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();
}
}
解决方法
您不应该使用HttpContext来获取ASP.NET Web API中的文件.看看这个由微软(
http://code.msdn.microsoft.com/ASPNET-Web-API-File-Upload-a8c0fb0d/sourcecode?fileId=67087&pathId=565875642)编写的例子.
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);
}
}
}
