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  1. Prepare
  2. Python
  3. Regex and Parsing
  4. Validating Credit Card Numbers
  5. Discussions

Validating Credit Card Numbers

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487 Discussions

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  • mohammedaymaned1
     + 0 comments

    here is my code without regex because i don't know what that is 

    `num = int(input())
i = 0
credits = [None] * num
while (i < num):
    credits[i] = input()
    i += 1

for credit in credits:
    if credit[0] not in "456":
        print ("Invalid")
        continue
    count = 0
    digit = False
    is_16_digit = False
    is_consecutive = False
    hyphen_count = 0
    for c in credit:
        if c.isdigit():
            count +=1
            digit = True
        elif c == "-" and count % 4 == 0:
            hyphen_count += 1
            continue
        else:
            digit = False
            break
        if count == 16:
            is_16_digit = True
    credit = credit.replace("-","")
    for c in credit:
        if c+c+c+c in credit:
            is_consecutive = True
            break
    if is_16_digit == False:
        print("Invalid")
        continue
     
    elif digit == False:
        print("Invalid")
        continue
    elif is_consecutive:
        print("Invalid")
        continue
    elif hyphen_count > 3:
        print("Invalid")
        continue
    else:
        print("Valid")

    `

  • dilajevolves
     + 0 comments
    import re
pattern = r'([456]\d{15}|[456]\d{3}(-\d{4}){3})'
for _ in range(int(input())):
    card_no = input()
    if not re.fullmatch(pattern,card_no):
        print("Invalid")
        continue
        
    card_no = card_no.replace("-","")
    
    if re.search(r'(\d)\1{3,}',card_no):
        print("Invalid")
    else:
        print("Valid")

  • mohamedhari317
     + 0 comments
    import re
def isvalid(a):
    pattern=r"^[456]\d{15}|[456]\d{3}(\-\d{4}){3}$"
    if not re.match(pattern,a):
        return False
    a=a.replace("-","")
    if re.findall(r"(\d)\1{3,}",a):
        return False
    return True

     
for _ in range(int(input())):
    a=input()
    print("Valid" if isvalid(a) else "Invalid")

  • maluisamrok
     + 0 comments

    my solution

    import re

patern =r'^[4|5|6](\d{15}|\d{3}(-\d{4}){3})'
duplicate=r'(\d)(-?\1){3}'
N= int(input())
for i in range(N):
    card= input()
    if len(card)<=19 and re.match(patern, card):
        if re.search(duplicate, card):
            print("Invalid")
        else:
            print("Valid")
    else:
        print("Invalid")

  • maxence_qnt
     + 0 comments

    Love this challenge ! It was very fast for me to check the 5 first conditions in regex, but so complex to check the "Not 4 or more consecutive repeated digits" condition.

    For those who come after, I think that a good tip is to create 2 regex patterns. Instead of doing 1 big pattern.

    In my logic, I check all conditions except the first one. Because if already one of them is not valid, I can print "Invalid".

    But if all conditions are valid, then I check the consecutive repeated digit condition.

    For that, I firstly remove "-" if necessary, because it's obviously most complex if not.

    Then I use the regex pattern "([0-9])\1{3,}".

    I didn't know the "back reference" in regex (\1, \2 etc...).

    \1 means : "re-use the last group"

    So ([0-9])\1 means : I capture a digit. And I want this exact same digit after.

    Obviously, by adding {3,} we want that this logic is applied 3 times or more.

    And if this condition is True, it's Invalid, because we don't want to have 4 consecutives repeated digits.

    Note: Be careful to use re.search() and not re.match() for this pattern. Because re.match() only search at the start, whereas re.search() search everywhere in the string.

    Here my code :

    import re

N = int(input())

pattern = r"(^([456])([0-9]{3})-([0-9]{4})-([0-9]{4})-([0-9]{4})$)|(^([456])([0-9]{15})$)"
# Easier to have a second regex pattern only for the repeated consecutive digit condition
pattern_repeated_digit = r"([0-9])\1{3,}"

for _ in range(N):
    current_credit_card = input()

    if bool(re.match(pattern, current_credit_card)):
        
        # Now that we have check all conditions except repeated digits, we will check this last condition. Before, we have to remove '-'. However, it will be too complex.
        if '-' in current_credit_card:
            current_credit_card = current_credit_card.replace('-', '')
        
        # If there is consecutive repeated digit : Invalid. Else : Valid.
        if bool(re.search(pattern_repeated_digit, current_credit_card)):
            print("Invalid")
        else:
            print("Valid")
    else:
        print("Invalid")