posting a file as part of a form

问题:

I am trying to post a file and some other data to a website in an App that I am writing using Delphi XE8 but it is not working. When I monitor the network traffic using "Microsoft Network Monitor 3.4", the file is only partially sent and none of the other data is sent. I have tried both Indy 10 and the new TNetHTTPClient and got the same result which tells me I am doing something wrong. please help.

Indy Way

procedure TMyClass.SendFile(aUrl: String);
var
  mCookies: TIdCookieManager;
  pHttp: TIdHTTP;
  PostStream: TIdMultiPartFormDataStream;
  ResponseStream: TStringStream;
  fName, mimeStr: String;
begin
  fName := 'Image000.jpg';
  mCookies := TIdCookieManager.Create(nil);
  pHttp := TIdHTTP.Create(nil);
  pHttp.CookieManager := mCookies;
  PostStream:= TIdMultiPartFormDataStream.Create();
  ResponseStream := TStringStream.Create('');
  mimeStr := GetMIMETypeFromFile(fieldValue); // This returns 'image/pjpeg' instead of 'image/jpeg'. I have manually fixed it and it did not change the result
  PostStream.AddFile('sourceFile', fName, mimeStr);
  PostStream.AddFormField('name1', 'value1');
  PostStream.AddFormField('name2', 'value2');
  PostStream.AddFormField('name3', 'value3');
  PostStream.AddFormField('name4', 'value4');
  PostStream.AddFormField('name5', 'value5');
  .
  .
  .
  pHttp.Request.ContentType := PostStream.RequestContentType;
  pHttp.Request.Accept := '*/*';
  pHttp.Request.AcceptLanguage := 'en-us,en';
  pHttp.Request.AcceptEncoding := 'gzip, deflate';
  pHttp.Post(aUrl, PostStream, ResponseStream); // Get a 500 error from server for bad data
  .
  .
  .
  PostStream.Free();
  ResponseStream.Free();
  mCookies.Free();
  pHttp.Free();
end;

by the way GetMIMETypeFromFile returns the wrong value but even if I hardcode the correct one, it does not make any different.

the new XE8 way

procedure TMyClass.SendFile(aUrl: String);
var
  mCookies: TCookieManager;
  pHttp: TNetHTTPClient;
  fName: String;
  mpFormData: TMultipartFormData;
  respData: IHTTPResponse;
begin
  fName := 'Image000.jpg';
  mCookies := TCookieManager.Create();
  pHttp := TNetHTTPClient.Create(nil);
  pHttp.CookieManager := mCookies;
  mpFormData := TMultipartFormData.Create();
  mpFormData.AddFile('sourceFile', fName);
  mpFormData.AddField('name1', 'value1');
  mpFormData.AddField('name2', 'value2');
  mpFormData.AddField('name3', 'value3');
  mpFormData.AddField('name4', 'value4');
  mpFormData.AddField('name5', 'value5');
  .
  .
  .
  pHttp.ContentType := 'multipart/form-data';
  pHttp.Accept := '*/*';
  pHttp.AcceptLanguage := 'en-us,en';
  pHttp.AcceptEncoding := 'gzip, deflate';
  mpFormData.Stream.Position := 0;
  respData := pHttp.Post(aUrl, mpFormData.Stream); //Same problem, error 500 here
  .
  .
  .
  mpFormData.Free();
  pHttp.Free();
  mCookies.Free();
end;

I know the server is working correctly because another application (written in Intel XDA) works just fine. the image is valid, and all the Get calls that I make before this works as well. I really need help. Thank you in advance

回答1:

When I monitor the network traffic using "Microsoft Network Monitor 3.4", the file is only partially sent and none of the other data is sent.

Are you sure the monitor is simply not displaying partial data in its UI? A packet sniffer like Wireshark would be more reliable (or attach a TIdLog... component to TIdHTTP), as it will show you everything that is actually being transmitted. The only possible way that posting a TIdMultipartFormDataStream would only send a portion of the file and skip the other fields altogether is if the socket is disconnected while transmitting the file.

mimeStr := GetMIMETypeFromFile(fieldValue); // This returns 'image/pjpeg' instead of 'image/jpeg'.

Internally, GetMIMETypeFromFile() builds its own list of hard-coded MIME types and then uses OS information to overwrite that list. The default value that is hard-coded for .jpg is image/jpeg. When it then queries the OS, it sees image/jpeg and image/pjpeg (in that order) are registered for .jpg, so image/pjpeg was the last MIME type seen for .jpg and that is what GetMIMETypeFromFile() ends up returning.

pHttp.Request.ContentType := PostStream.RequestContentType;

You do not need that, Post() handles that internally for you.

pHttp.Request.AcceptEncoding := 'gzip, deflate';

Do not do that at all. Post() handles that internally for you, based on whether TIdHTTP.Compressor is assigned and ready.