Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
584 views
in Technique[技术] by (71.8m points)

ruby - Luhn Algorithm- Returning True for a False Credit Card

Having a little issue with my code to see if a credit card number adheres to the Luhn Algorithm. The code is returning true when the Credit Card is divisible by 10, but also is returning true when the CC number is not divisible by 10. I have printed out the final sum to make sure the numbers were actually adding to the sum variable, and they seem to be.. Below is my code. I know it can be cleaner, but at this stage I would like to see it work first.

def check_card

   c_num= []

   sum=0

   s_numbers=@card_numbers.to_s.reverse.split("")

   s_numbers.each_slice(2) do |x| 
   c_num << (x.last.to_s.to_i*2)
   c_num << (x.first.to_s.to_i)
     end

  c_num.each do |num|
    if num.to_i > 9
      sum+= (num.to_i % 10)+1
    else 
      sum += num.to_i
    end
  end

sum % 10==0 

end

Here is how it is being called:

it 'returns false for a bad card' do
  card = CreditCard.new(4408041234567892)
  card.check_card.should eq false
end
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Now that another answer has appeared I will offer a suggested coding. This does not answer your question, but I thought it might be of interest and couldn't very well put it in a comment, because of formatting limitations.

def valid?(card)
  return false unless card  =~ /^d+$/ # Ensure card contains only digits
  arr = card.split('').reverse.each_with_index.map {|d, index| (index.odd? ? 2*(d.to_i) : d.to_i)}
  (arr.join.split('').inject(0) {|tot, d| tot + d.to_i}) % 10 == 0
end 

valid?("1234567890123456") => false

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...