Reid Posted June 1, 2014 Posted June 1, 2014 Not too sure how this would be done , some assistance would be nice osb2 btw
Eliot Posted June 1, 2014 Posted June 1, 2014 (edited) See Swizz's answer. Edited June 1, 2014 by Eliot
Deffiliate Posted June 1, 2014 Posted June 1, 2014 (edited) One basic way to do it (not update proof) is to simply check that position for the ID of the open door, there are better ways to do this, but this is the most basic (that I can think of). No ID required, use the name of the door. Just search for the doors containing w/e the name of the door is, then check if any of the positions of those doors are in the open position. Edited June 1, 2014 by Deffiliate
Eliot Posted June 1, 2014 Posted June 1, 2014 No ID required, use the name of the door. Just search for the doors containing w/e the name of the door is, then check if any of the positions of those doors are in the open position. How do you know if the door is open or closed?
Swizzbeat Posted June 1, 2014 Posted June 1, 2014 lol you guys make this so complex: Arrays.asList(door.getDefinition().getActions()).contains("Close") 1
Eliot Posted June 1, 2014 Posted June 1, 2014 lol you guys make this so complex: Arrays.asList(door.getDefinition().getActions()).contains("Close") Yes, do it this way.
Reid Posted June 1, 2014 Author Posted June 1, 2014 No ID required, use the name of the door. Just search for the doors containing w/e the name of the door is, then check if any of the positions of those doors are in the open position. worked like a charm thanks
Nezz Posted June 1, 2014 Posted June 1, 2014 You can also check the orientation. But Swizz's option is probably the best.