'How to convert text data into an Image?

Once you load and image into a component, I can see that Delphi store the image data on DFM, . Example:

object img1: TImage
  Left = 71
  Top = 2
  Width = 18
  Height = 18
  Picture.Data = {
    0954506E67496D61676589504E470D0A1A0A0000000D49484452000000100000
    001008060000001FF3FF610000000473424954080808087C0864880000000970
    485973000000750000007501E3C207650000001974455874536F667477617265
    007777772E696E6B73636170652E6F72679BEE3C1A000000EF4944415478DAAD
    923B0AC2401086B3D7100F6110B415FBD8AB60E323E62262632DF15158A8BDB6
    22D682E821C41B58C76F7003EB8A9A10073E36ECFCFB6766765514458E842258
    3A5083A2F38C136C6016C5422B94EC7336C7F7122A7081A3CE97A0000768A2BD
    BD1968F6E0428068FD2250AACE32863354ED4AE4701726D0B00F5B262BE8A199
    DA065BC893709D2F8189547045E7D906D2D79684FFC32064F1D0E5FE6E90B985
    CC434C738DF2F7BB7995691E521F163A1FC4262AE15396AA7650D6FBD2862F26
    EAC313B767A0741BE64DCD657E890C0C93500F3D8E616203C344CA6FEBAD5B2A
    03C364002D183D00658D8FCCCDEDEA100000000049454E44AE426082}
end

As you can see it's an small image. How it's possible to get this information without the DFM and make it an Image, for example:

procedure TForm12.btn2Click(Sender: TObject);
var
  img2: TImage;
  Loutput: TStream;
begin
  ObjectTextToBinary(TStringStream.Create(
     '0954506E67496D61676589504E470D0A1A0A0000000D49484452000000100000'
    + '001008060000001FF3FF610000000473424954080808087C0864880000000970'
    + '485973000000750000007501E3C207650000001974455874536F667477617265'
    + '007777772E696E6B73636170652E6F72679BEE3C1A000000EF4944415478DAAD'
    + '923B0AC2401086B3D7100F6110B415FBD8AB60E323E62262632DF15158A8BDB6'
    + '22D682E821C41B58C76F7003EB8A9A10073E36ECFCFB6766765514458E842258'
    + '3A5083A2F38C136C6016C5422B94EC7336C7F7122A7081A3CE97A0000768A2BD'
    + 'BD1968F6E0428068FD2250AACE32863354ED4AE4701726D0B00F5B262BE8A199'
    + 'DA065BC893709D2F8189547045E7D906D2D79684FFC32064F1D0E5FE6E90B985'
    + 'CC434C738DF2F7BB7995691E521F163A1FC4262AE15396AA7650D6FBD2862F26'
    + 'EAC313B767A0741BE64DCD657E890C0C93500F3D8E616203C344CA6FEBAD5B2A'
    + '03C364002D183D00658D8FCCCDEDEA100000000049454E44AE426082'), Loutput);
  img2 := TImage.Create(self);
  img2.Name := 'image2';
  img2.Left := 71;
  img2.Top := 30;
  img2.Width := 18;
  img2.Height := 18;
  img2.Picture.Graphic.LoadFromStream(Loutput);

  img1.Parent := Self;
end;

delphi

Solution 1:^[1]

You cannot access the TPicture.Graphic property until an image has been loaded into the TPicture first.

TPicture does not support loading data from a TStream (see QC #12434: Add LoadFromStream() method to TPicture), so you will have to stream the image data manually.

The Picture.Data property data starts with a UTF-8 encoded ShortString containing the name of the TGraphic-derived class that produced the image data. In your example, that class name is encoded as:

0954506E67496D616765

The first byte (hex 09) is the number of bytes in the class name (9), the following 9 bytes (hex 54 50 6E 67 49 6D 61 67 65) are the UTF-8 octets of the class name (TPngImage), and the remaining stream bytes are the actual PNG image data.

So, you need to:

extract the class name from the stream.
instantiate the specified TGraphic-derived class type.
load the remaining stream into the object.
assign the object to TPicture.

For example:

uses
  System.Classes,
  System.SysUtils,
  Vcl.Graphics,
  Vcl.Imaging.Jpeg,
  Vcl.Imaging.GIFImg,
  Vcl.Imaging.PngImage;

type
  TGraphicAccess = class(TGraphic)
  end;

procedure TForm12.btn2Click(Sender: TObject);
var
  Linput: String;
  Loutput: TMemoryStream;
  LclsName: ShortString;
  Lgraphic: TGraphic;
  img2: TImage;
begin
  Linput := '0954506E67496D61676589504E470D0A1A0A0000000D49484452000000100000'
          + '001008060000001FF3FF610000000473424954080808087C0864880000000970'
          + '485973000000750000007501E3C207650000001974455874536F667477617265'
          + '007777772E696E6B73636170652E6F72679BEE3C1A000000EF4944415478DAAD'
          + '923B0AC2401086B3D7100F6110B415FBD8AB60E323E62262632DF15158A8BDB6'
          + '22D682E821C41B58C76F7003EB8A9A10073E36ECFCFB6766765514458E842258'
          + '3A5083A2F38C136C6016C5422B94EC7336C7F7122A7081A3CE97A0000768A2BD'
          + 'BD1968F6E0428068FD2250AACE32863354ED4AE4701726D0B00F5B262BE8A199'
          + 'DA065BC893709D2F8189547045E7D906D2D79684FFC32064F1D0E5FE6E90B985'
          + 'CC434C738DF2F7BB7995691E521F163A1FC4262AE15396AA7650D6FBD2862F26'
          + 'EAC313B767A0741BE64DCD657E890C0C93500F3D8E616203C344CA6FEBAD5B2A'
          + '03C364002D183D00658D8FCCCDEDEA100000000049454E44AE426082';

  Loutput := TMemoryStream.Create;
  try
    Loutput.Size := Length(Linput) div 2;
    HexToBin(PChar(Linput), Loutput.Memory^, Loutput.Size);

    LclsName := PShortString(Loutput.Memory)^;

    Lgraphic := TGraphicClass(FindClass(UTF8Decode(LclsName))).Create;
    try
      Loutput.Position := 1 + Length(LclsName);
      TGraphicAccess(Lgraphic).ReadData(Loutput);

      img2 := TImage.Create(self);
      img2.Parent := Self;
      img2.Name := 'image2';
      img2.Left := 71;
      img2.Top := 30;
      img2.Width := 18;
      img2.Height := 18;
      img2.Picture.Assign(Lgraphic);
    finally
      Lgraphic.Free;
    end;
  finally
    Loutput.Free;
  end;
end;

initialization
  // this is not necessary for TPicture's own DFM streaming,
  // but it is necessary for manual streaming, unless you
  // implement your own classname lookups...
  // 
  RegisterClass(TMetafile);
  RegisterClass(TIcon);
  RegisterClass(TBitmap);
  RegisterClass(TWICImage);
  RegisterClass(TJpegImage);
  RegisterClass(TGifImage);
  RegisterClass(TPngImage);
  // and so on...
end.

Solution 2:^[2]

procedure TForm1.FormCreate(Sender: TObject);
const
  CONST_SIGN = '0954506E67496D616765';
var
  LString: String;
  LStart: Integer;
  LStringStream: TStringStream;
  LMem: TMemoryStream;
  R: TBytes;
begin
  LString :=
    '0954506E67496D61676589504E470D0A1A0A0000000D49484452000000100000'
    + '001008060000001FF3FF610000000473424954080808087C0864880000000970'
    + '485973000000750000007501E3C207650000001974455874536F667477617265'
    + '007777772E696E6B73636170652E6F72679BEE3C1A000000EF4944415478DAAD'
    + '923B0AC2401086B3D7100F6110B415FBD8AB60E323E62262632DF15158A8BDB6'
    + '22D682E821C41B58C76F7003EB8A9A10073E36ECFCFB6766765514458E842258'
    + '3A5083A2F38C136C6016C5422B94EC7336C7F7122A7081A3CE97A0000768A2BD'
    + 'BD1968F6E0428068FD2250AACE32863354ED4AE4701726D0B00F5B262BE8A199'
    + 'DA065BC893709D2F8189547045E7D906D2D79684FFC32064F1D0E5FE6E90B985'
    + 'CC434C738DF2F7BB7995691E521F163A1FC4262AE15396AA7650D6FBD2862F26'
    + 'EAC313B767A0741BE64DCD657E890C0C93500F3D8E616203C344CA6FEBAD5B2A'
    + '03C364002D183D00658D8FCCCDEDEA100000000049454E44AE426082';

  { Find and rid signature }

  LStart := Pos(CONST_SIGN, LString);

  if LStart = 0 then
    Exit;

  Delete(LString, LStart, Length(CONST_SIGN));

  { Main }

  LStringStream := TStringStream.Create(LString);
  LMem := TMemoryStream.Create;
  try
    { Prepare out array }
    SetLength(R, Length(LString) div SizeOf(Char));

    { Convert }
    HexToBin(PWideChar(LString), R, Length(LString) div SizeOf(Char));

    { Copy array to stream }
    LMem.WriteBuffer(R[0], Length(R));

    { Save stream with image as file }
    LMem.SaveToFile('xxx.png');

    { Load image from file }
    Image1.Picture.LoadFromFile('xxx.png');
  finally
    LStringStream.Free;
    LMem.Free;
  end;
end;

Solution 3:^[3]

I had cause to do this recently, later version of Delphi (not sure what version it was introduced) has the LoadFromStream function so the functions I created are as follows.

Function ImageToHex(Image:Timage; LineLen:integer):Tstringlist;
var ms:TmemoryStream; s:String; t:Ansistring;
begin
  ms:=tmemorystream.Create;
  try
    image.Picture.SaveToStream(ms);
    setlength(t,ms.Size*2);
    BinToHex(ms.Memory^,Pansichar(t),ms.Size);
    Result:=Tstringlist.create;
    repeat
      s:=copy(t,1,LineLen);
      Result.Add(s);
      delete(t,1,LineLen);
    until t='';
  finally
    ms.free
  end;
end;

procedure HexToImage(HexData:TstringList; var Image:Timage);
var ms:TmemoryStream; s:String;
begin
  ms:=TmemoryStream.Create;
  s:=HexData.Text;
  try
    ms.Size := Length(s) div 2;
    HexToBin(PChar(s), ms.Memory^, ms.Size);
    Image.Picture.LoadFromStream(ms);
  finally
    ms.free
  end;
end;

Solution 4:^[4]

I liked @remy-lebeau explanation, but once one knows that the PNG image is well preserved in the hex data, it was straightforward to save the .dfm file subset posted above into so2.dfm and use the following perl one-liner to convert it to binary.

perl -ne 'END{$PNG=index $b,"PNG"; die "PNG" if $PNG<1; print substr $b,$PNG-1; } $b.=pack "H*",$1 if ( /object img1:/i ... /}/ ) and (/Picture.Data = {/i ... /}/) and m/^\s*([0-9A-F]+)}?\s*$/ ' so2.dfm > img1.png

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1
Solution 2	Zam
Solution 3	Andy k
Solution 4	David Dyck

'How to convert text data into an Image?

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

Solution 4:[4]

Sources

Related Questions

Solution 1:^[1]

Solution 2:^[2]

Solution 3:^[3]

Solution 4:^[4]