我正在将文件上传到网络服务......网络服务的协议如下:
HTTP POST请求参数“USERNAME”,“PASSWORD”和“FILE”是必需的。
必须使用多部分表单提交,即使用'multipart / form-data'的FORM'enctype'属性。
和 当用于构造URL时,“&”字符应替换为“%26”,“”(空格)替换为“%20”。
我指的是帖子Upload files with HTTPWebrequest (multipart/form-data)来构建此网络请求...
我的代码如下......
string boundary =“---------------------------”+ DateTime.Now.Ticks.ToString(“x”); < / p>
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
NameValueCollection nvc = new NameValueCollection();
nvc.Add("USERNAME", "myusernammeee");
// nvc.Add(“PASSWORD”,“mS89_n3w”);
nvc.Add("PASSWORD", "mypasswordddd");
foreach (string key in nvc.Keys)
{
rs.Write(boundarybytes, 0, boundarybytes.Length);
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
rs.Write(formitembytes, 0, formitembytes.Length);
}
rs.Write(boundarybytes, 0, boundarybytes.Length);
string headerTemplate =
"Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
string header = string.Format(headerTemplate, "FILE", files[0], "xml");
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
StringBuilder inFile = new StringBuilder(1000);
//First of all lets read the contents of the file in.
using (TextReader streamReader = new StreamReader(files[0]))
{
while (streamReader.Peek() >= 0)
{
inFile.Append(streamReader.ReadLine());
}
}
string inputFile = inFile.ToString().Replace("&", "%26amp;");
byte[] fileBytes = Encoding.UTF8.GetBytes(inputFile);
rs.Write(fileBytes,0, fileBytes.Length);
byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
rs.Write(trailer, 0, trailer.Length);
rs.Close();
WebResponse wresp = null;
try
{
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
}
catch (Exception ex)
{
if (wresp != null)
{
wresp.Close();
wresp = null;
}
}
finally
{
wr = null;
}
但是当我这样做时,我收到错误“错误请求”。我不确定我在这里做错了什么..任何建议都将不胜感激。
谢谢