Firemonkey TBitmap em TBytes TakePhotoFromCameraAction1DidFinishTaking

23/03/2020

Delphi FireMonkey Amazon Web Services - AWS

Tenho uma fun��o que envia arquivos para o AWS S3
preciso pegar a imagem capturada no TakePhotoFromCameraAction1DidFinishTaking e transformar em Byte para enviar mais n�o sei por onde come�ar.

function TF_Cadsuspeito.UploadFile(pFile: TBytes; FileName, Bucket: string): boolean;
var
Service: TAmazonStorageService;
ConAmazon: TAmazonConnectionInfo;
begin
try
ConAmazon := TAmazonConnectionInfo.Create(nil);
ConAmazon.AccountKey := SecretKey;
ConAmazon.AccountName := AccessKey;
ConAmazon.QueueEndpoint := 'queue.amazonaws.com';
ConAmazon.StorageEndpoint := 's3-us-east-1.amazonaws.com';
ConAmazon.TableEndpoint := 'sdb.amazonaws.com';
ConAmazon.UseDefaultEndpoints := False;
Service := TAmazonStorageService.Create(ConAmazon);
Result := Service.UploadObject(Bucket, FileName, pFile, TRUE, nil, nil, amzbaPrivate, nil);
finally
ConAmazon.Free;
Service.Free;
end;
end;

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24/03/2020

Inarais Jer�nimo

A solu��o foi criar um memorystream do Tbitmap isso pode ser feito de duas formas:

 Bitmap.SaveToStream(Image_Stream);

ou como eu preferi usando um procedimento

procedure BitmapToStream( aBitmap: TBitmap; aStream: TStream );
var
  MS: TMemoryStream;
  S: int32;
begin

  MS := TMemoryStream.Create;
  try

    aBitmap.SaveToStream(MS);
    MS.Position := 0;

    if MS.Size>MaxInt then begin
      raise Exception.Create('TSpriteSheet.BitmapToStream: Image data too large.');
    end;
    S := MS.Size;
    aStream.Write(S,Sizeof(S));
    MS.SaveToStream(aStream);
  finally
    MS.Free;
  end;
end;

ao final minha rotina ficou assim


procedure BitmapToStream( aBitmap: TBitmap; aStream: TStream );
var
  MS: TMemoryStream;
  S: int32;
begin

  MS := TMemoryStream.Create;
  try

    aBitmap.SaveToStream(MS);
    MS.Position := 0;

    if MS.Size>MaxInt then begin
      raise Exception.Create('TSpriteSheet.BitmapToStream: Image data too large.');
    end;
    S := MS.Size;
    aStream.Write(S,Sizeof(S));
    MS.SaveToStream(aStream);
  finally
    MS.Free;
  end;
end;



procedure UploadImage();
var
   // ConAmazon: TAmazonConnectionInfo;
    Service: TAmazonStorageService;
    regions:TAmazonRegion;
    Response: TCloudResponseInfo;
    Image_Byte: TBytes;
    Image_Stream: TMemoryStream;
    bResult : Boolean;
begin
    try
        Image_Stream:= TMemoryStream.Create;

       //Form1.Image1.MultiResBitmap.Items[0].Bitmap.SaveToStream(Image_Stream);
       //Image_Stream:= TFileStream.Create('C:\\Users\\sinho\\OneDrive\\Imagens\\teste.jpg', 0);

       BitmapToStream(Form1.Image1.MultiResBitmap.Items[0].Bitmap,Image_Stream);
       SetLength(Image_Byte, Image_Stream.Size);
       Image_Stream.Position := 0;
       Image_Stream.Read(Image_Byte, Image_Stream.Size );


       Form1.ConAmazon := TAmazonConnectionInfo.Create(nil);
       Form1.ConAmazon.AccountKey := 'LRxxxxxxxxxxxxxxxxxxxx';
       Form1.ConAmazon.AccountName := 'AKIxxxxxxxxxxxxxxxxx';
       Form1.ConAmazon.Region:= TAmazonRegion.amzrUSEast1 ;
       Form1.ConAmazon.QueueEndpoint := 'queue.amazonaws.com';
       Form1.ConAmazon.TableEndpoint := 'sdb.amazonaws.com';
       Service := TAmazonStorageService.Create(Form1.ConAmazon);
       Response := TCloudResponseInfo.Create;
       bResult:= Service.UploadObject('romobr', 'teste.jpg', Image_Byte, False, nil, nil, amzbaPrivate, Response, regions);
       if bResult = True then
          ShowMessage('Upload success'+#13+IntToStr(Response.StatusCode)+' -> ' + Response.StatusMessage)
       else
          ShowMessage('Upload fails'+#13+IntToStr(Response.StatusCode)+' -> ' + Response.StatusMessage)
    Finally
        Image_Stream.Free;
        Service.Free;
        Form1.ConAmazon.Free;
    end;
end;

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