In my lua program, i want to stop and ask user for confirmation before proceeding with an operation. I'm not sure how to stop and wait for user input, how can it be done?
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6 Answers
local answer
repeat
io.write("continue with this operation (y/n)? ")
io.flush()
answer=io.read()
until answer=="y" or answer=="n"
2 Comments
Egor Skriptunoff
Does
io.read() impose automatic io.flush() when working with default stdin/out?
lhf
@EgorSkriptunoff, it might, but we can't be sure. I don't think ANSI C says anything about this.
Take a look at the io library, which by default has standard-input as the default input file:
Comments
I've worked with code like this. I will type this in a way it will work:
io.write("continue with this operation (y/n)?")
answer=io.read()
if answer=="y" then
--(put what you want it to do if you say y here)
elseif answer=="n" then
--(put what you want to happen if you say n)
end
Comments
I use:
print("Continue (y/n)?")
re = io.read()
if re == "y" or "Y" then
(Insert stuff here)
elseif re == "n" or "N" then
print("Ok...")
end
1 Comment
PaulR
The conditionals there are wrong, eg
re == "y" or "Y" should be re == "y" or re == "Y". It's probably good to be aware that we can check re:lower() == "y" too.try to use folowing code
m=io.read()
if m=="yes" then
(insert functions here)
end
Comments
print("Continue (y/n)?")
re = io.read()
if re == "y" or "Y" then
(Insert stuff here)
elseif re == "n" or "N" then
print("Ok...")
end
From the bit of lua that I've done (not a lot), I'm going to say that using both uppercase and lowercase letters is redundant if you use string.sub.
print("Continue? (y/n)")
local re = io.read()
--[[Can you get string.sub from a local var?
If so, this works. I'm unfamiliar with io(game
lua uses GUI elements and keypresses in place of the CLI.]]
if re.sub == "y" then
--do stuff
if re.sub == "n" then
--do other stuff
end
That should work.
1 Comment
Demur Rumed
re.sub will resolve to the function string.sub & always be unequal to "y" or "n". Besides, string matching is case sensitive. At best you can do re:match("[nN]") and re:match("[yY]")