[Solved] Input menu, go to a new result?
Posted: Sun May 24, 2020 7:22 am
Sorry for the vague title, not sure how to phrase it.
I already know which part breaks the code, but I don't know how to fix it... So lemme just show you and explain it.
label user123:
$ user1_input = renpy.input(prompt = "Input code. To leave, enter '0'", allow="1234567890")
$ user1_input = user1_input.lower().strip()
if user1_input == "": #This will happen if there's no input.
system "You need to actually write something..."
jump user123
if user1_input in ('0'):
jump user1
if user1_input == ('123') and x == False:
system "stuff is said"
$ x = True
elif user1_input == ('432'):
system "blergh"
elif user1_input == ('123','432'): # if you remove the '432' the code will work.
$ xyz = renpy.random.randint(1,3)
if xyz == 1:
system "more stuff is said"
if xyz == 2:
system "more stuff more stuff"
if xyz == 3:
system "this is more stuff being said"
jump user123
any help will be appreciated!
I already know which part breaks the code, but I don't know how to fix it... So lemme just show you and explain it.
label user123:
$ user1_input = renpy.input(prompt = "Input code. To leave, enter '0'", allow="1234567890")
$ user1_input = user1_input.lower().strip()
if user1_input == "": #This will happen if there's no input.
system "You need to actually write something..."
jump user123
if user1_input in ('0'):
jump user1
if user1_input == ('123') and x == False:
system "stuff is said"
$ x = True
elif user1_input == ('432'):
system "blergh"
elif user1_input == ('123','432'): # if you remove the '432' the code will work.
$ xyz = renpy.random.randint(1,3)
if xyz == 1:
system "more stuff is said"
if xyz == 2:
system "more stuff more stuff"
if xyz == 3:
system "this is more stuff being said"
jump user123
any help will be appreciated!