，我必须根据输入参数使用 c#（大小可以在 10mb - 400mb 之间变化）从服务器下载一个 zip 文件。例如，下载 userId = 10 和 year = 2012
的报告。网络服务器接受这两个参数并返回一个 zip 文件。如何使用 WebClient 类来实现这一点？
谢谢

Answer 1

您可以通过扩展 WebClient 类来做到这一点

class ExtWebClient : WebClient
    {

        public NameValueCollection PostParam { get; set; }

        protected override WebRequest GetWebRequest(Uri address)
        {
            WebRequest tmprequest = base.GetWebRequest(address);

            HttpWebRequest request = tmprequest as HttpWebRequest;

            if (request != null && PostParam != null && PostParam.Count > 0)
            {
                StringBuilder postBuilder = new StringBuilder();
                request.Method = "POST";
                //build the post string

                for (int i = 0; i < PostParam.Count; i++)
                {
                    postBuilder.AppendFormat("{0}={1}", Uri.EscapeDataString(PostParam.GetKey(i)),
                                             Uri.EscapeDataString(PostParam.Get(i)));
                    if (i < PostParam.Count - 1)
                    {
                        postBuilder.Append("&");
                    }
                }
                byte[] postBytes = Encoding.ASCII.GetBytes(postBuilder.ToString());
                request.ContentLength = postBytes.Length;
                request.ContentType = "application/x-www-form-urlencoded";

                var stream = request.GetRequestStream();
                stream.Write(postBytes, 0, postBytes.Length);
                stream.Close();
                stream.Dispose();

            }

            return tmprequest;
        }
    }

用法：如果你必须创建 POST 类型的请求

class Program
{
    private static void Main()
    {
        ExtWebClient webclient = new ExtWebClient();
        webclient.PostParam = new NameValueCollection();
        webclient.PostParam["param1"] = "value1";
        webclient.PostParam["param2"] = "value2";

        webclient.DownloadFile("http://www.example.com/myfile.zip", @"C:\myfile.zip");
    }
}

用法：对于 GET 类型的请求，你可以简单地使用普通的 webclient

class Program
{
    private static void Main()
    {
        WebClient webclient = new WebClient();

        webclient.DownloadFile("http://www.example.com/myfile.zip?param1=value1&param2=value2", @"C:\myfile.zip");
    }
}

Answer 2

string url = @"http://www.microsoft.com/windows8.zip";

WebClient client = new WebClient();

    client.DownloadFileCompleted +=    new AsyncCompletedEventHandler(client_DownloadFileCompleted);

    client.DownloadFileAsync(new Uri(url), @"c:\windows\windows8.zip");


void client_DownloadFileCompleted(object sender, AsyncCompletedEventArgs e)
{
    MessageBox.Show("File downloaded");
}