是否有任何类、库或一些代码可以帮助我使用HTTPWebrequest上传文件？
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，就像您将您的头像上传到论坛或通过 Web 应用程序中的表单上传文件一样。上传到使用 multipart/form-data 的表单。
编辑：
WebClient 不能满足我的要求，所以我正在寻找HTTPWebrequest的解决方案。
Slough
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435613 次
21 回答
259
采用上面的代码并进行修复，因为它会引发内部服务器错误 500。\r\n 定位错误和空格等存在一些问题。使用内存流应用重构，直接写入请求流。结果如下：
public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc) {
log.Debug(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");
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
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);
log.Debug(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
} catch(Exception ex) {
log.Error("Error uploading file", ex);
if(wresp != null) {
wresp.Close();
wresp = null;
}
} finally {
wr = null;
}
}
和示例用法：
NameValueCollection nvc = new NameValueCollection();
nvc.Add("id", "TTR");
nvc.Add("btn-submit-photo", "Upload");
HttpUploadFile("http://your.server.com/upload",
@"C:\test\test.jpg", "file", "image/jpeg", nvc);
它可以扩展为处理多个文件，或者为每个文件多次调用它。但是，它适合您的需求。
于 2010-06-08T11:39:08.080 回答
152
我正在寻找这样的东西，发现于： http ://bytes.com/groups/net-c/268661-how-upload-file-via-c-code （为了正确性而修改）：
public static string UploadFilesToRemoteUrl(string url, string[] files, NameValueCollection formFields = null)
{
string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
HttpWebRequest request = (HttpWebRequest) WebRequest.Create(url);
request.ContentType = "multipart/form-data; boundary=" +
boundary;
request.Method = "POST";
request.KeepAlive = true;
Stream memStream = new System.IO.MemoryStream();
var boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" +
boundary + "\r\n");
var endBoundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" +
boundary + "--");
string formdataTemplate = "\r\n--" + boundary +
"\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
if (formFields != null)
{
foreach (string key in formFields.Keys)
{
string formitem = string.Format(formdataTemplate, key, formFields[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
memStream.Write(formitembytes, 0, formitembytes.Length);
}
}
string headerTemplate =
"Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n" +
"Content-Type: application/octet-stream\r\n\r\n";
for (int i = 0; i < files.Length; i++)
{
memStream.Write(boundarybytes, 0, boundarybytes.Length);
var header = string.Format(headerTemplate, "uplTheFile", files[i]);
var headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
memStream.Write(headerbytes, 0, headerbytes.Length);
using (var fileStream = new FileStream(files[i], FileMode.Open, FileAccess.Read))
{
var buffer = new byte[1024];
var bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
memStream.Write(buffer, 0, bytesRead);
}
}
}
memStream.Write(endBoundaryBytes, 0, endBoundaryBytes.Length);
request.ContentLength = memStream.Length;
using (Stream requestStream = request.GetRequestStream())
{
memStream.Position = 0;
byte[] tempBuffer = new byte[memStream.Length];
memStream.Read(tempBuffer, 0, tempBuffer.Length);
memStream.Close();
requestStream.Write(tempBuffer, 0, tempBuffer.Length);
}
using (var response = request.GetResponse())
{
Stream stream2 = response.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
return reader2.ReadToEnd();
}
}
于 2009-02-19T22:11:13.060 回答
112
更新：使用 .NET 4.5（或通过添加来自 NuGet 的Microsoft.Net.Http包的 .NET 4.0），无需外部代码、扩展和“低级”HTTP 操作即可实现。这是一个例子：
// 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();
}
}
于 2013-06-04T18:40:38.840 回答
18
基于上面提供的代码，我添加了对多个文件的支持，并且还直接上传流而不需要本地文件。
要将文件上传到包含一些帖子参数的特定 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();
}
}
}
于 2013-04-20T18:39:59.007 回答
16
我的 ASP.NET 上传常见问题解答有一篇关于此的文章，带有示例代码：使用 RFC 1867 POST 请求和 HttpWebRequest/WebClient 上传文件。此代码不会将文件加载到内存中（与上面的代码相反），支持多个文件，并支持表单值、设置凭据和 cookie 等。
编辑：看起来 Axosoft 删除了该页面。多谢你们。
它仍然可以通过archive.org 访问。
于 2009-04-23T17:34:46.290 回答
13
像这样的东西很接近：（未经测试的代码）
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();
于 2009-02-19T18:06:55.697 回答
7
我认为您正在寻找更像WebClient的东西。
具体来说，UploadFile()。
于 2009-02-19T18:05:40.330 回答
6
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
于 2010-11-22T23:36:47.357 回答
6
接受上述内容并对其进行修改，以接受一些标头值和多个文件
NameValueCollection headers = new NameValueCollection();
headers.Add("Cookie", "name=value;");
headers.Add("Referer", "http://google.com");
NameValueCollection nvc = new NameValueCollection();
nvc.Add("name", "value");
HttpUploadFile(url, new string[] { "c:\\file1.txt", "c:\\file2.jpg" }, new string[] { "file", "image" }, new string[] { "application/octet-stream", "image/jpeg" }, nvc, headers);
public static void HttpUploadFile(string url, string[] file, string[] paramName, string[] contentType, NameValueCollection nvc, NameValueCollection headerItems)
{
//log.Debug(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");
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
foreach (string key in headerItems.Keys)
{
if (key == "Referer")
{
wr.Referer = headerItems[key];
}
else
{
wr.Headers.Add(key, headerItems[key]);
}
}
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
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 = "";
for(int i =0; i<file.Count();i++)
{
header = string.Format(headerTemplate, paramName[i], System.IO.Path.GetFileName(file[i]), contentType[i]);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(file[i], 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();
rs.Write(boundarybytes, 0, boundarybytes.Length);
}
rs.Close();
WebResponse wresp = null;
try
{
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
//log.Debug(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
}
catch (Exception ex)
{
//log.Error("Error uploading file", ex);
wresp.Close();
wresp = null;
}
finally
{
wr = null;
}
}
于 2012-01-10T04:05:09.943 回答
4
我最近不得不处理这个问题 - 另一种方法是使用 WebClient 是可继承的事实，并从那里更改底层 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，结果将如下所示：
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
于 2009-04-24T18:58:14.430 回答
3
我的一些评论还有另一个工作示例：
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;
}
还有使用的例子：
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!
于 2012-01-16T11:31:39.857 回答
2
我正在寻找文件上传并将一些参数添加到 VB.NET 中的 multipart/form-data 请求，而不是通过常规表单发布。感谢@JoshCodes 的回答，我得到了我正在寻找的方向。我正在发布我的解决方案，以帮助其他人找到一种方法来使用文件和参数执行帖子，与我尝试实现的 html 等效的是：html
<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>
由于我必须提供 apiKey 和签名（这是请求参数和 api 密钥连接字符串的计算校验和），我需要在服务器端进行。我需要在服务器端执行此操作的另一个原因是，可以通过指向服务器上已经存在的文件（提供路径）随时执行文件的发布，因此在表单期间不会手动选择文件post 因此表单数据文件将不包含文件流。否则我可以通过 ajax 回调计算校验和并使用 JQuery 通过 html post 提交文件。我使用的是 .net 4.0 版，在实际解决方案中无法升级到 4.5。所以我不得不使用 nuget cmd 安装 Microsoft.Net.Http
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
于 2015-03-27T12:19:08.137 回答
2
修改了@CristianRomanescu 代码以使用内存流，接受文件作为字节数组，允许空 nvc，返回请求响应并使用 Authorization-header。使用 Web Api 2 测试了代码。
private string HttpUploadFile(string url, byte[] file, string fileName, string paramName, string contentType, NameValueCollection nvc, string authorizationHeader)
{
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 = "POST";
wr.Headers.Add("Authorization", authorizationHeader);
wr.KeepAlive = true;
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
if (nvc != null)
{
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, fileName, contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
rs.Write(file, 0, file.Length);
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 response = reader2.ReadToEnd();
return response;
}
catch (Exception ex)
{
if (wresp != null)
{
wresp.Close();
wresp = null;
}
return null;
}
finally
{
wr = null;
}
}
测试代码：
[HttpPost]
[Route("postformdata")]
public IHttpActionResult PostFormData()
{
// Check if the request contains multipart/form-data.
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
var provider = new MultipartMemoryStreamProvider();
try
{
// Read the form data.
var result = Request.Content.ReadAsMultipartAsync(provider).Result;
string response = "";
// This illustrates how to get the file names.
foreach (var file in provider.Contents)
{
var fileName = file.Headers.ContentDisposition.FileName.Trim('\"');
var buffer = file.ReadAsByteArrayAsync().Result;
response = HttpUploadFile("https://localhost/api/v1/createfromfile", buffer, fileName, "file", "application/pdf", null, "AuthorizationKey");
}
return Ok(response);
}
catch (System.Exception e)
{
return InternalServerError();
}
}
于 2016-04-26T06:15:54.933 回答
1
对我来说，以下作品（主要受以下所有答案的启发），我从 Elad 的答案开始并修改/简化事情以满足我的需要（删除不是文件表单输入，只有一个文件，......）。
希望它可以帮助某人:)
（PS：我知道异常处理没有实现，它假设它是在一个类中编写的，所以我可能需要一些集成工作......）
private void uploadFile()
{
Random rand = new Random();
string boundary = "----boundary" + rand.Next().ToString();
Stream data_stream;
byte[] header = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"file_path\"; filename=\"" + System.IO.Path.GetFileName(this.file) + "\"\r\nContent-Type: application/octet-stream\r\n\r\n");
byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
// Do the request
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(MBF_URL);
request.UserAgent = "My Toolbox";
request.Method = "POST";
request.KeepAlive = true;
request.ContentType = "multipart/form-data; boundary=" + boundary;
data_stream = request.GetRequestStream();
data_stream.Write(header, 0, header.Length);
byte[] file_bytes = System.IO.File.ReadAllBytes(this.file);
data_stream.Write(file_bytes, 0, file_bytes.Length);
data_stream.Write(trailer, 0, trailer.Length);
data_stream.Close();
// Read the response
WebResponse response = request.GetResponse();
data_stream = response.GetResponseStream();
StreamReader reader = new StreamReader(data_stream);
this.url = reader.ReadToEnd();
if (this.url == "") { this.url = "No response :("; }
reader.Close();
data_stream.Close();
response.Close();
}
于 2013-03-07T09:47:59.007 回答
1
不确定这是否之前发布过，但我使用 WebClient 进行了此操作。我阅读了 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 ("") 并且 address 不包含绝对 URI，则 address 必须是与 BaseAddress 组合以形成所请求数据的绝对 URI 的相对 URI。如果 QueryString 属性不是空字符串，则将其附加到地址。
所以我所做的只是 wc.QueryString.Add("source", generatedImage) 添加不同的查询参数，并以某种方式将属性名称与我上传的图像匹配。希望能帮助到你
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();
}
于 2014-02-04T04:48:15.377 回答
1
当进行多部分表单上传时，我使用 WebClient 编写了一个类。
http://ferozedaud.blogspot.com/2010/03/multipart-form-upload-helper.html
///
/// MimePart
/// Abstract class for all MimeParts
///
abstract class MimePart
{
public string Name { get; set; }
public abstract string ContentDisposition { get; }
public abstract string ContentType { get; }
public abstract void CopyTo(Stream stream);
public String Boundary
{
get;
set;
}
}
class NameValuePart : MimePart
{
private NameValueCollection nameValues;
public NameValuePart(NameValueCollection nameValues)
{
this.nameValues = nameValues;
}
public override void CopyTo(Stream stream)
{
string boundary = this.Boundary;
StringBuilder sb = new StringBuilder();
foreach (object element in this.nameValues.Keys)
{
sb.AppendFormat("--{0}", boundary);
sb.Append("\r\n");
sb.AppendFormat("Content-Disposition: form-data; name=\"{0}\";", element);
sb.Append("\r\n");
sb.Append("\r\n");
sb.Append(this.nameValues[element.ToString()]);
sb.Append("\r\n");
}
sb.AppendFormat("--{0}", boundary);
sb.Append("\r\n");
//Trace.WriteLine(sb.ToString());
byte [] data = Encoding.ASCII.GetBytes(sb.ToString());
stream.Write(data, 0, data.Length);
}
public override string ContentDisposition
{
get { return "form-data"; }
}
public override string ContentType
{
get { return String.Empty; }
}
}
class FilePart : MimePart
{
private Stream input;
private String contentType;
public FilePart(Stream input, String name, String contentType)
{
this.input = input;
this.contentType = contentType;
this.Name = name;
}
public override void CopyTo(Stream stream)
{
StringBuilder sb = new StringBuilder();
sb.AppendFormat("Content-Disposition: {0}", this.ContentDisposition);
if (this.Name != null)
sb.Append("; ").AppendFormat("name=\"{0}\"", this.Name);
if (this.FileName != null)
sb.Append("; ").AppendFormat("filename=\"{0}\"", this.FileName);
sb.Append("\r\n");
sb.AppendFormat(this.ContentType);
sb.Append("\r\n");
sb.Append("\r\n");
// serialize the header data.
byte[] buffer = Encoding.ASCII.GetBytes(sb.ToString());
stream.Write(buffer, 0, buffer.Length);
// send the stream.
byte[] readBuffer = new byte[1024];
int read = input.Read(readBuffer, 0, readBuffer.Length);
while (read > 0)
{
stream.Write(readBuffer, 0, read);
read = input.Read(readBuffer, 0, readBuffer.Length);
}
// write the terminating boundary
sb.Length = 0;
sb.Append("\r\n");
sb.AppendFormat("--{0}", this.Boundary);
sb.Append("\r\n");
buffer = Encoding.ASCII.GetBytes(sb.ToString());
stream.Write(buffer, 0, buffer.Length);
}
public override string ContentDisposition
{
get { return "file"; }
}
public override string ContentType
{
get {
return String.Format("content-type: {0}", this.contentType);
}
}
public String FileName { get; set; }
}
///
/// Helper class that encapsulates all file uploads
/// in a mime part.
///
class FilesCollection : MimePart
{
private List files;
public FilesCollection()
{
this.files = new List();
this.Boundary = MultipartHelper.GetBoundary();
}
public int Count
{
get { return this.files.Count; }
}
public override string ContentDisposition
{
get
{
return String.Format("form-data; name=\"{0}\"", this.Name);
}
}
public override string ContentType
{
get { return String.Format("multipart/mixed; boundary={0}", this.Boundary); }
}
public override void CopyTo(Stream stream)
{
// serialize the headers
StringBuilder sb = new StringBuilder(128);
sb.Append("Content-Disposition: ").Append(this.ContentDisposition).Append("\r\n");
sb.Append("Content-Type: ").Append(this.ContentType).Append("\r\n");
sb.Append("\r\n");
sb.AppendFormat("--{0}", this.Boundary).Append("\r\n");
byte[] headerBytes = Encoding.ASCII.GetBytes(sb.ToString());
stream.Write(headerBytes, 0, headerBytes.Length);
foreach (FilePart part in files)
{
part.Boundary = this.Boundary;
part.CopyTo(stream);
}
}
public void Add(FilePart part)
{
this.files.Add(part);
}
}
///
/// Helper class to aid in uploading multipart
/// entities to HTTP web endpoints.
///
class MultipartHelper
{
private static Random random = new Random(Environment.TickCount);
private List formData = new List();
private FilesCollection files = null;
private MemoryStream bufferStream = new MemoryStream();
private string boundary;
public String Boundary { get { return boundary; } }
public static String GetBoundary()
{
return Environment.TickCount.ToString("X");
}
public MultipartHelper()
{
this.boundary = MultipartHelper.GetBoundary();
}
public void Add(NameValuePart part)
{
this.formData.Add(part);
part.Boundary = boundary;
}
public void Add(FilePart part)
{
if (files == null)
{
files = new FilesCollection();
}
this.files.Add(part);
}
public void Upload(WebClient client, string address, string method)
{
// set header
client.Headers.Add(HttpRequestHeader.ContentType, "multipart/form-data; boundary=" + this.boundary);
Trace.WriteLine("Content-Type: multipart/form-data; boundary=" + this.boundary + "\r\n");
// first, serialize the form data
foreach (NameValuePart part in this.formData)
{
part.CopyTo(bufferStream);
}
// serialize the files.
this.files.CopyTo(bufferStream);
if (this.files.Count > 0)
{
// add the terminating boundary.
StringBuilder sb = new StringBuilder();
sb.AppendFormat("--{0}", this.Boundary).Append("\r\n");
byte [] buffer = Encoding.ASCII.GetBytes(sb.ToString());
bufferStream.Write(buffer, 0, buffer.Length);
}
bufferStream.Seek(0, SeekOrigin.Begin);
Trace.WriteLine(Encoding.ASCII.GetString(bufferStream.ToArray()));
byte [] response = client.UploadData(address, method, bufferStream.ToArray());
Trace.WriteLine("----- RESPONSE ------");
Trace.WriteLine(Encoding.ASCII.GetString(response));
}
///
/// Helper class that encapsulates all file uploads
/// in a mime part.
///
class FilesCollection : MimePart
{
private List files;
public FilesCollection()
{
this.files = new List();
this.Boundary = MultipartHelper.GetBoundary();
}
public int Count
{
get { return this.files.Count; }
}
public override string ContentDisposition
{
get
{
return String.Format("form-data; name=\"{0}\"", this.Name);
}
}
public override string ContentType
{
get { return String.Format("multipart/mixed; boundary={0}", this.Boundary); }
}
public override void CopyTo(Stream stream)
{
// serialize the headers
StringBuilder sb = new StringBuilder(128);
sb.Append("Content-Disposition: ").Append(this.ContentDisposition).Append("\r\n");
sb.Append("Content-Type: ").Append(this.ContentType).Append("\r\n");
sb.Append("\r\n");
sb.AppendFormat("--{0}", this.Boundary).Append("\r\n");
byte[] headerBytes = Encoding.ASCII.GetBytes(sb.ToString());
stream.Write(headerBytes, 0, headerBytes.Length);
foreach (FilePart part in files)
{
part.Boundary = this.Boundary;
part.CopyTo(stream);
}
}
public void Add(FilePart part)
{
this.files.Add(part);
}
}
}
class Program
{
static void Main(string[] args)
{
Trace.Listeners.Add(new ConsoleTraceListener());
try
{
using (StreamWriter sw = new StreamWriter("testfile.txt", false))
{
sw.Write("Hello there!");
}
using (Stream iniStream = File.OpenRead(@"c:\platform.ini"))
using (Stream fileStream = File.OpenRead("testfile.txt"))
using (WebClient client = new WebClient())
{
MultipartHelper helper = new MultipartHelper();
NameValueCollection props = new NameValueCollection();
props.Add("fname", "john");
props.Add("id", "acme");
helper.Add(new NameValuePart(props));
FilePart filepart = new FilePart(fileStream, "pics1", "text/plain");
filepart.FileName = "1.jpg";
helper.Add(filepart);
FilePart ini = new FilePart(iniStream, "pics2", "text/plain");
ini.FileName = "inifile.ini";
helper.Add(ini);
helper.Upload(client, "http://localhost/form.aspx", "POST");
}
}
catch (Exception e)
{
Trace.WriteLine(e);
}
}
}
这适用于所有版本的 .NET 框架。
于 2016-09-18T05:20:03.700 回答
0
我永远无法让示例正常工作，将其发送到服务器时总是收到 500 错误。
但是我在这个网址中遇到了一种非常优雅的方法
它易于扩展，并且显然可以与二进制文件以及 XML 一起使用。
你用类似的东西来称呼它
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);
}
}
}
于 2010-06-07T11:02:04.860 回答
0
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);
于 2012-09-08T10:17:23.180 回答
0
客户端使用转换文件到ToBase64String，使用xml发布到服务器调用后，该服务器使用File.WriteAllBytes(path,Convert.FromBase64String(dataFile_Client_sent))。
好幸运！
于 2017-02-14T09:20:02.000 回答
0
此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;
}
于 2020-04-01T09:13:25.730 回答
-1
我意识到这可能真的很晚了，但我正在寻找相同的解决方案。我从 Microsoft 代表那里找到了以下回复
private void UploadFilesToRemoteUrl(string url, string[] files, string logpath, NameValueCollection nvc)
{
long length = 0;
string boundary = "----------------------------" +
DateTime.Now.Ticks.ToString("x");
HttpWebRequest httpWebRequest2 = (HttpWebRequest)WebRequest.Create(url);
httpWebRequest2.ContentType = "multipart/form-data; boundary=" +
boundary;
httpWebRequest2.Method = "POST";
httpWebRequest2.KeepAlive = true;
httpWebRequest2.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream memStream = new System.IO.MemoryStream();
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
foreach(string key in nvc.Keys)
{
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
memStream.Write(formitembytes, 0, formitembytes.Length);
}
memStream.Write(boundarybytes,0,boundarybytes.Length);
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";
for(int i=0;i<files.Length;i++)
{
string header = string.Format(headerTemplate,"file"+i,files[i]);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
memStream.Write(headerbytes,0,headerbytes.Length);
FileStream fileStream = new FileStream(files[i], FileMode.Open,
FileAccess.Read);
byte[] buffer = new byte[1024];
int bytesRead = 0;
while ( (bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0 )
{
memStream.Write(buffer, 0, bytesRead);
}
memStream.Write(boundarybytes,0,boundarybytes.Length);
fileStream.Close();
}
httpWebRequest2.ContentLength = memStream.Length;
Stream requestStream = httpWebRequest2.GetRequestStream();
memStream.Position = 0;
byte[] tempBuffer = new byte[memStream.Length];
memStream.Read(tempBuffer,0,tempBuffer.Length);
memStream.Close();
requestStream.Write(tempBuffer,0,tempBuffer.Length );
requestStream.Close();
WebResponse webResponse2 = httpWebRequest2.GetResponse();
Stream stream2 = webResponse2.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
webResponse2.Close();
httpWebRequest2 = null;
webResponse2 = null;
}
于 2009-12-17T21:46:00.017 回答