javascript - 嵌套数组并访问javascript中的所有其他索引
问题描述
// 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();
我需要关于 codecademy 的一个小项目的帮助来创建信用卡验证器。这个想法是从右到左检查数组中的数字。之后,如果信用卡号上的所有其他索引在加倍减 9 后大于 9,我需要将其乘以 2。卡中的所有元素都应该等于 100。如果它们是,那么卡号是有效的。我试图用嵌套的 for 循环来做到这一点,但我正在努力检查每个其他索引的数字。
解决方案
代码应该是不言自明的。由于您是初学者,您可以在此处找到标准数组方法:
// 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) )
})
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