How do I turn off one color in a bundled cable?

is there a way to turn off only one cable?

GAMR DUD3, on 10 April 2015 - 01:11 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 )

KingofGamesYami, on 10 April 2015 - 01:22 AM said:

GAMR DUD3, on 10 April 2015 - 01:11 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:

1. shell.run("clear")
2. textutils.slowPrint("Welcome to *******'s light system!")
3. textutils.slowPrint("")
4. textutils.slowPrint("Red–Blue")
5. textutils.slowWrite("Color: ")
6. input = read()
7. if (input) == ("red") then
8. redstone.testBundledInput("back", colors.red)
9. if (redstone.testBundledInput("back", colors.red)) == false then
10. textutils.slowPrint("Status: Off")
11. textutils.slowWrite("Turn on?: ")
12. input = read()
13. if (input) == ("yes") then
14. redstone.setBundledOutput("back", colors.red)
15. sleep(1)
16. shell.run("a")
17. else shell.run("a")
18. end
19. elseif (redstone.testBundledInput("back", colors.red)) == true then
20. textutils.slowPrint("Status: On")
21. textutils.slowWrite("Turn off?: ")
22. input = read()
23. if (input) == ("yes") then
24. redstone.setBundledOutput("back", 0)
25. sleep(1)
26. shell.run("a")
27. end
28. end
29. elseif (input) == ("blue") then
30. redstone.testBundledInput("back", colors.blue)
31. if (redstone.testBundledInput("back", colors.blue)) == false then
32. textutils.slowPrint("Status: Off")
33. textutils.slowWrite("Turn on?: ")
34. input = read()
35. if (input) == ("yes") then
36. redstone.setBundledOutput("back", colors.blue)
37. sleep(1)
38. shell.run("a")
39. else shell.run("a")
40. end
41. elseif (redstone.testBundledInput("back", colors.blue)) == true then
42. textutils.slowPrint("Status: On")
43. textutils.slowWrite("Turn off?: ")
44. input = read()
45. if (input) == ("yes") then
46. redstone.setBundledOutput("back", 0)
47. sleep(1)
48. shell.run("a")
49. else shell.run("a")
50. end
51. end
52. else textutils.slowPrint("Sorry, not an option.")
53. sleep(2)
54. shell.run("a")
55. end

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

Removes 1 color.

… 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* ))

Bomb Bloke, on 10 April 2015 - 08:59 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.