Need help for code to let user try again in Python

Question

Basically, I am creating this game which consists of the user answering some multiple choice questions. I want it so when the user gets the answer wrong once, they try again and if they get the answer wrong twice, the correct answer shows up. But right now, if the user gets the answer wrong, it skips to the next question. How can I do this? Any help is appreciated. Here is my code:

score = 0
def correct():
  print("Congratulations! You got a point!")
  global score
  score = score + 1
  print("Points: ",score)
  
def incorrect():
  print("Wrong, try again!")
  print("Points: ",score)
  
  
def question(q,a):
  print(q)
  answer = input("Answer: ")
  if answer == a:
    correct()
  else:
    incorrect()

question("What is the capital city of England?\n a. London\n b. Paris\n c. York", 'a')
question("What is the capital city of France?\n a. Rome\n b. Paris\n c. Jersey", 'b')
question("What is the capital city of Germany?\n a. Accra\n b. Weimar\n c. Berlin", 'c')
question("What is the capital city of The Netherlands?\n a. Moscow\n b. Amsterdam\n c. Cape Town", 'b')

Can you add only the necessary code on here, instead of the link. — de_classified, Aug 05 '20 at 20:55
after `incorrect()` You call the question function again `question(q,a)` — RaoufM, Aug 05 '20 at 20:55

alani · Answer 1 · 2020-08-05T21:14:40.803

1

Use a for loop. If a correct answer is given, return early from the function.

def question(q,a):
    print(q)
    for attempt in range(2):
        answer = input("Answer: ")
        if answer == a:
            correct()
            return
        else:
            incorrect()
    print("The answer is:", a)

Note: based on the code in the question, the call to incorrect() will always write "try again". You might not want it to print "try again" after a second failed attempt, because the player does not get to try again in that case. You could change this by passing in the attempt number as a parameter to the incorrect function and put a suitable if condition in that function to decide whether to print "try again" or not.

edited Aug 05 '20 at 21:14

answered Aug 05 '20 at 20:58

alani

12,573
2
13
23

this is good too, but an alternate would be to make use of `while True:` and then break at the end after `correct()`. this leaves out ambiguity for needing a `range()` of some sort. – de_classified Aug 05 '20 at 21:01
@FishingCode the user only gets two attempts - if there are two wrong answers, it moves onto the next question. – alani Aug 05 '20 at 21:02
Ah, missed the bit about the correct answer shows up. I'll add that... – alani Aug 05 '20 at 21:03
ah yup. that's right only 2 attempts. – de_classified Aug 05 '20 at 21:04

score 1 · Answer 2 · answered Aug 05 '20 at 21:01

1

You could modify question function:

def question(q,a):
  print(q)
  answer = input("Answer: ")
  if answer == a:
    correct()
  else:
    print(q)
    answer = input("Answer: ")
    if answer == a:
      correct()
    else:
      incorrect()

answered Aug 05 '20 at 21:01

Martin Lim

21
3

this is only repeating the user input, and keeps checking again if the answer is incorrect, more of a hard-code. – de_classified Aug 05 '20 at 21:07
1

true solution by @alaniwi is more elegant – Martin Lim Aug 05 '20 at 21:15

Need help for code to let user try again in Python

2 Answers2