for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency - .1
end
wait(1)
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency + .1
end
wait(1)
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency - .1
end
wait(1)
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency +.1
end
wait(1)
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency - .1
end
✓ ANSWER ACCEPTED
Is there a way to make this code possibly shorter?
1 answer
✓ Accepted answer
Sure! Just use a while loop. This will keep it going forever,
while true do
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency - .1
end
wait(1)
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency + .1
end
end
To keep it looping a certain amount of times, just create a value and make it increase after each loop.
local LoopBreak = 0
while true do
if LoopBreak == 5 then --Once the LoopBreak reaches 5, it will break the loop.
break
end
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency - .1
end
wait(1)
for i = 1, 10 do
wait(0.01)
pMenu.loading.ImageTransparency = pMenu.loading.ImageTransparency + .1
end
LoopBreak = LoopBreak + 1 --Increases the value by one everytime it loops.
end
4 comments
0
Thanks, this is perfect. But how would i make it loop for about 5 times?
0
I'll edit the question for ya.
0
K thanks
0
Alright Thanks For Helping :)
This discussion is preserved as it was. New answers, comments, and votes are closed.
1 comment
use the i parameter stop wasting your computer's resources wasting calculations on the i parameter already