我知道那里有几篇类似的文章,但我几乎都尝试过。我正在尝试使用HttpWebRequest在网站上上传图像。 代码运行没有错误,服务器的响应与手动上传的响应相同。问题是它代替了照片,而是显示了类似于下图的黑十字
我不知道出了什么问题以及什么原因导致了问题,我最好的猜测是问题可能出在我在服务器上读写文件的方式上。 这是我的代码
public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
{
MessageBox.Show(string.Format("Uploading {0} to {1}", file, url));
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
CookieContainer cookieContainer = LoginToUrl();//this is to just pass the cookies;
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.CookieContainer = cookieContainer;
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
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, paramName, file, contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();
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);
MessageBox.Show(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
}
catch (Exception ex)
{
MessageBox.Show(string.Format("Error uploading file", ex));
if (wresp != null)
{
wresp.Close();
wresp = null;
}
}
finally
{
wr = null;
}
}
这就是我所说的:
private void UploadFile()
{
string url="https://portal.com/index.php?r=location/file";
string file = "C:\\Vbn.jpg";
string paramName= "file";
string contentType = "image/jpeg";
NameValueCollection nvc = new NameValueCollection();
nvc.Add("rowid", "");
nvc.Add("model", "Photo");
nvc.Add("category", "LeadIn");
nvc.Add("row", "");
nvc.Add("oper", "new");
nvc.Add("model_id", "2262");
nvc.Add("max_image", "0");
nvc.Add("type", "[Lead-in]Building Lead-in (from inside showing no of lead-in and cable currently housed)");
HttpUploadFile(url, file, paramName, contentType, nvc);
}