Display submenu item based on gedcom file id

[George Scott]

I have a test such as the following;
// Fourth sub-menu (visible only to users with site Admin rights)
if (PGV_USER_IS_ADMIN) {
$submenu = new Menu("About This Data", "custom link #2");
$submenu->addClass("submenuitem$ff", "submenuitem_hover$ff");
$menu->addSubmenu($submenu);
}
I would like to test instead if I'm currently using the gedfile Scott.ged.  What is the proper if statement?

[Gerry Kroll]

The following constants might help:
PGV_GED_ID : the currently active internal GEDCOM ID
PGV_GEDCOM : file name of that curretnly active GEDCOM

The following function might also help:
get_gedcom_from_id($ged_id)

Input is any valid GEDCOM number.  Output is the GEDCOM file name to match.

[George Scott]

Thanks!  The following works great:
if (PGV_GEDCOM=="HestonTest.ged"){
$submenu = new Menu("Link to Scott GEDCOM", "http://www.hestonscottgen.com/phpgedview/individual.php?pid=I581&ged=ScottMC.ged");
$submenu->addClass("submenuitem$ff", "submenuitem_hover$ff");
$menu->addSubmenu($submenu);
}