请问如何让弹出窗体像QQ一样慢慢从屏幕右上角向下展开？？？

Question

要保证每次在不同机器上都是从屏幕右上角开始展开？！

Answer 1

你用个Timer不就可以实现了吗？

Answer 2

我收集过一个人家写的控件，可以参考一下：
unit EaPopup;interfaceuses
  Windows, Messages, SysUtils, Classes, Graphics, Controls, Forms, Dialogs,PoPoMsg;type
  TEaPopup = class(TComponent)
  private
    { Private declarations }
    FPopMsg: TPoPMsg;
    FTitle: string;
    FMessages: string;
    FStayTime: Integer;
    procedure SetStayTime(const Value: Integer);
  protected
    { Protected declarations }
  public
    { Public declarations }
    constructor Create(AOwner: TComponent);override;
    procedure PopUp;overload;
    procedure PopUp(const msgTitle, msg: string; msgStayTime:Integer);overload;
  published
    { Published declarations }
    property  Title: string read FTitle write FTitle;
    property  Messages: string read FMessages write FMessages;
    property  StayTime: Integer read FStayTime write SetStayTime;
  end;procedure Register;implementationprocedure Register;
begin
  RegisterComponents('EaPack', [TEaPopup]);
end;{ TEaPopup }
procedure TEaPopup.PopUp;
begin
  PopUp(FTitle,Messages,FStayTime);
end;constructor TEaPopup.Create(AOwner: TComponent);
begin
  inherited;
  FTitle := '消息';
  FMessages := 'Easoft Popup';
  FStayTime := 5;
end;procedure TEaPopup.PopUp(const msgTitle, msg: string; msgStayTime:Integer);
begin
  if Assigned(FPoPMsg) then
    FreeAndNil(FPopMsg);
  FPopMsg := TPoPMsg.Create(Self);
  
  with FPopMsg do
  begin
    Title.Caption := msgTitle;
    MessageText.Caption := msg;
    StayTime := msgStayTime;
    if StayTime < 1 then
      StayTime := 5;
    ScrollShow;
  end;
end;
procedure TEaPopup.SetStayTime(const Value: Integer);
begin
  FStayTime := Value;
  if FStayTime <1 then
    FStayTime := 5;
end;
end.