咨询区

• dr. evil

我不希望通过 WebDAV 文件夹的方式上传文件，我想通过相关的 HttpWebRequest 类，类库或者代码片段来帮我模拟浏览器行为来上传文件，请问是否有好的解决方案？

回答区

• Joshcodes

如果你用的是基于 .NET 4.5 以上的版本，直接从 NuGet 上引用 Microsoft.Net.Http 工具包即可，不需要任何扩展代码，外部代码或者 HTTP 低级别的代码，参考如下例子：

private async Task<System.IO.Stream> UploadAsync(string url, string filename, Stream fileStream, byte [] fileBytes)
{// Convert each of the three inputs into HttpContent objectsHttpContent stringContent = new StringContent(filename);// examples of converting both Stream and byte [] to HttpContent objects// representing input type fileHttpContent 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 successif (!response.IsSuccessStatusCode){return null;}return await response.Content.ReadAsStreamAsync();}
}

• Stefan

关于文件上传，除了单文件，我的类库中还支持多文件上传，无本地文件的流上传，参考下面代码：

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 fileusing(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 streamwhile((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();}}
}

然后像下面这样使用。

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 } },});

点评区

这是一个老问题了，不过谈灵活莫过于 HttpWebRequest。