有没有任何一个类、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文件夹或类似的地方。我想模拟一个浏览器，就像你将你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无法满足我的要求，因此我正在寻找一个使用HTTPWebrequest的解决方案。
使用HTTPWebrequest上传文件（multipart/form-data）
321
- Slough
3
12这里有一篇关于此事的绝佳博客文章 - http://www.paraesthesia.com/archive/2009/12/16/posting-multipartform-data-using-.net-webrequest.aspx - hwiechers
1@hwiechers：这对我真的很有用，不像其他答案。我只需要添加我的cookies。 - Tim Schmelter
1如果您使用的是.NET >= 4.0，则可以查看此答案以获取不需要自定义对象的解决方案。 - Joshcodes
22个回答
263
参考了上述代码并进行了修改，因为它会抛出500内部服务器错误。其中存在\r\n位置错误和空格等问题。使用内存流进行重构，直接向请求流中写入。以下是结果：
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;
}
}
并且示例用法:
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，或者针对每个文件调用多次。但是根据您的需求进行选择。
- Cristian Romanescu
16
6我尝试了这段代码，但它无法上传JPEG文件，也没有报错？这怎么可能。 - Orhan Cinar
2我添加了一个wr.CookieContainer来保留之前调用的cookies。 - JoaquinG
8如果你想要将其扩展以处理多个文件，请注意：只有最后一个边界会获得额外的两个破折号：" \r\n--" + boundary + "--\r\n"。否则其他附加文件将被截断。 - Peter Drier
1+1. 附注：如果您在使用紧凑框架时使用此功能，则需要将Web请求AllowWriteStreamBuffering设置为true; - Mitch Wheat
1这个程序能正确地转义吗？特别是
name=\"{0}\" 部分。如果我有一个带引号的字符串，这个程序还能正常工作吗？例如 <input name='string"with"quotes' />不过一般来说，name部分通常很简单，很少会出现引号。所以对于大多数正常情况，不会有任何问题。 - mithuntnt显示剩余11条评论
156
我正在寻找像这样的东西，找到了：http://bytes.com/groups/net-c/268661-how-upload-file-via-c-code（已修改以确保正确性）：
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();
}
}
- Slough
13
9FYI...你可以重构掉中间的MemoryStream，直接写入请求流。关键是确保在完成后关闭请求流，这将自动设置请求体的内容长度！ - John Clayton
5当我删除了一个额外的空格后，那个方法对我有效了。"r\n Content-Type: application/octet-stream" 需要改为 "\r\nContent-Type: application/octet-stream"。 - Karl B
1我还发现在头部末尾的双重\r\n可能会导致问题。删除其中一个就解决了我的问题。 - Hugo Estrada
2好的，这段代码对我来说不起作用，但是Christian的代码第一次就完美地运行了 - https://dev59.com/NHRB5IYBdhLWcg3wn4jc#2996904 -- 我是在http://cgi-lib.berkeley.edu/ex/fup.html上进行测试的。 - CVertex
3我知道这是一个早已回答的问题，但我本周才尝试做这个。使用当前的.NET框架，您只需要3行代码即可完成所有操作...WebClient client = new WebClient();
byte[] responseBinary = client.UploadFile(url, file);
string result = Encoding.UTF8.GetString(responseBinary); - soutarm
显示剩余8条评论
118
这是可能的，而无需外部代码、扩展或“低级”HTTP操作（只需要从NuGet获取Microsoft.Net.Http包）。以下是一个示例：
// 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();
}
}
- Joshcodes
18
2可以使用Microsoft.Net.Http NuGet包与4.0一起使用。请参见：https://dev59.com/qGgu5IYBdhLWcg3wso-j。 - amolbk
2这最终成为了一种非常简单的方法，可以做一些相当强大的事情，包括为每个表单部分设置自定义标头。 - JasonRShaver
11@php-jquery-programmer，这是通用的示例代码，因此参数具有通用名称。将“param1”视为“your_well_named_param_here”，请重新考虑您的-1。 - Joshcodes
3您建议使用param1的替代方案是什么？ - Joshcodes
2给我一个建议，将其更改为“bro”。“文件名”对你有用吗？ - Joshcodes
显示剩余13条评论
18
根据上面提供的代码，我添加了对多个文件的支持，并且可以直接上传流而不需要本地文件。
要将文件上传到特定的URL（包括一些POST参数），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 } },
});
为了进一步提高效果，可以从给定的文件本身确定文件名和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();
}
}
}
- Stefan
2
如果内容类型为multipart/related，会有任何改变吗？ - Som Bhattacharyya
非常好的答案。我使用相同的代码成功上传了音频文件。谢谢伙计。 - Naveed Khan
16
我的ASP.NET上传常见问题解答中有一篇文章涵盖了这个话题，包括示例代码：使用HttpWebRequest/WebClient进行RFC 1867 POST请求上传文件。与上面的代码相比，这段代码不会将文件加载到内存中，支持上传多个文件、支持设置表单值、设置凭据和cookie等。
编辑：看起来Axosoft已经将该页面下线了。谢谢你们。
可以通过archive.org继续访问该页面。
- Chris Hynes
4
谢谢你的链接，Chris。我已经将另一个链接实现到我的库中，并添加了支持（除了内存之外）。同时已经转换为VB.NET :) - dr. evil
谢谢，Chris。这帮了我很多！ - flipdoubt
第一流的解决方案！非常感谢。 - Bob Denny
谢谢你提醒我！不幸的是，我无法控制那个网站。我在archive.org上找到了该页面（包括源代码），并相应地更新了链接。 - Chris Hynes
13
类似这样的代码已经接近实现:（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();
- Moose
1
谢谢，但我不需要WebDAV或类似的解决方案，我已经澄清了我的回答。请查看编辑。 - dr. evil
8
VB示例（从另一篇帖子上转换的C#示例）：
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
- Keith Walton
1
使用这段代码在向Net Core控制器发帖时，总是得到一个空文件。 public async Task<IActionResult> Savefile(IFormFile file) - mrapi
7
我认为你需要的更像是WebClient。
具体来说，是UploadFile()。
- John T
1
2它应该使用HTTPWebrequest, 我知道WebClient，但对于这个项目来说不好用。 - dr. evil
6
取上述内容并进行修改，接受一些标题值和多个文件。
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;
}
}
- mopenstein
1
直到我修改了最后一个边界条目，这才对我起作用。确保最后一个文件之后的边界以两个破折号结尾
\r\n--" + boundary + "--\r\n。 - Ben Ripley4
我最近也遇到了这个问题 - 另一种处理方法是利用WebClient是可继承的事实，并从那里更改基础的WebRequest：
http://msdn.microsoft.com/en-us/library/system.net.webclient.getwebrequest(VS.80).aspx
我更喜欢C#，但如果你被迫使用VB，结果可能会像这样：
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
- Chris
1
+1 仍然，Web客户端过于不可定制化，因此实现它会很尴尬，但这是一个非常有趣的方法，我不知道它是可能的。 - dr. evil
网页内容由stack overflow 提供, 点击上面的可以查看英文原文，
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