public:howto:htmlforms [Ararat Synapse]

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====== How To Send More Complex HTML Forms By HTTP Protocol ======

I have adapted the HTTPpostFile to allow you to upload the file and form fields together, instead of just the file. If you wish, add it to HTTPSend.pas

<code delphi>
function HttpPostFileForm(const URL, FieldName, FileName: string;
  const Data: TStream; const FormFields:TStrings; const ResultData:
TStrings): Boolean;
var
  HTTP: THTTPSend;
  Bound, s: string;
  i:integer;
const
  FIELD_MASK = CRLF + '--%s' + CRLF +
              'Content-Disposition: form-data; name="%s"' + CRLF + CRLF
+
              '%s';
begin
  Bound := IntToHex(Random(MaxInt), 8) + '_Synapse_boundary';
  HTTP := THTTPSend.Create;
  try
    s := '--' + Bound + CRLF;
    s := s + 'content-disposition: form-data; name="' + FieldName +
'";';
    s := s + ' filename="' + FileName +'"' + CRLF;
    s := s + 'Content-Type: Application/octet-string' + CRLF + CRLF;
    HTTP.Document.Write(Pointer(s)^, Length(s));
    HTTP.Document.CopyFrom(Data, 0);
    // Include formfield
    for i:=0 to FormFields.Count-1 do
      begin
        S:= Format(FIELD_MASK,[Bound, FormFields.Names[I],
FormFields.Values[FormFields.Names[I]]]);
        HTTP.Document.Write(Pointer(S)^, Length(S));
      end;
    s := CRLF + '--' + Bound + '--' + CRLF;
    HTTP.Document.Write(Pointer(s)^, Length(s));
    HTTP.MimeType := 'multipart/form-data, boundary=' + Bound;
    Result := HTTP.HTTPMethod('POST', URL);
    ResultData.LoadFromStream(HTTP.Document);
  finally
    HTTP.Free;
  end;
end;
</code>
 
Regards,
 Eric Paschoalick Chaves