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@p3kfinger
Last active June 6, 2021 17:43
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Credit Card Checker
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
// Add your functions below:
const validateCred = (array) => {
let s = 0;
let doubleDigit = false;
//loop from the end of the array
for(let i = array.length - 1; i >= 0; i--) {
let digit = +array[i];
if (doubleDigit) {
digit *= 2;
if (digit > 9)
digit -= 9;
}
s += digit;
doubleDigit = !doubleDigit;
}
return s % 10 == 0;
}
console.log('---------------------------------');
console.log(validateCred(valid1))
console.log(validateCred(valid2))
console.log(validateCred(valid3))
console.log(validateCred(valid4))
console.log(validateCred(valid5))
console.log('---------------------------------');
console.log(validateCred(invalid1))
console.log(validateCred(invalid2))
console.log(validateCred(invalid3))
console.log(validateCred(invalid4))
console.log(validateCred(invalid5))
console.log('---------------------------------');
console.log(validateCred(mystery1))
console.log(validateCred(mystery2))
console.log(validateCred(mystery3))
console.log(validateCred(mystery4))
console.log(validateCred(mystery5))
console.log('----------------------------------');
console.log('---------INVALID CARDS------------');
const findInvalidCards = (creditCards) => {
let newArray = creditCards.filter(card => validateCred(card) === false);
return newArray;
}
console.log(findInvalidCards(batch));
console.log('---------------------------------');
console.log('----------VALID CARDS------------');
const findValidCards = (creditCards) => {
let newArray2 = creditCards.filter(card => validateCred(card) === true);
return newArray2;
}
console.log(findValidCards(batch));
console.log('----------------------------------');
console.log('--------Invalid Cards Type--------');
const idInvalidCardCompanies = (array) => {
let companies = [];
for(let i = 0; i < array.length; i++) {
switch (array[i][0]) {
case 3:
if (companies.indexOf('Amex') === -1) {
companies.push('Amex');
}
break
case 4:
if (companies.indexOf('Visa') === -1) {
companies.push('Visa');
}
break
case 5:
if (companies.indexOf('Mastercard') === -1) {
companies.push('Mastercard');
}
break
case 6:
if (companies.indexOf('Discover') === -1) {
companies.push('Discover');
}
break
default:
console.log('Company not found');
}
}
return companies;
}
console.log(idInvalidCardCompanies([invalid1]));
console.log(idInvalidCardCompanies([invalid2]));
console.log(idInvalidCardCompanies([invalid3]));
console.log(idInvalidCardCompanies([invalid4]));
console.log(idInvalidCardCompanies([invalid5]));
console.log('----------------------------------');
console.log(idInvalidCardCompanies(batch));
console.log('----------------------------------');
// Other cards been tested will return something out of the switch case, because we are checking index of very 1st index in the card number... we need to test a bit more advanced way.
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