服务器端:
public HttpResponseMessage Post([FromUri]string machineName)
{
HttpResponseMessage result = null;
var httpRequest = HttpContext.Current.Request;
if (httpRequest.Files.Count > 0 && !String.IsNullOrEmpty(machineName))
...
客户端:
public static void PostFile(string url, string filePath)
{
if (String.IsNullOrWhiteSpace(url) || String.IsNullOrWhiteSpace(filePath))
throw new ArgumentNullException();
if (!File.Exists(filePath))
throw new FileNotFoundException();
using (var handler = new HttpClientHandler { Credentials= new NetworkCredential(AppData.UserName, AppData.Password, AppCore.Domain) })
using (var client = new HttpClient(handler))
using (var content = new MultipartFormDataContent())
using (var ms = new MemoryStream(File.ReadAllBytes(filePath)))
{
var fileContent = new StreamContent(ms);
fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
{
FileName = Path.GetFileName(filePath)
};
content.Add(fileContent);
content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
var result = client.PostAsync(url, content).Result;
result.EnsureSuccessStatusCode();
}
}
在服务器端,httpRequest.Files集合始终为空。但标题(内容长度等)是正确的。
答案 0 :(得分:9)
您不应该使用HttpContext来获取ASP.NET Web API中的文件。看看这个由Microsoft编写的示例(http://code.msdn.microsoft.com/ASPNET-Web-API-File-Upload-a8c0fb0d/sourcecode?fileId=67087&pathId=565875642)。
public class UploadController : ApiController
{
public async Task<HttpResponseMessage> PostFile()
{
// Check if the request contains multipart/form-data.
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = HttpContext.Current.Server.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
try
{
StringBuilder sb = new StringBuilder(); // Holds the response body
// Read the form data and return an async task.
await Request.Content.ReadAsMultipartAsync(provider);
// This illustrates how to get the form data.
foreach (var key in provider.FormData.AllKeys)
{
foreach (var val in provider.FormData.GetValues(key))
{
sb.Append(string.Format("{0}: {1}\n", key, val));
}
}
// This illustrates how to get the file names for uploaded files.
foreach (var file in provider.FileData)
{
FileInfo fileInfo = new FileInfo(file.LocalFileName);
sb.Append(string.Format("Uploaded file: {0} ({1} bytes)\n", fileInfo.Name, fileInfo.Length));
}
return new HttpResponseMessage()
{
Content = new StringContent(sb.ToString())
};
}
catch (System.Exception e)
{
return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
}
}
}
答案 1 :(得分:0)
试试这个方法:
public void UploadFilesToRemoteUrl()
{
string[] files = { @"your file path" };
string url = "Your url";
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}";
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]);
string header = string.Format(headerTemplate, "uplTheFile", 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;
}
答案 2 :(得分:0)
如果其他人遇到同样的问题:请确保您的边界字符串有效,例如: 不这样做:
using (var content =
new MultipartFormDataContent("Upload----" + DateTime.Now.ToString(CultureInfo.InvariantCulture)))
{
...
}
由于边界字符无效,可能是“/”日期分隔符。至少这解决了我在Context.Request.Files
控制器中访问Nancy
时遇到的问题(始终为空)。
最好使用类似DateTime.Now.Ticks.ToString("x")
的内容。
答案 3 :(得分:0)
除了内容类型multipart/form-data
之外,代码中的所有内容都很好看。请尝试更改代码以反映正确的内容类型:
content.Headers.ContentType = new MediaTypeHeaderValue("multipart/form-data");
您可能需要参考this帖子,了解为什么将内容类型设置为application/octet-stream
在客户端没有意义。