Problem of the Day
A new programming or logic puzzle every Mon-Fri

Credit Card Type

To continue on from yesterday's credit card validation we also need a way to send over what type of card the user has entered. Rather than asking the user for the type of card we handle this automagically. Write a function to determine what type of credit card (amex, visa, etc) the user has entered.

Note: part of the challenging is figuring out the formula to how this is done.

Permalink: http://problemotd.com/problem/credit-card-type/

Comments:

  • Anonymous - 10 years, 5 months ago

    I extended my solution for yesterday's problem

    main :: IO ()
main = do
            putStrLn "Credit Card Number: "
            i <- readLn
            let k = main' i 
            if k
                then putStrLn $ checkCardTyp $ intToList i
                else putStrLn "Invalid"


checkCardTyp :: [Int] -> String
checkCardTyp (x:xs) 
            | x == 4 = "Visa"
            | x == 5 && mc xs = "Mastercard"
            | x == 3 && ae xs = "American Express"
            | x == 6 && dis xs = "Discover"
            | x == 3 && dc xs = "Diners Club"
            | (x == 2 && jcb2 xs) || (x == 1 && jcb1 xs) || (x == 3 && jcb3 xs) = "JCB"
            | otherwise = "No Credit Card"
                where
                    mc (y:ys) = y >= 1 && y <= 5
                    ae (y:ys) = y == 4 || y == 7
                    dis (y:ys) = y == 5 || (y == 0 && dis' ys)
                    dis' (y1:y2:ys) = y1 == 1 && y2 == 1
                    dc (y:ys) = y == 6 || y == 8 || (y == 0 && dc' ys)
                    dc' (y:ys) = y >= 0 && y <= 5
                    jcb1 (y1:y2:y3:ys) = y1 == 8 && y2 == 0 && y3 == 0
                    jcb2 (y1:y2:y3:ys) = y1 == 1 && y2 == 3 && y3 == 1
                    jcb3 (y:ys) = y == 5

    where the main' is the main function of yesterday (but this time it takes only 1 number (full credit card number))

    i tested it with my credit card and it worked... maybe there are some typos, but these are the rules i found on the net

    reply permalink

  • bumbleguppy - 10 years, 5 months ago

    javascript

    function creditCardType(num){
    var regxp;
    var creditCardTypes =   [['^4[0-9]{6,}$','Visa'],
                            ['^5[1-5][0-9]{5,}$','MasterCard'],
                            ['^3[47][0-9]{5,}$', 'American Express'],
                            ['^3(?:0[0-5]|[68][0-9])[0-9]{4,}$', 'Diners Club'],
                            ['^6(?:011|5[0-9]{2})[0-9]{3,}$', 'Discover'],
                            ['^(?:2131|1800|35[0-9]{3})[0-9]{3,}$', 'JCB']];

    num = num.toString().replace(/\D/, '');

    for(var i = 0, len = creditCardTypes.length; i < len; i++){
        regxp = new RegExp(creditCardTypes[i][0]);
        if(regxp.test(num)){
            return creditCardTypes[i][1];
        }
    }

    return 'Unknown';
}

    reply permalink

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