如何获得一个目录下最迟生成的那个文件的文件名?

Question

急~~~如题~

Answer 1

用FindFirst、FindNext遍历目录下所有文件，比较创建日期，最晚创建的那个就是你要的文件了

Answer 2

procedure TForm1.Button1Click(Sender: TObject);
var ss:Tsearchrec;
    filepath:string;
    fileTime:integer;
    fileName:string;
begin
   filepath:='c:\';
   if findfirst(filepath+'*.*',faAnyFile,ss)=0 then
   begin
      if not ((ss.Attr and fadirectory)=fadirectory) then
      begin
          fileTime:=ss.Time;
          fileName:=ss.Name;
      end;
      while findnext(ss)=0 do
      begin
          if not ((ss.attr and fadirectory)=fadirectory) then
          begin
              if ss.Time<fileTime then
              begin
                  fileTime:=ss.Time;
                  fileName:=ss.Name;
              end;
          end;
      end;
      findclose(ss);
      showmessage(filename);
   end;
end;

Answer 3

有点小毛病，没有初始化fileTimer的值
把这句家在开头就可以了
fileTime:=2147483647;