嵌套数组并访问javascript中的所有其他索引

Question

// 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:
let i = batch.length - 1
const validateCred = () => {
    for(i; i >= 0; i--) {
        for (var j = batch[i].length - 1; j >= 0; j--) {
           console.log(batch[i][j]);
        }
    }
}
validateCred();

我需要在codecademy上的一个小项目上获得帮助，以创建信用卡验证程序。 The idea is to check the digits from right to left in the arrays。之后，如果将信用卡号上的所有其他索引乘以9后再大于9，则需要将其乘以2。如果卡中的所有元素均等于100，则索引号就是有效的。我试图使用嵌套的for循环来执行此操作，但是我正在努力检查每个其他索引的数量。

Answer 1

该代码应不言自明。由于您是初学者，因此可以找到标准的数组方法here：

// All valid credit card numbers
const valids =[ 
 [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8],
 [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9],
 [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6],
 [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5],
 [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6]
]
// All invalid credit card numbers
const invalids = [ 
  [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5],
  [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3],
  [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4],
  [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5],
  [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4]
]

// Can be either valid or invalid
const mysterious = [
  [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4],
  [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9],
  [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3],
  [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3],
  [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3]
]

function validate_credit_card( code ) {
  result = code.reverse();
  result = result.map( double_odd_positions );
  result = result.map( substract_nine_from_numbers_above_nine );
  result = result.reduce( add, 0);
  result = result % 10 === 0;
  return result;
}

function double_odd_positions( number, index ) {
  if( index % 2 === 0 ) return number;
  else return number * 2;
}

function substract_nine_from_numbers_above_nine (number) {
  if( number > 9 ) return number - 9;
  else return number;
}
function add(number, accumulator) {
  return accumulator + number;
}

console.log('valids:');
valids.forEach( code => {
  console.log( validate_credit_card(code) )
})

console.log('invalids:');
invalids.forEach( code => {
  console.log( validate_credit_card(code) )
})
console.log('mysterious:');
mysterious.forEach( code => {
  console.log( validate_credit_card(code) )
})