Lua 2nd nested if statement is ignored

Question

I am writing a tic tac toe game for my assignment, and I was stuck with the nested if statement. Attached part of my code below:

if name[1] == "AI" then 
     print("test string")
     playerId = 1
     if level == 1 or level == 2 then
        print("AI lvl 1/2")
        easy()
     elseif level == 3 then
        print("AI lvl 3")
        hard()
     end
elseif name[1] == "player" then
    print("test string2")
    Runtime:addEventListener("touch", fill)
end

while this executed, test string was printed on console, and playerId = 1, but inside 2nd if statement was completely ignored. Not even print on console. Could someone help me solve it please? Or tell me what is wrong with my code. Thanks

Check what the value of level is. It sounds like it's something other than 1, 2 or 3. Where are you getting it from? If you post the code where level is defined we might be able to give you some more specific help. — Jack Taylor
– Jack Taylor, Commented Nov 5, 2017 at 3:34
in this scene: local level = composer.getVariable("level"), in the difficulty level scene, composer.setVariable( "level", "1" ), level represents difficulty level, only 1,2,3 — JINEN LIN
– JINEN LIN, Commented Nov 5, 2017 at 4:13
after adding print("level"..level) below playerId =1 of above codes, it prints level 1/2/3 while I selected related level. — JINEN LIN
– JINEN LIN, Commented Nov 5, 2017 at 4:27

user8816383 · Accepted Answer · 2017-11-05 05:33:21Z

3

Add something like print(level, type(level)) to see what level is. Remember, lua can convert strings like '123' to numbers and vice versa when you want to do some arithmetics or string concatenations, but you can't compare strings with numbers like that: 123 == '123' will return false.

So, if level is a string, either replace 1, 2 and 3 in your code with '1', '2' and '3', or convert level to a number: level = tonumber(level)

answered Nov 5, 2017 at 5:33

user8816383

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