Estoy tratando de enviar un archivo usando POST en datos de múltiples partes / formularios a través de Indy 10.5.8. Estoy usando Delphi XE2 y he estado intentando enviar un archivo a un servidor. Este es el primer momento que lo he intentado y, como mi experiencia con Indy es bastante limitada, tomé el siguiente fragmento de código:

unit MsMultiPartFormData;

interface

uses
  SysUtils, Classes;

const
  CONTENT_TYPE = 'multipart/form-data; boundary=';
  CRLF = #13#10;
  CONTENT_DISPOSITION = 'Content-Disposition: form-data; name="%s"';
  FILE_NAME_PLACE_HOLDER = '; filename="%s"';
  CONTENT_TYPE_PLACE_HOLDER = 'Content-Type: %s' + crlf + crlf;
  CONTENT_LENGTH = 'Content-Length: %d' + crlf;

type
  TMsMultiPartFormDataStream = class(TMemoryStream)
  private
    FBoundary: string;
    FRequestContentType: string;
    function GenerateUniqueBoundary: string;
  public
    procedure AddFormField(const FieldName, FieldValue: string);
    procedure AddFile(const FieldName, FileName, ContentType: string; FileData: TStream); overload;
    procedure AddFile(const FieldName, FileName, ContentType: string); overload;
    procedure PrepareStreamForDispatch;
    constructor Create;
    property Boundary: string read FBoundary;
    property RequestContentType: string read FRequestContentType;
  end;

implementation

{ TMsMultiPartFormDataStream }

constructor TMsMultiPartFormDataStream.Create;
begin
  inherited;
  FBoundary := GenerateUniqueBoundary;
  FRequestContentType := CONTENT_TYPE + FBoundary;
end;

procedure TMsMultiPartFormDataStream.AddFile(const FieldName, FileName,
  ContentType: string; FileData: TStream);
var
  sFormFieldInfo: string;
  Buffer: PChar;
  iSize: Int64;
begin
  iSize := FileData.Size;
  sFormFieldInfo := Format(CRLF + '--' + Boundary + CRLF + CONTENT_DISPOSITION +
    FILE_NAME_PLACE_HOLDER + CRLF + CONTENT_LENGTH +
      CONTENT_TYPE_PLACE_HOLDER, [FieldName, FileName, iSize, ContentType]);
  {so: boundary + crlf + content-disposition+file-name-place-holder}

  Write(Pointer(sFormFieldInfo)^, Length(sFormFieldInfo));
  FileData.Position := 0;
  GetMem(Buffer, iSize);
  try
    FileData.Read(Buffer^, iSize);
    Write(Buffer^, iSize);
  finally
    FreeMem(Buffer, iSize);
  end;
end;

procedure TMsMultiPartFormDataStream.AddFile(const FieldName, FileName,
  ContentType: string);
var
  FileStream: TFileStream;
begin
  FileStream := TFileStream.Create(FileName, fmOpenRead or fmShareDenyWrite);
  try
    AddFile(FieldName, FileName, ContentType, FileStream);
  finally
    FileStream.Free;
  end;
end;

procedure TMsMultiPartFormDataStream.AddFormField(const FieldName,
  FieldValue: string);
var
  sFormFieldInfo: string;
begin
  sFormFieldInfo := Format(CRLF + '--' + Boundary + CRLF + CONTENT_DISPOSITION + CRLF + CRLF +
    FieldValue, [FieldName]);
  Write(Pointer(sFormFieldInfo)^, Length(sFormFieldInfo));
end;

function TMsMultiPartFormDataStream.GenerateUniqueBoundary: string;
begin
  Result := '---------------------------' + FormatDateTime('mmddyyhhnnsszzz', Now);
end;

procedure TMsMultiPartFormDataStream.PrepareStreamForDispatch;
var
  sFormFieldInfo: string;
begin
  sFormFieldInfo := CRLF + '--' + Boundary + '--' + CRLF;
  Write(Pointer(sFormFieldInfo)^, Length(sFormFieldInfo));
  Position := 0;
end;

end.

Estoy llamando al código así:

function PostFile(filename, apikey: string): boolean;
var
  ResponseStream: TMemoryStream;
  MultiPartFormDataStream: TMsMultiPartFormDataStream;
begin
//  Form5.IdHTTP1.HandleRedirects := true;
  Form5.idHTTP1.ReadTimeout := 0;
//  Form5.idHTTP1.IOHandler.LargeStream           := True;
  Result := false;
  MultiPartFormDataStream := TMsMultiPartFormDataStream.Create;
  ResponseStream := TMemoryStream.Create;
  try
    try
    Form5.IdHttp1.Request.ContentType := MultiPartFormDataStream.RequestContentType;
    MultiPartFormDataStream.AddFormField('apikey', apikey);
    MultiPartFormDataStream.AddFile('file', filename, 'multipart/form-data');

    MultiPartFormDataStream.PrepareStreamForDispatch;
    MultiPartFormDataStream.Position := 0;
    Form5.IdHTTP1.Post('http://www.updserver.tld/api//file/save', MultiPartFormDataStream, ResponseStream);
    MultiPartFormDataStream.SaveToFile(ExtractFilePath(Application.ExeName) + 'a.txt');
    Result := true;
    except
 on E:Exception do
   begin
    Form5.Close;
    ShowMessage('Upload failed! '+E.Message);
   end;

    end;
  finally
    MultiPartFormDataStream.Free;
    ResponseStream.Free;
  end;
end;

El archivo se envía, pero es rechazado por el servidor. Una inspección más cercana de los datos enviados revela que los datos se corrompen un poco (sospecho que hay problemas de codificación), lo que veo es:

POST /api/file/save HTTP/1.0
Connection: keep-alive
Content-Type: multipart/form-data; boundary=---------------------------071312151405662
Content-Length: 11040172
Host: www.updserver.tld
Accept: text/html, */*
Accept-Encoding: identity
User-Agent: Mozilla/3.0 (compatible; Indy Library)




.
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.
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Los encabezados normales, enviados desde un cliente Python en funcionamiento, se ven así:

POST https://updserver.tld/api/file/save HTTP/1.0
content-type: multipart/form-data; boundary=---------------------------071312151405662
content-length: 6613364

---------------------------071312151405662
Content-Disposition: form-data; name="apikey"
ac36fae9a406596[rest-of-api-key-goes-here]17966c42b60c8c4cd
---------------------------071312151405662
Content-Disposition: form-data; name="file"; filename="C:\Users\User\Desktop\Project1.exe"
Content-Type: application/octet-stream

¿Alguna idea de lo que estoy haciendo mal?

Gracias por adelantado.