请问怎样让应用程序无论何时都可接收处理消息?
我希望不论应用程序运行到哪一步,都可以响应其它程序发来的消息,应该怎样做到?
例,新建一个Delphi工程:
program Project1;
uses
Forms,
Windows,
Messages,
Dialogs,
Unit1 in 'Unit1.pas' {Form1};
const UserMessageStr = 'testmsg';
{$R *.res}
var
UserMessage:Uint;
Function ChooseFile:Boolean;
var
OpenDialog:TOpenDialog;
Begin
Result:=False;
OpenDialog:=TOpenDialog.Create(nil);
OpenDialog.Title := 'test';
OpenDialog.Filter:='test'+'(*.txt)|*.txt|';
OpenDialog.Options:=[ofHideReadOnly,ofAllowMultiSelect];
OpenDialog.FileName:='';
If OpenDialog.Execute then
Result:=True;
End;
begin
UserMessage:=RegisterWindowMessage(UserMessageStr);
Application.Initialize;
ChooseFile;
Application.CreateForm(TForm1, Form1);
Application.Run;
end.
这个程序在运行的时候,其它程序会发广播消息SendMessage(HWND_BROADCAST,UserMessage,0,0);我希望这个程序无论何时,只要收到消息UserMessage,就马上弹出消息框('Success'),请问应怎样做到。