#python : I want the correction of this exercise | SoloLearn: Learn to code for FREE!

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#python : I want the correction of this exercise

Loop Write a program that asks the user for a 4-digit code and leaves him 3 attempts to find the correct password (the password to find is password = “2021"

3/1/2021 9:30:53 PM

Mamadou Drame

12 Answers

New Answer

+3

isnt an attempt required to be corrected?

+2

attempts = 0 start = True while start: attempts += 1 in = int(input()) if in == 2021: print("correct") break if attempts == 3: start = False

+1

Hint: While loops .isnumeric ()

+1

Mamadou Drame I'm not writing the code to improve yourself. I'm just hinting.😉 1. Use loop 2. Ask Password 3. Compare input with Password 4. Before loop, define a variable that holds the number of attempts 5. Increase this variable by 1 each time and compare with 3 attempts 6. If the variable is 3, print a warning and end the loop. I wish you success😊

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where is the code to correct?

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Mamadou Drame I saw this same question with a code attempt. Is this a duplicate? EDIT: 👍 maybe it was someone else that has a similar similar class. In addition to ∆BH∆Y comment a counter variable to test and branch elsewhere after 3 attempts.

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I have an exam that our teacher gave us

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what did u tried so far?

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Hello ! well I have tried beautiful and there is no rush

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There is a small modification print ('password combo to continue') count = 0 while count <3: password = input ('Enter password:') if password == '2021' print ('Access granted') break

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#this is an advanced code, hopefully you will understand it and learn smthg from it import re trials = 0 pwd = 2021 pat = "\d{4}" while trials <= 2: try: t = int(input("Enter password: ")) if t == pwd: print("access granted") break elif re.search(pat, str(t)) == None: print("only 4 digits allowed!") trials += 1 print("Trial N°", trials) else: print("Access denied") trials += 1 print("Trial N°", trials) except ValueError: print("Invalid input! only digits allowed!") trials += 1 print("Trial N°", trials)

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for i in "21": if input() == "2021": print("Access Granted!") exit() else: print("Attempt Failed!\n{} attempt{} left".format(i, "s" * (i > "1"))) raise ValueError("INTRUDER ALERT!") # Hope this helps