L
lbt
Unregistered / Unconfirmed
GUEST, unregistred user!
我用OpenPictureDialog1打开图像文件,将其复制到我的软件目录下面,我是这样实现的
var
ypath,mbpath:string;
ypath:=OpenPictureDialog1.FileName;
mbpath:=extractfilepath(application.ExeName);
copyfile(ypath,mbpath,true);
编译时他提示:
incompatible type:'String' and 'Pansichar'
我又将ypath,mbpath定义成Pansichar类型的函数,可是他们又不能获取源路径和目标
路径;
请问各位高手我要怎么办啊;
var
ypath,mbpath:string;
ypath:=OpenPictureDialog1.FileName;
mbpath:=extractfilepath(application.ExeName);
copyfile(ypath,mbpath,true);
编译时他提示:
incompatible type:'String' and 'Pansichar'
我又将ypath,mbpath定义成Pansichar类型的函数,可是他们又不能获取源路径和目标
路径;
请问各位高手我要怎么办啊;