简单问题!

Question

if OpenPictureDialog1.Execute() then
       begin
       filename:=OpenPictureDialog1.FileName   ;
 Image1.Picture.LoadFromFile(filename);
            a:=ExtractFilePath(OpenPictureDialog1.FileName);            b:=ExtractFilePath(Application.Name)+OpenPictureDialog1.FileName;
            CopyFile(pchar(a),pchar(b),false)
图片复制我这样写的,但是程序不能复制,我哪里错  了,我就是想打开的图片复制到程序的文件里去啊!

Answer 1

你可以ShowMessage(a和b)看一下，你的路径，是不是又把原图存到原路径里去了。程序当前路径是：ExtractFilePath(Application.ExeName)

Answer 2

a:=OpenPictureDialog1.FileName;b:=ExtractFilePath(Application.ExeName)+ExtractFileName(OpenPictureDialog1.FileName);