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

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.

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

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• 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';
  }
  

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