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
客户端使用转换文件到ToBase64String,使用Xml发布后 到服务器调用,这个服务器使用File.WriteAllBytes(path,Convert.FromBase64String(dataFile_Client_sent))。
好幸运!
其他回答
我最近不得不处理这个问题——另一种方法是使用WebClient是可继承的这一事实,并从那里改变底层的WebRequest:
http://msdn.microsoft.com/en-us/library/system.net.webclient.getwebrequest (VS.80) . aspx
我更喜欢c#,但如果你坚持使用VB,结果将是这样的:
Public Class BigWebClient
Inherits WebClient
Protected Overrides Function GetWebRequest(ByVal address As System.Uri) As System.Net.WebRequest
Dim x As WebRequest = MyBase.GetWebRequest(address)
x.Timeout = 60 * 60 * 1000
Return x
End Function
End Class
'Use BigWebClient here instead of WebClient
类似这样的代码很接近:(未测试的代码)
byte[] data; // data goes here.
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
request.Credentials = userNetworkCredentials;
request.Method = "PUT";
request.ContentType = "application/octet-stream";
request.ContentLength = data.Length;
Stream stream = request.GetRequestStream();
stream.Write(data,0,data.Length);
stream.Close();
response = (HttpWebResponse)request.GetResponse();
StreamReader reader = new StreamReader(response.GetResponseStream());
temp = reader.ReadToEnd();
reader.Close();
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);
}
}
}
不确定这是否张贴之前,但我得到了这个工作与WebClient。我读了WebClient的文档。他们提出的一个关键点是
如果BaseAddress属性不是空字符串("")和address 不包含绝对URI,地址必须是相对URI那 与BaseAddress结合形成所请求的URI的绝对URI 数据。如果QueryString属性不是空字符串,那么它就是空字符串 附于地址。
我所做的就是wc。querystring。添加(“源”,generatedImage)来添加不同的查询参数,以某种方式将属性名称与我上传的图像匹配。希望能有所帮助
public void postImageToFacebook(string generatedImage, string fbGraphUrl)
{
WebClient wc = new WebClient();
byte[] bytes = System.IO.File.ReadAllBytes(generatedImage);
wc.QueryString.Add("source", generatedImage);
wc.QueryString.Add("message", "helloworld");
wc.UploadFile(fbGraphUrl, generatedImage);
wc.Dispose();
}
这不需要外部代码、扩展和“低级”HTTP操作(只需要NuGet中的Microsoft.Net.Http包)。这里有一个例子:
// Perform the equivalent of posting a form with a filename and two files, in HTML:
// <form action="{url}" method="post" enctype="multipart/form-data">
// <input type="text" name="filename" />
// <input type="file" name="file1" />
// <input type="file" name="file2" />
// </form>
private async Task<System.IO.Stream> UploadAsync(string url, string filename, Stream fileStream, byte [] fileBytes)
{
// Convert each of the three inputs into HttpContent objects
HttpContent stringContent = new StringContent(filename);
// examples of converting both Stream and byte [] to HttpContent objects
// representing input type file
HttpContent fileStreamContent = new StreamContent(fileStream);
HttpContent bytesContent = new ByteArrayContent(fileBytes);
// Submit the form using HttpClient and
// create form data as Multipart (enctype="multipart/form-data")
using (var client = new HttpClient())
using (var formData = new MultipartFormDataContent())
{
// Add the HttpContent objects to the form data
// <input type="text" name="filename" />
formData.Add(stringContent, "filename", "filename");
// <input type="file" name="file1" />
formData.Add(fileStreamContent, "file1", "file1");
// <input type="file" name="file2" />
formData.Add(bytesContent, "file2", "file2");
// Invoke the request to the server
// equivalent to pressing the submit button on
// a form with attributes (action="{url}" method="post")
var response = await client.PostAsync(url, formData);
// ensure the request was a success
if (!response.IsSuccessStatusCode)
{
return null;
}
return await response.Content.ReadAsStreamAsync();
}
}