我创建了一个API，它将文件作为输入并进行处理。 样本是这样的......
[HttpPost]
public string ProfileImagePost(HttpPostedFile HttpFile)
{
//rest of the code
}
然后我创建了一个客户端来消费它，如下所示......
string path = @"abc.csv";
FileStream rdr = new FileStream(path, FileMode.Open, FileAccess.Read);
byte[] inData = new byte[rdr.Length];
rdr.Read(inData, 0, Convert.ToInt32(rdr.Length));
HttpWebRequest req = (HttpWebRequest)WebRequest.Create("http://localhost/abc/../ProfileImagePost");
req.KeepAlive = false;
req.ContentType = "multipart/form-data";
req.Method = "POST";
req.ContentLength = rdr.Length;
req.AllowWriteStreamBuffering = true;
Stream reqStream = req.GetRequestStream();
reqStream.Write(inData, 0, Convert.ToInt32(rdr.Length));
reqStream.Close();
HttpWebResponse TheResponse = (HttpWebResponse)req.GetResponse();
string TheResponseString1 = new StreamReader(TheResponse.GetResponseStream(), Encoding.ASCII).ReadToEnd();
TheResponse.Close();
但是我在客户端遇到500错误。 帮帮我吧。
提前完成了......
答案 0 :(得分：5)
ASP.NET Web API不适用于HttpPostedFile。相反，您应该使用following tutorial中显示的MultipartFormDataStreamProvider。
您的客户端电话也是错误的。您已将ContentType设置为multipart/form-data，但您根本不尊重此编码。您只是将文件写入请求流。
让我们举一个例子：
public class UploadController : ApiController
{
public Task<HttpResponseMessage> Post()
{
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = HostingEnvironment.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
// Read the form data
return Request.Content.ReadAsMultipartAsync(provider).ContinueWith(t =>
{
// This illustrates how to get the file names.
foreach (MultipartFileData file in provider.FileData)
{
Trace.WriteLine(file.Headers.ContentDisposition.FileName);
Trace.WriteLine("Server file path: " + file.LocalFileName);
}
return Request.CreateResponse(HttpStatusCode.OK);
}, TaskScheduler.FromCurrentSynchronizationContext());
}
}
然后您可以使用HttpClient来调用此API：
using System;
using System.IO;
using System.Net.Http;
using System.Net.Http.Headers;
class Program
{
static void Main(string[] args)
{
using (var client = new HttpClient())
using (var content = new MultipartFormDataContent())
{
client.BaseAddress = new Uri("http://localhost:16724/");
var fileContent = new ByteArrayContent(File.ReadAllBytes(@"c:\work\foo.txt"));
fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
{
FileName = "Foo.txt"
};
content.Add(fileContent);
var result = client.PostAsync("/api/upload", content).Result;
Console.WriteLine(result.StatusCode);
}
}
}
答案 1 :(得分：1)
而不是HttpPostedFile，直接从客户端发送文件作为流，在服务器端以文件形式读取文件，如：
//read uploaded csv file at server side
Stream csvStream = HttpContext.Current.Request.Files[0].InputStream;
//Send file from client
public static void PostFile()
{
string[] files = { @"C:\Test.csv" };
string url = "http://localhost/abc/../test.xml";
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)
您应该坚持使用webapi路由命名约定来节省一些麻烦。
所有名为PostSomething（）的方法都将被接受为HttpPost。他们的参数签名将告诉他们你需要两个同名的。
因此，调用您的WebApi方法PostProfileImage（HttpPostedFile文件），然后您可以将该文件作为数据发布到http：/// api //。
在WebApi中获取错误的路由基础是HTTP / 500的根本原因，并且会导致异常。
答案 3 :(得分：0)
这是我发布截图的代码 客户端
public void PostMethod()
{
ImageConverter converter = new ImageConverter();
var bytes = (byte[])converter.ConvertTo(bmpScreenshot, typeof(byte[]));
StringConverter s = new StringConverter();
string uri = "http://localhost:3844/api/upload";
byte[] postBytes = bytes;
string str = Properties.Settings.Default.token.ToString(); //after login user receives a response token, it is stored in the application settings. All Posts save in db with a this token
byte[] bA = ASCIIEncoding.ASCII.GetBytes(str);
MultipartFormDataContent multiPartData = new MultipartFormDataContent();
ByteArrayContent byteArrayContent = new ByteArrayContent(postBytes);
ByteArrayContent bAC = new ByteArrayContent(bA);
multiPartData.Add(bAC, "token");
multiPartData.Add(byteArrayContent,"picture");
HttpRequestMessage requestMessage = new HttpRequestMessage(HttpMethod.Post, uri);
requestMessage.Content = multiPartData;
HttpClient httpClient = new HttpClient();
Task<HttpResponseMessage> httpRequest = httpClient.SendAsync(requestMessage);
HttpResponseMessage httpResponse = httpRequest.Result;
HttpContent responseContent = httpResponse.Content;
}
服务器端
public class UploadController : ApiController
{
public masterEntities context = new masterEntities();
public ImgEntitySet imgEntity = new ImgEntitySet();
public async Task<HttpResponseMessage> PostRawBufferManual()
{
MultipartFormDataStreamProvider streamProvider = new MultipartFormDataStreamProvider("~/App_Data");
MultipartFileStreamProvider dataContent = await Request.Content.ReadAsMultipartAsync(streamProvider);
foreach (HttpContent data in dataContent.Contents)
{
string fileName = data.Headers.ContentDisposition.Name;
byte[] n = await data.ReadAsByteArrayAsync();
string m = Encoding.ASCII.GetString(n);
int z = int.Parse(m);
imgEntity.UID = z;
break;
}
foreach (HttpContent data in dataContent.Contents)
{
string fileNamePicture = data.Headers.ContentDisposition.Name;
if (fileNamePicture == "picture")
{
byte[] b = await data.ReadAsByteArrayAsync();
imgEntity.Image = b;
}
}
context.ImgEntitySet.Add(imgEntity);
context.SaveChanges();
return Request.CreateResponse(HttpStatusCode.OK );
}
}