How can I fix the "input ==1:" on line 19 problem such that if i input "1" it must have an output of
A-C,2-C,3-C,4-C,5-C,6-C,7-C,8-C,9-C,10-C,J-C,Q-C,K-C
This is my trial and error code for this problem
from typing import List, Any
import random
def printmenu():
print("Menu")
print("1. Show first 13 cards")
print("2. Shuffle and show first 13 cards")
print("3. Play 2 card poker")
print("4. Stack the deck and show first 13 cards")
print("5. Exit")
_input: int = int(input())
return options(_input)
def options(_input: int):
suits = ['C', 'D', 'H', 'S']
value = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K']
deck = []
if _input == 1:
first13cards = []
if(deck == []):
for i in suits:
for j in value:
print(j+"-"+i, end=",")
elif deck != []:
deck.clear()
first13cards.clear()
for i in range(52):
deck.append(value[random.randint(0, 12)] + suits[random.randint(0, 3)])
for i in range(13):
first13cards.append(deck[i])
return printmenu()
elif _input == 2:
for i in range(13):
deck.append(value[random.randint(0, 12)] + suits[random.randint(0, 3)])
print(deck)
return printmenu()
elif _input == 3:
#play 2 cards poker
return printmenu()
elif _input == 4:
#stack the deck & show first13 cards
return printmenu()
elif _input == 5:
exit() # exit the program
for j in value:
suit = suits[random.randint(0, 3)]
_temp = j + suit + ""
deck.append(_temp)
print(deck)
return deck
if __name__ == '__main__':
while True:
printmenu()
if printmenu() == 5:
break
This is my trial and error code for this set
For the string value that you're looking for, you could do something like
where each suit will be concatenated with
-C, then joined together into one string with each element separated by a comma.