用Request.Files["name"]获取值，客户端后台应该怎么传

Question

C#ASP.NETRequestRequest.Files

Answer 1

Files?,那你的上传控件(html)的name属性必须为  name="name"

Answer 2

那就看你控件的名字是不是name了

Answer 3

模拟POST            string url = "";
            HttpWebRequest request = HttpWebRequest.Create(url) as HttpWebRequest;            request.ContentType = "application/x-www-form-urlencoded";
            request.Method = "POST";
            Stream stream = request.GetRequestStream();
            string name = "";
            string strFormData = string.Format("name={0}", name);
            byte[] buff = System.Text.Encoding.Default.GetBytes(strFormData);
            stream.Write(buff, 0, buff.Length);
            HttpWebResponse response = request.GetResponse() as HttpWebResponse;
            Stream streamRes = response.GetResponseStream();
            StreamReader streamReader = new StreamReader(streamRes, System.Text.Encoding.Default);
            string resultHtml = streamReader.ReadToEnd();
            response.Close();
            streamReader.Close();模拟Get            string url = "";
            string name = "";
            HttpWebRequest request = HttpWebRequest.Create(url + "?name=" + name) as HttpWebRequest;
            HttpWebResponse response = request.GetResponse() as HttpWebResponse;
            Stream streamRes = response.GetResponseStream();
            StreamReader streamReader = new StreamReader(streamRes, System.Text.Encoding.Default);
            string resultHtml = streamReader.ReadToEnd();
            response.Close();
            streamReader.Close();

Answer 4

普通的模拟Request["name"]可以取到参数，Request.Files["name"]不行
我要问的是接收端Request.Files["name"]里的name怎么传

Answer 5

参考:
Upload files with HTTPWebrequest (multipart/form-data)
http://stackoverflow.com/questions/566462/upload-files-with-httpwebrequest-multipart-form-data