5 replies to this topic

[MoRBiiD Legacy]

is there a way to turn off only one cable?

[KingofGamesYami]

GAMR DUD3, on 10 April 2015 - 12:11 AM, said:

is there a way to turn off only one cable?

Set the output to the cables you want to remain on, eg.

rs.setBundledOutput( 2^2 + 2^8 )
--#now turn off one
rs.setBundledOutput( 2^2 )

[MoRBiiD Legacy]

KingofGamesYami, on 10 April 2015 - 12:22 AM, said:

GAMR DUD3, on 10 April 2015 - 12:11 AM, said:

is there a way to turn off only one cable?

Set the output to the cables you want to remain on, eg.

rs.setBundledOutput( 2^2 + 2^8 )
--#now turn off one
rs.setBundledOutput( 2^2 )

how about when you want all lights to be managed independently? (for example: turn on red, turn on blue, turn off blue, turn on red. not turn on red, turn on blue (deactivates red), turn off blue) this is what happens when i use my code:

• shell.run("clear")
  
• textutils.slowPrint("Welcome to *******'s light system!")
  
• textutils.slowPrint("")
  
• textutils.slowPrint("Red--Blue")
  
• textutils.slowWrite("Color: ")
  
• input = read()
  
• if (input) == ("red") then
  
• redstone.testBundledInput("back", colors.red)
  
• if (redstone.testBundledInput("back", colors.red)) == false then
  
• textutils.slowPrint("Status: Off")
  
• textutils.slowWrite("Turn on?: ")
  
• input = read()
  
• if (input) == ("yes") then
  
• redstone.setBundledOutput("back", colors.red)
  
• sleep(1)
  
• shell.run("a")
  
• else shell.run("a")
  
• end
  
• elseif (redstone.testBundledInput("back", colors.red)) == true then
  
• textutils.slowPrint("Status: On")
  
• textutils.slowWrite("Turn off?: ")
  
• input = read()
  
• if (input) == ("yes") then
  
• redstone.setBundledOutput("back", 0)
  
• sleep(1)
  
• shell.run("a")
  
• end
  
• end
  
• elseif (input) == ("blue") then
  
• redstone.testBundledInput("back", colors.blue)
  
• if (redstone.testBundledInput("back", colors.blue)) == false then
  
• textutils.slowPrint("Status: Off")
  
• textutils.slowWrite("Turn on?: ")
  
• input = read()
  
• if (input) == ("yes") then
  
• redstone.setBundledOutput("back", colors.blue)
  
• sleep(1)
  
• shell.run("a")
  
• else shell.run("a")
  
• end
  
• elseif (redstone.testBundledInput("back", colors.blue)) == true then
  
• textutils.slowPrint("Status: On")
  
• textutils.slowWrite("Turn off?: ")
  
• input = read()
  
• if (input) == ("yes") then
  
• redstone.setBundledOutput("back", 0)
  
• sleep(1)
  
• shell.run("a")
  
• else shell.run("a")
  
• end
  
• end
  
• else textutils.slowPrint("Sorry, not an option.")
  
• sleep(2)
  
• shell.run("a")
  
• end

[KingofGamesYami]

rs.setBundledOutput( "back", rs.getBundledOutput( "back" ) - *your color here* )

Removes 1 color.

[Bomb Bloke]

... though that's a bit risky if the line might be executed when the colour is already disabled. Hence colours.subtract(), which can handle such a scenario correctly:

rs.setBundledOutput( "back", colours.subtract(rs.getBundledOutput( "back" ), *your color here* ))

[The Lone Wolfling]

Bomb Bloke, on 10 April 2015 - 07:59 AM, said:

... though that's a bit risky if the line might be executed when the colour is already disabled. Hence colours.subtract(), which can handle such a scenario correctly:

rs.setBundledOutput( "back", colours.subtract(rs.getBundledOutput( "back" ), *your color here* ))

I have two functions to do basically this - add or subtract (a set of) color(s). I find it a whole lot more intuitive to use.