lua code for get the seasons

Developer's Corner General Dev Discussions
1

Is there a simple lua code i could use to deterime which seaon it is?
So if it is winter, spring, summer or autum (at my location)?

I want to use the season to set some lighting and temp control…

How can i do this simply?

2

This may help:
Programming in Lua : 22.1

3

See http://forum.micasaverde.com/index.php?topic=19596.0

4

thx but not really, i still have to make rules to determine which season it is…

I meant more like a app or code like: isSummer() isWinter() isAutum() isSpring() which return me a true or false
Or a code like WhichSeason() which returns the actual season ‘spring’…

5

function WhichSeason() local tNow = os.date("*t") local dayofyear = tNow.yday local season if (dayofyear >= 79) and (dayofyear < 172) then season = "Spring" elseif (dayofyear >= 172) and (dayofyear < 266) then season = "Summer" elseif (dayofyear >= 266) and (dayofyear < 355) then season = "Autumn" else season = "Winter" end return season end

Note: This assumes that you are in the Northern hemisphere!

6

Or:

[code]function isSpring()
local tNow = os.date(“*t”)
local dayofyear = tNow.yday
return (dayofyear >= 79) and (dayofyear < 172)
end

function isSummer()
local tNow = os.date(“*t”)
local dayofyear = tNow.yday
return (dayofyear >= 172) and (dayofyear < 266)
end

function isAutumn()
local tNow = os.date(“*t”)
local dayofyear = tNow.yday
return (dayofyear >= 266) and (dayofyear < 355)
end

function isWinter()
local tNow = os.date(“*t”)
local dayofyear = tNow.yday
return (dayofyear >= 355) or (dayofyear < 79)
end[/code]

Note: The dayofyear conditions are only appropriate for the Northern hemisphere.