✓ ANSWER ACCEPTED

Help with part position?

Asked Sep 3, 2015215 views3 answers

I want to make a door that goes up in the "Y" axis, I think it's the Y axis. (Make it going up) I want it to go 1 stud up every 1 second.

Door = script.Parent

wait(9)

Door.Position = Vector3.new(-57.48, 7.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 8.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 9.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 10.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 11.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 12.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 13.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 14.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 15.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 16.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 17.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 18.2, 7)
wait(1)
Door.Position = Vector3.new(-57.48, 19.2, 7)
wait(1)

AnonymousSep 3, 2015

3 answers

✓ Accepted answer

Well a more efficient way to do this would be to use a for loop. To do this, we'll need to set up the loop like so:

for i = 1,12 do
	--Code
end

So now, i is the variable that stores the current iteration we're on. This'll make the 'code' part happen 12 times in a row.

To apply this to your code, we'll use this to our advantage, and just add 1 to the y value per loop:

Door = script.Parent

wait(9)

for i = 1,12 do
	Door.CFrame = CFrame.new(Door.Position) * CFrame.new(0,1,0) --Adds 1 to position every loop
	wait(1)
end

So now your code should work. If you have any further problems/questions, please leave a comment below, and I'll see what I can do. Hope I helped :P

AnonymousSep 3, 2015Edited Sep 3, 2015Permalink

Ok, a much more efficient way to do what you're describing is using a for loop and defining the position of the door and adding 1 to the y axis of the door.

Example:

local Door = script.Parent -- Don't forget your locals

wait(9)
for i =1,12 do -- This loop will start at 1 and keep running until it has ran 10 times. With i as the number of iterations
	Door.Position = Door.Position - Vector3.new(0,1,0) -- This is much more efficient then listing each vector3 position for the door
wait(1) -- Will wait one second
end

AnonymousSep 3, 2015Edited Sep 3, 2015Permalink

1 local Door = script.Parent -- Don't forget your locals 2

3 wait(9) 4 for i =1,12 do -- This loop will start at 1 and keep running until it has ran 10 times. With i as the number of iterations 5 Door.Position = Door.Position - Vector3.new(0,1,0) -- This is much more efficient then listing each vector3 position for the door 6 wait(1) -- Will wait one second 7 end

AnonymousSep 4, 2015Permalink

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