关于图片写入数据库的问题？

Question

我这么写有问题请问如何用OpenPictureDialog1将图片显示在image1中？
然后再如何将其写入数据库中？？？
procedure Tstudangai.SpeedButton5Click(Sender: TObject);
var ss:string;
begin
OpenPictureDialog1.Execute;
ss:=OpenPictureDialog1.GetNamePath;
image1.Picture:=ss;
end;

Answer 1

存到数据库
var
pjpg:tjpegimage;
pic:TMemoryStream;
begin
pjpg:=tjpegimage.Create;
pjpg.LoadFromFile(F:\JPG.jpg);
pic:=TMemoryStream.Create;
pjpg.SaveToStream(pic);
pic.Position:=0;
TBlobField(dataset.FieldByName('Img')).LoadFromStream(pic);
dataset.post;
pic.Free;
pjpg.free;
end;
从数据库取出
var
pjpg:tjpegimage;
pic:TMemoryStream;
begin
pjpg:=tjpegimage.Create;
pic:=TMemoryStream.Create;
TBlobField(dataset.FieldByName('Img')).savetoStream(pic);
pic.positon:=0;
pjpeg.LoadFromStream(pic);
image.graphi.bmp.assign(pjpeg);
end;end;

Answer 2

procedure Tstudangai.SpeedButton5Click(Sender: TObject);
var ss:string;
begin
 if OpenPictureDialog1.Execute then
begin
  ss:=OpenPictureDialog1.FIlename;
  if ss<>'' then image1.Picture:=ss;
end;
end;

Answer 3

image1.Picture//好象是LOADFROMFILE；

Answer 4

var 
  MyJPEG : TJPEGImage;
  MS: TMemoryStream;
begin
  openpicturedialog1.Execute;
  if openpicturedialog1.FileName<>'' then
  begin
    image.Picture.LoadFromFile(openpicturedialog1.FileName);
    openpicturedialog1.FileName:='';
    MyJPEG := TJPEGImage.Create;
    try
      with MyJPEG do
      begin
        Assign(Image.Picture.Graphic);
        MS:=TMemoryStream.create;
        SaveToStream(MS);
        MS.Position:=0;
        adotable1.Append;
        TBlobField(adotable1.FieldbyName('photo')).LoadFromStream(MS);
        adotable1.Post;
        messagebox(getactivewindow(),'图像保存完毕！','保存',mb_ok);
        image.Picture.CleanupInstance;
      end;
    finally
      MyJPEG.Free;
    end;
  end;
显示图片
var
  tempstream:TStringStream;
  tempjpeg:TJPEGImage;
begin
  try
    tempstream:=TStringStream.Create(' ');
    TBlobField(adodataset1.FieldByName('photo')).SaveToStream(tempstream);
    tempstream.Position:=0;
    tempjpeg:=TJPEGImage.Create;
    tempjpeg.LoadFromStream(tempstream);
    DBImage1.Picture.Bitmap.Assign(tempjpeg);
  finally
    tempstream.Free;
    tempjpeg.Free;
  end;
end;

Answer 5

angle097113(深思不解)大哥，这句话不能运行呀！
image.Picture.LoadFromFile(openpicturedialog1.FileName);