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
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
其他回答
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();
}
}
}
我的ASP。NET上传常见问题解答中有一篇关于这方面的文章,有示例代码:使用HttpWebRequest/WebClient的RFC 1867 POST请求上传文件。此代码不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:看起来好像是Axosoft把这个页面删除了。谢谢你的家伙。
它仍然可以通过archive.org访问。
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);
}
}
}
该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;
}
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);
http://mytoolkit.codeplex.com/wikipage?title=Http