Delphi程序只能打开一个的方法.doc

工作上，避免我的Upload在一个电脑上被打开多个。1、新建一个unit名字是 保存为 once.pas ，内容如下：unit once;interfaceusesWindows ,Messages, SysUtils, Classes, Forms;implementationconstSTR_UNIQUE = 2BE6D96E-827F-4BF9-B33E-8740412CDE96;MI_ACTIVEAPP =1; 激活应用程序MI_GETHANDLE =2; 取得句柄variMessageID : Integer;OldWProc : TFNWndProc;MutHandle : THandle;BSMRecipients : DWORD;function NewWndProc(Handle: HWND; Msg: Integer; wParam, lParam:Longint):Longint; stdcall;beginResult := 0;if Msg = iMessageID thenbegincase wParam ofMI_ACTIVEAPP: 激活应用程序if lParam0 thenbegin收到消息的激活前一个实例为什么要在另一个程序中激活?因为在同一个进程中SetForegroundWindow并不能把窗体提到最前if IsIconic(lParam) thenOpenIcon(lParam)elseSetForegroundWindow(lParam);Application.Terminate; 终止本实例end;MI_GETHANDLE: 取得程序句柄beginPostMessage(HWND(lParam), iMessageID, MI_ACTIVEAPP,Application.Handle);end;end;endelseResult :=CallWindowProc(OldWProc, Handle, Msg, wParam, lParam);end;procedure InitInstance;begin取代应用程序的消息处理OldWProc := TFNWndProc(SetWindowLong(Application.Handle, GWL_WNDPROC,Longint(NewWndProc);打开互斥对象MutHandle := OpenMutex(MUTEX_ALL_ACCESS, False,STR_UNIQUE);if MutHandle = 0 thenbegin建立互斥对象MutHandle := CreateMutex(nil,False, STR_UNIQUE);endelse beginApplication.ShowMainForm := False;已经有程序实例,广播消息取得实例句柄BSMRecipients :=BSM_APPLICATIONS;BroadCastSystemMessage(BSF_IGNORECURRENTTASK or BSF_POSTMESSAGE,BSMRecipients, iMessageID, MI_GETHANDLE,Application.Handle);end;end;initialization注册消息iMessageID :=RegisterWindowMessage(STR_UNIQUE);InitInstance;finalization还原消息处理过程if OldWProc Nil thenSetWindowLong(Application.Handle, GWL_WNDPROC,LongInt(OldWProc);关闭互斥对象if MutHandle 0then CloseHandle(MutHandle);end.2、把这个once.pas考到 upload（我的工作程序里），然后打开UpLoad程序（也可以是你自己的程序）