1 思路:
WebClient.UploadFile()方法可以上传文件;UploadData()方法可以上传数据参数;如何合二为一既上传文件又上传参数呢?可将文件也当做参数,调用UploadData()方法
2 客户端
FileStream fs = new FileStream(“需上传文文件路径”, FileMode.Open, FileAccess.Read);
byte[] byteFile = new byte[fs.Length];
fs.Read(byteFile, 0, Convert.ToInt32(fs.Length));
fs.Close();
string postData = "param1=pwd&FileName=file1.xml&UploadFile=" + HttpUtility.UrlEncode(Convert.ToBase64String(byteFile));
var webclient = new WebClient();
webclient.Headers.Add("Content-Type", "application/x-www-form-urlencoded");
byte[] byteArray = Encoding.UTF8.GetBytes(postData);
byte[] buffer = webclient.UploadData(“远程ashx URL”, "POST", byteArray);
var msg = Encoding.UTF8.GetString(buffer);
3 服务端
string param1= context.Request["param1"].ToString();
FileStream fs = new FileStream(“需要保存文件的路径”, FileMode.Create, FileAccess.Write);
fs.Write(Convert.FromBase64String(context.Request["UploadFile"].ToString()), 0, Convert.FromBase64String(context.Request["UploadFile"].ToString()).Length);
fs.Flush();
fs.Close();