防止窗口重复创建的问题

Question

要求:
点击菜单后,窗口用show的方式打开.各窗口之间可切换.再次点击该菜单.将把先前已打开的该窗口重新击活显示在最上方,而不是重新创建.
这是几个api函数的问题,哪位熟的帮帮忙.

Answer 1

"防止窗口重复创建"??mdi窗体的问题？？procedure OpenChildForm(FormClass: TFormClass; var Fm; AOwner:TComponent);
var
  i: integer;
  Child:TForm;
begin
  for i := 0 to Screen.FormCount -1 do
      if Screen.Forms[i].ClassType=FormClass then
      begin
        Child:=Screen.Forms[i];
        if Child.WindowState=wsMinimized then
           ShowWindow(Child.handle,SW_SHOWNORMAL)
        else
           ShowWindow(Child.handle,SW_SHOWNA);
        if (not Child.Visible) then Child.Visible:=True;
        Child.BringToFront;
        Child.Setfocus;
        TForm(Fm):=Child;
        exit;
      end;
  Child:=TForm(FormClass.NewInstance);
  TForm(fm):=Child;
  Child.Create(AOwner);
end;
//调用
OpenChildForm(TChildForm1,ChildForm1, Self);

Answer 2

if assigned(ChildForm) then
begin
  ChildForm.show;
  ChildForm.BringToFront;
end;

Answer 3

If Application.FindComponent('FormName')=nil then
  创建窗体;
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