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
对于我来说,下面的作品(主要是受到以下所有回答的启发),我从Elad的回答开始,修改/简化事情以符合我的需要(删除不是文件形式输入,只有一个文件,……)
希望它能帮助到一些人:)
(PS:我知道异常处理没有实现,并且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();
}
其他回答
不确定这是否张贴之前,但我得到了这个工作与WebClient。我读了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("")和address 不包含绝对URI,地址必须是相对URI那 与BaseAddress结合形成所请求的URI的绝对URI 数据。如果QueryString属性不是空字符串,那么它就是空字符串 附于地址。
我所做的就是wc。querystring。添加(“源”,generatedImage)来添加不同的查询参数,以某种方式将属性名称与我上传的图像匹配。希望能有所帮助
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();
}
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();
}
}
}
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();
}
}
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);
}
}
}
我想你在寻找更像WebClient的东西。
具体来说,还是()。