从路径名中提取文件名称

ExtractFileName(OpenDialog1.FileName)

C:\Documents and Settings\Administrator\桌面\1.txt 结果为 1.txt

 

引申：

function ExtractFilePath(const FileName: string): string;

function ExtractFileName(const FileName: string): string;

返回指定文件的文件名及扩展名

var

s:string;

begin

s:=ExtractFileName('D:\Program Files\Borland\Delphi7\Projects\Unit1.dcu');

showmessage(s);\\显示:Unit1.dcu

end;

var

s:string;

begin

s:=ExtractFilepath('D:\Program Files\Borland\Delphi7\Projects\Unit1.dcu');

showmessage(s);\\显示:'D:\Program Files\Borland\Delphi7\Projects

end;

ExtractFilePath(Application.ExeName)//获取应程序文件的路径