如何让程序在系统启动时就运行？

Question

先uses registry; var Reg:Tregistry;
begin
  Reg:=Tregistry.Create;
  Reg.RootKey:=HKEY_LOCAL_MACHINE;
  Reg.OpenKey(Software\Microsoft\Windows\CurrentVersion\Run',False);
  Reg.WriteString('sysExplr','d:\SthVCD\SysExplr.EXE');//换成你的
  Reg.CloseKey;
  Reg.Free;
end;

Answer 1

在注册表中加入HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Run

Answer 2

还是放到Config.sys文件里优先级最高

Answer 3

写在HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Run
或RUNSERVICE下

Answer 4

在注册表的启动项中加入程序的执行路径就可以了 
 先uses  registry;  var  Reg:Tregistry; 
begin 
    Reg:=Tregistry.Create; 
    Reg.RootKey:=HKEY_LOCAL_MACHINE; 
    Reg.OpenKey(Software\Microsoft\Windows\CurrentVersion\Run',False); 
    Reg.WriteString('sysExplr','d:\SthVCD\SysExplr.EXE');//换成你的 
    Reg.CloseKey; 
    Reg.Free; 
end;

Answer 5

HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Run
HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\RUNSERVICE
启动

Answer 6

要是在9X下，用传统的autoexec.bat也行