S
songyuanhai
Unregistered / Unconfirmed
GUEST, unregistred user!
我在一个上下浏览按钮中写了代码,这个代码是“上一条”按钮,我想在上下移动时,把每条记录的图片显示出来,但是却出现 “jpeg error #42”错误,请问是什么错误?代码如下:procedure Tbysllform.Button4Click(Sender: TObject);var myjpeg2:tjpegimage; ms2:tmemorystream;begin if not dmunit.adobys.Bof then dmunit.adobys.Prior; if not dmunit.adobys.Active then dmunit.adobys.Open; myjpeg2:=tjpegimage.Create ; ms2:=tmemorystream.Create ; tblobfield(dmunit.adobys.fields[29]).savetostream(ms2); ms2.Position:=0; myjpeg2.LoadFromStream(ms2); image1.Picture.Assign(myjpeg2); myjpeg2.Free; ms2.Free;end;