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

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();
    }
}

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);
        }
    }
}

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

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!