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1

I have this chunk of code:

[AcceptVerbs("POST")]
[Route("api/doc/ImageUpload")]
public HttpResponseMessage ImageUpload()

 try
 
 if (!Request.Content.IsMimeMultipartContent())
 
 throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
 

 string root = HttpContext.Current.Server.MapPath("~/imageUpload");
 var provider = new MultipartFormDataStreamProvider(root);

 Request.Content.ReadAsMultipartAsync(provider);

 foreach (MultipartFileData file in provider.FileData)
 fileName.Contains(@""))
 
 fileName = Path.GetFileName(fileName);
 

 using (FileStream stream = File.Open(file.LocalFileName, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
 
 //do something
 
 
 

And this Postman call:

I can see the file is being generated in the folder:

But when trying to actually use the file in file stream and save it as proper jpeg or png I'm getting the IOException. Also I'm noticing the file saved is 32kb but the file I'm trying to upload is 120kb.

asp.net-web-api asp.net-web-api2 multipartform-data ioexception form-data

share|improve this question

edited Mar 30 at 0:09

halfer

14.7k759116

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196

add a comment | 

1

I have this chunk of code:

[AcceptVerbs("POST")]
[Route("api/doc/ImageUpload")]
public HttpResponseMessage ImageUpload()

 try
 
 if (!Request.Content.IsMimeMultipartContent())
 
 throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
 

 string root = HttpContext.Current.Server.MapPath("~/imageUpload");
 var provider = new MultipartFormDataStreamProvider(root);

 Request.Content.ReadAsMultipartAsync(provider);

 foreach (MultipartFileData file in provider.FileData)
 fileName.Contains(@""))
 
 fileName = Path.GetFileName(fileName);
 

 using (FileStream stream = File.Open(file.LocalFileName, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
 
 //do something
 
 
 

And this Postman call:

I can see the file is being generated in the folder:

But when trying to actually use the file in file stream and save it as proper jpeg or png I'm getting the IOException. Also I'm noticing the file saved is 32kb but the file I'm trying to upload is 120kb.

asp.net-web-api asp.net-web-api2 multipartform-data ioexception form-data

share|improve this question

edited Mar 30 at 0:09

halfer

14.7k759116

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196

add a comment | 

I have this chunk of code:

[AcceptVerbs("POST")]
[Route("api/doc/ImageUpload")]
public HttpResponseMessage ImageUpload()

 try
 
 if (!Request.Content.IsMimeMultipartContent())
 
 throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
 

 string root = HttpContext.Current.Server.MapPath("~/imageUpload");
 var provider = new MultipartFormDataStreamProvider(root);

 Request.Content.ReadAsMultipartAsync(provider);

 foreach (MultipartFileData file in provider.FileData)
 fileName.Contains(@""))
 
 fileName = Path.GetFileName(fileName);
 

 using (FileStream stream = File.Open(file.LocalFileName, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
 
 //do something
 
 
 

And this Postman call:

I can see the file is being generated in the folder:

But when trying to actually use the file in file stream and save it as proper jpeg or png I'm getting the IOException. Also I'm noticing the file saved is 32kb but the file I'm trying to upload is 120kb.

asp.net-web-api asp.net-web-api2 multipartform-data ioexception form-data

share|improve this question

edited Mar 30 at 0:09

halfer

14.7k759116

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196

[AcceptVerbs("POST")]
[Route("api/doc/ImageUpload")]
public HttpResponseMessage ImageUpload()

 try
 
 if (!Request.Content.IsMimeMultipartContent())
 
 throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
 

 string root = HttpContext.Current.Server.MapPath("~/imageUpload");
 var provider = new MultipartFormDataStreamProvider(root);

 Request.Content.ReadAsMultipartAsync(provider);

 foreach (MultipartFileData file in provider.FileData)
 fileName.Contains(@""))
 
 fileName = Path.GetFileName(fileName);
 

 using (FileStream stream = File.Open(file.LocalFileName, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
 
 //do something
 
 
 

share|improve this question

edited Mar 30 at 0:09

halfer

14.7k759116

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196

share|improve this question

edited Mar 30 at 0:09

halfer

14.7k759116

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196

edited Mar 30 at 0:09

halfer

14.7k759116

edited Mar 30 at 0:09

halfer

14.7k759116

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196

asked Mar 19 at 17:41

LazialeLaziale

3,06635105196