Develop Reference

How can I create a credit card validator using Luhn's algorithm?

I am trying to create a simple credit card validator using Luhn's algorithm. If the check digit matches the last inputted number, then it should alert the user that it is valid. Else, say that it isn't valid. Currently, I am getting an error with my sum (total) coming up as NaN. I assume that is the only problem with the code.
<input type="number" id="creditCard" placeholder="0000 0000 0000 0000">
<input type="submit" id="checkButton" value="CHECK VALIDITY" onclick="checkNumber()">
function checkNumber() {
let number = document.getElementById("creditCard").value;
let multiplier = "212121212121212";
let multipliedNumber;
let multipliedString;
if (number.length != 16) {
alert("Please enter a Credit Card number that is 16 digits in length.");
} else {
for (count = 0; count < number.length - 1; count++) {
multipliedNumber = number[count] * multiplier[count];
console.log(multipliedNumber);
if (multipliedNumber > 9) {
multipliedNumber = multipliedNumber[0] + multipliedNumber[1];
multipliedString = multipliedString + multipliedNumber;
} else {
multipliedString = multipliedString + multipliedNumber;
}
}
console.log(multipliedString);
let checkDigit = 10 - (multipliedString % 10);
if (checkDigit == number[15]) {
alert(`${number} is a valid Credit Card number.`);
} else {
alert(`${number} is not a valid Credit Card number.`);
}
}
}

There are several issues:
multipliedNumber is a product, so it is a number type. Therefore accessing properties like [0] or [1] on it, will just evaluate to undefined. Either turn that value to string first, or (better) use arithmetic to extract the two digits:
multipliedNumber = multipliedNumber % 10 + Math.floor(multipliedNumber/10);
multipliedString is not initialised, so adding things to it will not give the desired outcome. Secondly, you define it as a string, but it should be a number, as with Luhn's algorithm you are supposed to sum up the resulting digits, not concatenate them. So initialise a variable like this:
sum = 0;
... and use it like you did -- although you could benefit from the += operator, and since the operation is the same for both cases, you can do it outside of the if..else blocks.
The calculation of the check digit is wrong when the modulo operation evaluates to 0: 10 - (multipliedString % 10) then returns 10, but in that case the check digit is supposed to be 0. It is much easier to just treat that last digit also in the loop and then check that you have reached a multiple of 10. This is also how the algorithm is explained on Wikipedia
Corrected version:
function checkNumber() {
let number = document.getElementById("creditCard").value;
let multiplier = "2121212121212121"; // One more character added...
let multipliedNumber;
let sum = 0 // Initialise it as a number.
if (number.length != 16) {
console.log("Please enter a Credit Card number that is 16 digits in length.");
} else {
for (count = 0; count < number.length; count++) { // Include last digit in loop
multipliedNumber = number[count] * multiplier[count];
if (multipliedNumber > 9) {
// Use arithmetic to add the two digits
multipliedNumber = multipliedNumber % 10 + Math.floor(multipliedNumber/10);
}
sum += multipliedNumber;
}
let check = sum % 10; // Simpler now because all digits were processed
if (check == 0) { // Sum is multiple of 10
console.log(`${number} is a valid Credit Card number.`);
} else {
console.log(`${number} is not a valid Credit Card number.`);
}
}
}
<input type="number" id="creditCard" placeholder="0000 0000 0000 0000">
<input type="submit" id="checkButton" value="CHECK VALIDITY" onclick="checkNumber()">

wrong error message output in function

I am trying to validate a Poperty value and down payment. The conditions are as follows:
property value:
Must be present
Must be numeric - positive - whole number
Must be at least 65,000 dollars more that the down payment.
down payment:
Must be present
Must be numeric - positive - whole number
Must be at least 20% of the value of the property (propValue)
My function is (sort of) working. It doesn't pass all the validation tests. If someone can point me in the right direction as to how to improve this it would be greatly appreciated. My 2 functions for the down pay and value:
function propValueValidation(errMessages){
var propValueLength = document.mortgage.propValue.value.length;
var propValueNumber = isNaN(document.mortgage.propValue.value);
var propValue = document.mortgage.propValue.value;
var downPayPlus = document.mortgage.downPay.value + 65000;
if (!propValueLength) {
errMessages += " Property Value is a required field";
return errMessages;
}
else if (typeof propValue === 'number') {
var remainder = (propValue % 1);
if(remainder != 0){
errMessages += "Property Value must be a positive whole number";
return errMessages;
}
}
else if (propValue < downPayPlus){
errMessages += "Property Value must be at least 65,000 greater than the down payment";
return errMessages;
}
return errMessages;
}
//validate down pay
function downPayValidation(errMessages){
var downPayLength = document.mortgage.downPay.value.length;
var downPay = document.mortgage.downPay.value;
var propValueMin = document.mortgage.propValue.value * 0.2;
if (!downPayLength) {
errMessages += "Down Payment is a required field";
return errMessages;
}
else if (typeof downPay === 'number') {
var remainder = (downPay % 1);
if(remainder != 0){
errMessages += "Down Payment must be a positive whole number";
return errMessages;
}
}
else if (downPay < propValueMin){
errMessages +="Down Payment must be at least 20% of the property value";
return errMessages;
}
return errMessages;
}
HTML:
<label class="label">Property Value </label>
<input type="text" name="propValue" id="propValue" size="7" maxlength="6" >
<br>
<label class="label">Down Payment </label>
<input type="text" name="downPay" id="downPay" size="7" maxlength="6" >
when downpay is "1nn1" it will still submit the form for example. Thanks!

you can use parseInt() to confirm its a number.
var tempVal = document.mortgage.propValue.value;
var propValue = parseInt(tempVal); // this will try to extract an integer from tempVal
if (tempVal != propValue.toString()) // if true, there were non-number chars or value is NaN
{
errMessages += "Bad value, please enter an integer";
}

You should not use isNaN on string values (which input values are). Instead first convert such a string to number with Number() (or parseFloat, but Number will require the whole of the input to parse as number, while parseFloat or parseInt will accept strings that start with a number). Then call isNaN on that.
There is also a problem with the if else logic, because you have one branch for numeric data (where it will never get, because the value property of input elements is always a string), and an else on that to compare the amount with another amount. Yet for that last test the value must be numeric. So that test is in the wrong place.
Here is your first function's code with some alterations made:
function propValueValidation(errMessages){
var propValue = document.mortgage.propValue.value;
var propValueNumber = Number(propValue);
var downPayPlus = propValueNumber + 65000;
var genericMsg = ' property value was provided. Please provide a positive' +
' whole number, at least 65,000 greater than the down payment.\n';
if (!propValue.length) {
errMessages += 'No' + genericMsg;
} else if (isNaN(propValueNumber)) {
errMessages += 'A non-numerical' + genericMsg;
} else if (propValueNumber % 1) {
errMessages += 'A fractional' + genericMsg;
} else if (propValueNumber < downPayPlus){
errMessages += 'A too small' + genericMsg;
}
return errMessages;
}

Javascript - String of a Byte with all combinations possible

i have a sting with a byte in it ("00001011") and now id like to get a array with all possible combinations of the 1 (acitve) "bits" in it also as a "byte string"
so from
var bString = "00001011"; //outgoing string
to a array with all string in it with all possible combinations of this "byte string" like - "00000001", "00000011", "00000010" and so on
is that possible?
thank you in advance

function combinations( input ){
var number = parseInt( input, 2 );
var combinations = [];
var zeroes = (new Array(input.length)).join(0);
for(var i=1;i<=number;i++){
if((i&number) == i){ combinations.push( i ) }
}
return combinations.map( function(dec){
return (zeroes + dec.toString(2)).substr( -zeroes.length-1 );
});
}
http://jsfiddle.net/jkf7pfxn/3/
console.log( combinations("00001011") );
// ["00000001", "00000010", "00000011", "00001000", "00001001", "00001010", "00001011"]
The idea goes as follows: iterate all numbers from 1 to the input number. If current number AND input number return the current number then both have 1 bits in the same place.
On a smaller number, "0101" (which is 5) it works as follows:
1 & 5 == 1, (0001 & 0101) push 1 to the matches.
2 & 5 == 0, (0010 & 0101) no match.
3 & 5 == 1, (0011 & 0101) no match.
4 & 5 == 4, (0100 & 0101) push 4 to the matches.
5 & 5 == 5, (0101 & 0101) push 5 to the matches.
So the combinations for 0101 are 1 (0001), 2 (0010), 4 (0100) and 5 (0101).
Then there's this little trick to pad numbers with zeroes:
var zeroes = (new Array(input.length)).join(0); // gives a long enough string of zeroes
then
// convert to base 2, add the zeroas at the beginning,
// then return the last n characters using negative value for substring
return (zeroes + dec.toString(2)).substr( -1 * zeroes.length);

Since 11111111 is 255 so just loop all values and convert them to binary
$(document).ready(function() {
for (var i = 0; i < 256; i++) {
$('#core').append('<div>' + dec2bin(i) + '</div>');
}
function dec2bin(dec) {
return ('00000000' + (dec >>> 0).toString(2)).slice(-8);
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id='core'></div>

If you want to enumerate all combinations of binary numbers where 1 can only be in the place of your pattern, you can write a simple recursive function:
var input = "00010111";
var current = [];
function combinations()
{
if (input.length === current.length)
{
var output = current.join('');
if (parseInt(output, 2) !== 0) // exclude all-zeroes case
document.body.innerHTML += output + "<br/>";
return;
}
current.push('0');
combinations();
current.pop();
if (input[current.length - 1] === '1')
{
current.push('1');
combinations();
current.pop();
}
}
combinations();
This algorithm works well for input of any length.
Although it is a recursion, it has a linear time complexity.

Advice on how to code Luhn Credit Card validation with Javascript?

I'm pretty awful at Javascript as I've just started learning.
I'm doing a Luhn check for a 16-digit credit card.
It's driving me nuts and I'd just appreciate if someone looked over it and could give me some help.
<script>
var creditNum;
var valid = new Boolean(true);
creditNum = prompt("Enter your credit card number: ");
if((creditNum==null)||(creditNum=="")){
valid = false;
alert("Invalid Number!\nThere was no input.");
}else if(creditNum.length!=16){
valid = false;
alert("Invalid Number!\nThe number is the wrong length.");
}
//Luhn check
var c;
var digitOne;
var digitTwo;
var numSum;
for(i=0;i<16;i+2){
c = creditNum.slice(i,i+1);
if(c.length==2){
digitOne = c.slice(0,1);
digitTwo = c.slice(1,2);
numSum = numSum + (digitOne + digitTwo);
}else{
numSum = numSum + c;
}
}
if((numSum%10)!=0){
alert("Invalid Number!");
}else{
alert("Credit Card Accepted!");
}
</script>

The immediate problem in your code is your for loop. i+2 is not a proper third term. From the context, you're looking for i = i + 2, which you can write in shorthand as i += 2.
It seems your algorithm is "take the 16 digits, turn them into 8 pairs, add them together, and see if the sum is divisible by 10". If that's the case, you can massively simplify your loop - you never need to look at the tens' place, just the units' place.
Your loop could look like this and do the same thing:
for (i = 1; i < 16; i +=2) {
numSum += +creditNum[i];
}
Also, note that as long as you're dealing with a string, you don't need to slice anything at all - just use array notation to get each character.
I added a + in front of creditNum. One of the issues with javascript is that it will treat a string as a string, so if you have string "1" and string "3" and add them, you'll concatenate and get "13" instead of 4. The plus sign forces the string to be a number, so you'll get the right result.
The third term of the loop is the only blatant bug I see. I don't actually know the Luhn algorithm, so inferred the rest from the context of your code.
EDIT
Well, it would have helped if you had posted what the Luhn algorithm is. Chances are, if you can at least articulate it, you can help us help you code it.
Here's what you want.
// Luhn check
function luhnCheck(sixteenDigitString) {
var numSum = 0;
var value;
for (var i = 0; i < 16; ++i) {
if (i % 2 == 0) {
value = 2 * sixteenDigitString[i];
if (value >= 10) {
value = (Math.floor(value / 10) + value % 10);
}
} else {
value = +sixteenDigitString[i];
}
numSum += value;
}
return (numSum % 10 == 0);
}
alert(luhnCheck("4111111111111111"));
What this does is go through all the numbers, keeping the even indices as they are, but doubling the odd ones. If the doubling is more than nine, the values of the two digits are added together, as per the algorithm stated in wikipedia.
FIDDLE
Note: the number I tested with isn't my credit card number, but it's a well known number you can use that's known to pass a properly coded Luhn verification.

My below solution will work on AmEx also. I submitted it for a code test a while ago. Hope it helps :)
function validateCard(num){
var oddSum = 0;
var evenSum = 0;
var numToString = num.toString().split("");
for(var i = 0; i < numToString.length; i++){
if(i % 2 === 0){
if(numToString[i] * 2 >= 10){
evenSum += ((numToString[i] * 2) - 9 );
} else {
evenSum += numToString[i] * 2;
}
} else {
oddSum += parseInt(numToString[i]);
}
}
return (oddSum + evenSum) % 10 === 0;
}
console.log(validateCard(41111111111111111));
Enjoy - Mitch from https://spangle.com.au

#Spangle, when you're using even and odd here, you're already considering that index 0 is even? So you're doubling the digits at index 0, 2 and so on and not the second position, fourth and so on.. Is that intentional? It's returning inconsistent validations for some cards here compared with another algorithm I'm using. Try for example AmEx's 378282246310005.

get the number of n digit in a 2+ digit number

For example, getting "5" in "256". The closest I've gotten is Math.floor(256/10)), but that'll still return the numbers in front. Is there any simple way to get what I want or would I have to make a big function for it? Also, for clarity: "n digit" would be defined. Example, getDigit(2,256) would return 5 (second digit)

Math.floor((256 / 10) % 10)
or more generally:
Math.floor(N / (Math.pow(10, n)) % 10)
where N is the number to be extracted, and n is the position of the digit. Note that this counts from 0 starting from the right (i.e., the least significant digit = 0), and doesn't account for invalid values of n.

how about
(12345 + "")[3]
or
(12345 + "").charAt(3)
to count from the other end
[length of string - digit you want] so if you want the 2 it's:
5 - 4 = 1
(12345 + "")[1] = "2"
function getNumber (var num, var pos){
var sNum = num + "";
if(pos > sNum.length || pos <= 0){return "";}
return sNum[sNum.length - pos];
}

First, you need to cast the number to a string, then you can access the character as normal:
var num = 256;
var char = num.toString()[1]; // get the 2nd (0-based index) character from the stringified version of num
Edit: Note also that, if you want to access it without setting the number as a variable first, you need a double dot .. to access the function:
var char = 256..toString()[1];
The first dot tells the interpreter "this is a number"; the second accesses the function.

Convert to string and substring(2,2)?

This should do it:
function getDigit ( position, number ) {
number = number + ""; // convert number to string
return number.substr ( position + 1, 1 ); // I'm adding 1 to position, since 0 is the position of the first character and so on
}

Try this, last line is key:
var number = 12345;
var n = 2;
var nDigit = parseInt((number + '').substr(1,1));

If you want to try to do everything mathematically:
var number = 256;
var digitNum = 2;
var digit = ((int)(number/(Math.pow(10,digitNum-1))%10;
This code counts the digit from the right starting with 1, not 0. If you wish to change it to start at 0, delete the -1 portion in the call.
If you wish to count from the left, it gets more complicated and similar to other solutions:
var number = 256;
var digitNum = 2;
var digit = ((int)(number/(Math.pow(10,number.tostring().length-digitNum))%10;
edit:
Also, this assumes you want base 10 for your number system, but both of those will work with other bases. All you need to do is change instances of 10 in the final line of code to the number representing the base for the number system you'd like to use. (ie. hexadecimal =16, binary = 2)

// You do not say if you allow decimal fractions or negative numbers-
// the strings of those need adjusting.
Number.prototype.nthDigit= function(n){
var s= String(this).replace(/\D+/g,'');
if(s.length<=n) return null;
return Number(s.charAt(n))
}

use variable "count" to control loop
var count = 1; //starting 1
for(i=0; i<100; i++){
console.log(count);
if(i%10 == 0) count++;
}
output will fill
1
2
3
4
5
6
7
8
9