All you would need is 3 booleans in the blueprint, on, off, and dim. When the light is off set the off boolean to true and the other 2 to false. Do the same for the other 2 booleans. Then you would simply need to check which boolean is true and change the light accordingly.
So I have a Spotlight blueprint that in the details Visible is unchecked, so then it starts off as off. Inside the blueprint, I have the light to toggle whenever the player enters or exits the trigger.
I want to add a third mode where when the light is off, you press ‘E’ and the light turns on but only dim, and if you press ‘E’ again, the light turns on full brightness, all while only when the character is inside the trigger. I want the light to turn off fully whenever the character leaves the box trigger (no matter if it’s dim or on full brightness.)
At EventBeginPlay I have set a bool variable: Off.
Now I have an idea on how to make this work:
When first hit ‘E’, the spotlight’s Attenuation Radius is set to 500 and variable Dim is set. The second time ‘E’ is pressed, the Attenuation Radius is set to 100 and variable On is set…I just don’t know how to make the light tell the difference between Dim and On. Toggle Visibility won’t work, so I know I need to use Set Visibility but I don’t know how to tell the light which variable is set and what kind of variables to use (Bool shouldn’t work right, because there’s three choices not two. So make the variables strings or use Branches?)
Maybe I’m thinking of this the wrong way, any advice would be appreciated, sorry for the long read. I’ll continue to experiment in the meantime.
Ah I see, okay I’ll try that. It’s jsut hard for me to visualize what the code is doing in blueprint. I could have made some simple If blocks of code had this been Java, and I don’t know C++…thanks for the answers.