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

VB示例(从另一篇文章中的c#示例转换而来):

Private Sub HttpUploadFile( _
    ByVal uri As String, _
    ByVal filePath As String, _
    ByVal fileParameterName As String, _
    ByVal contentType As String, _
    ByVal otherParameters As Specialized.NameValueCollection)

    Dim boundary As String = "---------------------------" & DateTime.Now.Ticks.ToString("x")
    Dim newLine As String = System.Environment.NewLine
    Dim boundaryBytes As Byte() = Text.Encoding.ASCII.GetBytes(newLine & "--" & boundary & newLine)
    Dim request As Net.HttpWebRequest = Net.WebRequest.Create(uri)

    request.ContentType = "multipart/form-data; boundary=" & boundary
    request.Method = "POST"
    request.KeepAlive = True
    request.Credentials = Net.CredentialCache.DefaultCredentials

    Using requestStream As IO.Stream = request.GetRequestStream()

        Dim formDataTemplate As String = "Content-Disposition: form-data; name=""{0}""{1}{1}{2}"

        For Each key As String In otherParameters.Keys

            requestStream.Write(boundaryBytes, 0, boundaryBytes.Length)
            Dim formItem As String = String.Format(formDataTemplate, key, newLine, otherParameters(key))
            Dim formItemBytes As Byte() = Text.Encoding.UTF8.GetBytes(formItem)
            requestStream.Write(formItemBytes, 0, formItemBytes.Length)

        Next key

        requestStream.Write(boundaryBytes, 0, boundaryBytes.Length)

        Dim headerTemplate As String = "Content-Disposition: form-data; name=""{0}""; filename=""{1}""{2}Content-Type: {3}{2}{2}"
        Dim header As String = String.Format(headerTemplate, fileParameterName, filePath, newLine, contentType)
        Dim headerBytes As Byte() = Text.Encoding.UTF8.GetBytes(header)
        requestStream.Write(headerBytes, 0, headerBytes.Length)

        Using fileStream As New IO.FileStream(filePath, IO.FileMode.Open, IO.FileAccess.Read)

            Dim buffer(4096) As Byte
            Dim bytesRead As Int32 = fileStream.Read(buffer, 0, buffer.Length)

            Do While (bytesRead > 0)

                requestStream.Write(buffer, 0, bytesRead)
                bytesRead = fileStream.Read(buffer, 0, buffer.Length)

            Loop

        End Using

        Dim trailer As Byte() = Text.Encoding.ASCII.GetBytes(newLine & "--" + boundary + "--" & newLine)
        requestStream.Write(trailer, 0, trailer.Length)

    End Using

    Dim response As Net.WebResponse = Nothing

    Try

        response = request.GetResponse()

        Using responseStream As IO.Stream = response.GetResponseStream()

            Using responseReader As New IO.StreamReader(responseStream)

                Dim responseText = responseReader.ReadToEnd()
                Diagnostics.Debug.Write(responseText)

            End Using

        End Using

    Catch exception As Net.WebException

        response = exception.Response

        If (response IsNot Nothing) Then

            Using reader As New IO.StreamReader(response.GetResponseStream())

                Dim responseText = reader.ReadToEnd()
                Diagnostics.Debug.Write(responseText)

            End Using

            response.Close()

        End If

    Finally

        request = Nothing

    End Try

End Sub

其他回答

我想你在寻找更像WebClient的东西。

具体来说,还是()。

基于上面提供的代码,我添加了对多个文件的支持,也可以直接上传流,而不需要有本地文件。

要将文件上传到包含一些post参数的特定url,请执行以下操作:

RequestHelper.PostMultipart(
    "http://www.myserver.com/upload.php", 
    new Dictionary<string, object>() {
        { "testparam", "my value" },
        { "file", new FormFile() { Name = "image.jpg", ContentType = "image/jpeg", FilePath = "c:\\temp\\myniceimage.jpg" } },
        { "other_file", new FormFile() { Name = "image2.jpg", ContentType = "image/jpeg", Stream = imageDataStream } },
    });

为了进一步增强这一点,可以从给定的文件本身确定名称和mime类型。

public class FormFile 
{
    public string Name { get; set; }

    public string ContentType { get; set; }

    public string FilePath { get; set; }

    public Stream Stream { get; set; }
}

public class RequestHelper
{

    public static string PostMultipart(string url, Dictionary<string, object> parameters) {

        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
        request.ContentType = "multipart/form-data; boundary=" + boundary;
        request.Method = "POST";
        request.KeepAlive = true;
        request.Credentials = System.Net.CredentialCache.DefaultCredentials;

        if(parameters != null && parameters.Count > 0) {

            using(Stream requestStream = request.GetRequestStream()) {

                foreach(KeyValuePair<string, object> pair in parameters) {

                    requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
                    if(pair.Value is FormFile) {
                        FormFile file = pair.Value as FormFile;
                        string header = "Content-Disposition: form-data; name=\"" + pair.Key + "\"; filename=\"" + file.Name + "\"\r\nContent-Type: " + file.ContentType + "\r\n\r\n";
                        byte[] bytes = System.Text.Encoding.UTF8.GetBytes(header);
                        requestStream.Write(bytes, 0, bytes.Length);
                        byte[] buffer = new byte[32768];
                        int bytesRead;
                        if(file.Stream == null) {
                            // upload from file
                            using(FileStream fileStream = File.OpenRead(file.FilePath)) {
                                while((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
                                    requestStream.Write(buffer, 0, bytesRead);
                                fileStream.Close();
                            }
                        }
                        else {
                            // upload from given stream
                            while((bytesRead = file.Stream.Read(buffer, 0, buffer.Length)) != 0)
                                requestStream.Write(buffer, 0, bytesRead);
                        }
                    }
                    else {
                        string data = "Content-Disposition: form-data; name=\"" + pair.Key + "\"\r\n\r\n" + pair.Value;
                        byte[] bytes = System.Text.Encoding.UTF8.GetBytes(data);
                        requestStream.Write(bytes, 0, bytes.Length);
                    }
                }

                byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
                requestStream.Write(trailer, 0, trailer.Length);
                requestStream.Close();
            }
        }

        using(WebResponse response = request.GetResponse()) {
            using(Stream responseStream = response.GetResponseStream())
            using(StreamReader reader = new StreamReader(responseStream))
                return reader.ReadToEnd();
        }


    }
}

我想在VB中做文件上传和添加一些参数到multipart/form-data请求。NET而不是通过正规的表单发布。 感谢@JoshCodes的回答,我找到了我一直在寻找的方向。 我发布我的解决方案是为了帮助其他人找到一种方法来使用文件和参数来执行帖子 html等价于我试图实现的是: 超文本标记语言

<form action="your-api-endpoint" enctype="multipart/form-data" method="post"> 
<input type="hidden" name="action" value="api-method-name"/> 
<input type="hidden" name="apiKey" value="gs1xxxxxxxxxxxxxex"/> 
<input type="hidden" name="access" value="protected"/> 
<input type="hidden" name="name" value="test"/> 
<input type="hidden" name="title" value="test"/> 
<input type="hidden" name="signature" value="cf1d4xxxxxxxxcd5"/> 
<input type="file" name="file"/> 
<input type="submit" name="_upload" value="Upload"/> 
</form>

Due to the fact that I have to provide the apiKey and the signature (which is a calculated checksum of the request parameters and api key concatenated string), I needed to do it server side. The other reason I needed to do it server side is the fact that the post of the file can be performed at any time by pointing to a file already on the server (providing the path), so there would be no manually selected file during form post thus form data file would not contain the file stream.Otherwise I could have calculated the checksum via an ajax callback and submitted the file through the html post using JQuery. I am using .net version 4.0 and cannot upgrade to 4.5 in the actual solution. So I had to install the Microsoft.Net.Http using nuget cmd

PM> install-package Microsoft.Net.Http

Private Function UploadFile(req As ApiRequest, filePath As String, fileName As String) As String
    Dim result = String.empty
    Try
        ''//Get file stream
        Dim paramFileStream As Stream = File.OpenRead(filePath)
        Dim fileStreamContent As HttpContent = New  StreamContent(paramFileStream)
        Using client = New HttpClient()
            Using formData = New MultipartFormDataContent()
                ''// This adds parameter name ("action")
                ''// parameter value (req.Action) to form data
                formData.Add(New StringContent(req.Action), "action")
                formData.Add(New StringContent(req.ApiKey), "apiKey")
                For Each param In req.Parameters
                    formData.Add(New StringContent(param.Value), param.Key)
                Next
                formData.Add(New StringContent(req.getRequestSignature.Qualifier), "signature")
                ''//This adds the file stream and file info to form data
                formData.Add(fileStreamContent, "file", fileName)
                ''//We are now sending the request
                Dim response = client.PostAsync(GetAPIEndpoint(), formData).Result
                ''//We are here reading the response
                Dim readR = New StreamReader(response.Content.ReadAsStreamAsync().Result, Encoding.UTF8)
                Dim respContent = readR.ReadToEnd()

                If Not response.IsSuccessStatusCode Then
                    result =  "Request Failed : Code = " & response.StatusCode & "Reason = " & response.ReasonPhrase & "Message = " & respContent
                End If
                result.Value = respContent
            End Using
        End Using
    Catch ex As Exception
        result = "An error occurred : " & ex.Message
    End Try

    Return result
End Function

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!

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();