是否有任何类,库或一些代码片段,将帮助我上传文件与HTTPWebrequest?
编辑2:
我不想上传到WebDAV文件夹或类似的东西。我想模拟一个浏览器,就像你上传你的头像到一个论坛或通过一个web应用程序中的表单上传一个文件。上传到一个使用multipart/form-data的表单。
编辑:
WebClient不覆盖我的需求,所以我正在寻找一个解决方案与HTTPWebrequest。
是否有任何类,库或一些代码片段,将帮助我上传文件与HTTPWebrequest?
编辑2:
我不想上传到WebDAV文件夹或类似的东西。我想模拟一个浏览器,就像你上传你的头像到一个论坛或通过一个web应用程序中的表单上传一个文件。上传到一个使用multipart/form-data的表单。
编辑:
WebClient不覆盖我的需求,所以我正在寻找一个解决方案与HTTPWebrequest。
当前回答
我永远不能让例子正常工作,我总是收到一个500错误时,把它发送到服务器。
然而,我在这个url中遇到了一个非常优雅的方法
它很容易扩展,显然可以处理二进制文件和XML。
你可以用类似的方法来称呼它
class Program
{
public static string gsaFeedURL = "http://yourGSA.domain.com:19900/xmlfeed";
static void Main()
{
try
{
postWebData();
}
catch (Exception ex)
{
}
}
// new one I made from C# web service
public static void postWebData()
{
StringDictionary dictionary = new StringDictionary();
UploadSpec uploadSpecs = new UploadSpec();
UTF8Encoding encoding = new UTF8Encoding();
byte[] bytes;
Uri gsaURI = new Uri(gsaFeedURL); // Create new URI to GSA feeder gate
string sourceURL = @"C:\FeedFile.xml"; // Location of the XML feed file
// Two parameters to send
string feedtype = "full";
string datasource = "test";
try
{
// Add the parameter values to the dictionary
dictionary.Add("feedtype", feedtype);
dictionary.Add("datasource", datasource);
// Load the feed file created and get its bytes
XmlDocument xml = new XmlDocument();
xml.Load(sourceURL);
bytes = Encoding.UTF8.GetBytes(xml.OuterXml);
// Add data to upload specs
uploadSpecs.Contents = bytes;
uploadSpecs.FileName = sourceURL;
uploadSpecs.FieldName = "data";
// Post the data
if ((int)HttpUpload.Upload(gsaURI, dictionary, uploadSpecs).StatusCode == 200)
{
Console.WriteLine("Successful.");
}
else
{
// GSA POST not successful
Console.WriteLine("Failure.");
}
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
}
}
其他回答
该方法适用于同时上传多张图片
var flagResult = new viewModel();
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
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 = method;
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream rs = wr.GetRequestStream();
string path = @filePath;
System.IO.DirectoryInfo folderInfo = new DirectoryInfo(path);
foreach (FileInfo file in folderInfo.GetFiles())
{
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.FullName, 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);
var result = reader2.ReadToEnd();
var cList = JsonConvert.DeserializeObject<HttpViewModel>(result);
if (cList.message=="images uploaded!")
{
flagResult.success = true;
}
}
catch (Exception ex)
{
//log.Error("Error uploading file", ex);
if (wresp != null)
{
wresp.Close();
wresp = null;
}
}
finally
{
wr = null;
}
return flagResult;
}
我永远不能让例子正常工作,我总是收到一个500错误时,把它发送到服务器。
然而,我在这个url中遇到了一个非常优雅的方法
它很容易扩展,显然可以处理二进制文件和XML。
你可以用类似的方法来称呼它
class Program
{
public static string gsaFeedURL = "http://yourGSA.domain.com:19900/xmlfeed";
static void Main()
{
try
{
postWebData();
}
catch (Exception ex)
{
}
}
// new one I made from C# web service
public static void postWebData()
{
StringDictionary dictionary = new StringDictionary();
UploadSpec uploadSpecs = new UploadSpec();
UTF8Encoding encoding = new UTF8Encoding();
byte[] bytes;
Uri gsaURI = new Uri(gsaFeedURL); // Create new URI to GSA feeder gate
string sourceURL = @"C:\FeedFile.xml"; // Location of the XML feed file
// Two parameters to send
string feedtype = "full";
string datasource = "test";
try
{
// Add the parameter values to the dictionary
dictionary.Add("feedtype", feedtype);
dictionary.Add("datasource", datasource);
// Load the feed file created and get its bytes
XmlDocument xml = new XmlDocument();
xml.Load(sourceURL);
bytes = Encoding.UTF8.GetBytes(xml.OuterXml);
// Add data to upload specs
uploadSpecs.Contents = bytes;
uploadSpecs.FileName = sourceURL;
uploadSpecs.FieldName = "data";
// Post the data
if ((int)HttpUpload.Upload(gsaURI, dictionary, uploadSpecs).StatusCode == 200)
{
Console.WriteLine("Successful.");
}
else
{
// GSA POST not successful
Console.WriteLine("Failure.");
}
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
}
}
我想你在寻找更像WebClient的东西。
具体来说,还是()。
这里有另一个我的评论的工作示例:
List<MimePart> mimeParts = new List<MimePart>();
try
{
foreach (string key in form.AllKeys)
{
StringMimePart part = new StringMimePart();
part.Headers["Content-Disposition"] = "form-data; name=\"" + key + "\"";
part.StringData = form[key];
mimeParts.Add(part);
}
int nameIndex = 0;
foreach (UploadFile file in files)
{
StreamMimePart part = new StreamMimePart();
if (string.IsNullOrEmpty(file.FieldName))
file.FieldName = "file" + nameIndex++;
part.Headers["Content-Disposition"] = "form-data; name=\"" + file.FieldName + "\"; filename=\"" + file.FileName + "\"";
part.Headers["Content-Type"] = file.ContentType;
part.SetStream(file.Data);
mimeParts.Add(part);
}
string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
req.ContentType = "multipart/form-data; boundary=" + boundary;
req.Method = "POST";
long contentLength = 0;
byte[] _footer = Encoding.UTF8.GetBytes("--" + boundary + "--\r\n");
foreach (MimePart part in mimeParts)
{
contentLength += part.GenerateHeaderFooterData(boundary);
}
req.ContentLength = contentLength + _footer.Length;
byte[] buffer = new byte[8192];
byte[] afterFile = Encoding.UTF8.GetBytes("\r\n");
int read;
using (Stream s = req.GetRequestStream())
{
foreach (MimePart part in mimeParts)
{
s.Write(part.Header, 0, part.Header.Length);
while ((read = part.Data.Read(buffer, 0, buffer.Length)) > 0)
s.Write(buffer, 0, read);
part.Data.Dispose();
s.Write(afterFile, 0, afterFile.Length);
}
s.Write(_footer, 0, _footer.Length);
}
return (HttpWebResponse)req.GetResponse();
}
catch
{
foreach (MimePart part in mimeParts)
if (part.Data != null)
part.Data.Dispose();
throw;
}
这里有一个使用的例子:
UploadFile[] files = new UploadFile[]
{
new UploadFile(@"C:\2.jpg","new_file","image/jpeg") //new_file is id of upload field
};
NameValueCollection form = new NameValueCollection();
form["id_hidden_input"] = "value_hidden_inpu"; //there is additional param (hidden fields on page)
HttpWebRequest req = (HttpWebRequest)WebRequest.Create(full URL of action);
// set credentials/cookies etc.
req.CookieContainer = hrm.CookieContainer; //hrm is my class. i copied all cookies from last request to current (for auth)
HttpWebResponse resp = HttpUploadHelper.Upload(req, files, form);
using (Stream s = resp.GetResponseStream())
using (StreamReader sr = new StreamReader(s))
{
string response = sr.ReadToEnd();
}
//profit!
查看MyToolkit库:
var request = new HttpPostRequest("http://www.server.com");
request.Data.Add("name", "value"); // POST data
request.Files.Add(new HttpPostFile("name", "file.jpg", "path/to/file.jpg"));
await Http.PostAsync(request, OnRequestFinished);
http://mytoolkit.codeplex.com/wikipage?title=Http