设置和比较日期以检查信用卡有效性

Posted 2023-02-22

【中文标题】设置和比较日期以检查信用卡有效性

【英文标题】：Set and compare dates to check credit card validity

【发布时间】：2018-12-23 22:21:17

【问题描述】：

我有客户注册了 12 个月的分期付款选项。我需要知道他们的信用卡是否至少在分期付款选项的期限内有效 12 个月。

使用下面收到的对象中的card.exp_year 和card.exp_month，我将如何计算这张信用卡在一年后是否有效。我假设我必须使用 Date() 函数。

  "id": "tok_xxxxxxxx",
  "object": "token",
  "card": 
    "id": "card_xxxxxxxx",
    "object": "card",
    "address_city": null,
    "address_country": "BE",
    "address_line1": null,
    "address_line1_check": null,
    "address_line2": null,
    "address_state": null,
    "address_zip": null,
    "address_zip_check": null,
    "brand": "Visa",
    "country": "CA",
    "cvc_check": "unchecked",
    "dynamic_last4": null,
    "exp_month": 7,
    "exp_year": 2019,
    "funding": "credit",
    "last4": "3086",
    "metadata": ,
    "name": "John Doe ",
    "tokenization_method": null
  ,
  "client_ip": "149.xxx.xxx.xxx",
  "created": 1530xxxxxx,
  "livemode": true,
  "type": "card",
  "used": false

【问题讨论】：

你试过什么？如果你应该在纸上这样做，你会在现实生活中怎么做？

How do I get the current date in javascript?的可能重复

取一个日期，即当前日期，年份递增一次，然后比较卡到期日的月份和年份

@Luca 好主意，感谢指导，可能为我节省了一些时间

【参考方案1】：

谢谢，卢卡！最终的工作解决方案！

var cardYear = new Date();
cardYear.setMonth(12-1); // jan = 0 so minus 1
cardYear.setYear(2019);

console.log(cardYear);

var nextYearFromNow = new Date(new Date().setFullYear(new Date().getFullYear() + 1));
console.log(nextYearFromNow);

console.log(compareTime(cardYear, nextYearFromNow));

function compareTime(time1, time2) 
    return new Date(time1) > new Date(time2); // true if time1 is later

或

var year = user.attributes.token.card.exp_year;
var month = user.attributes.token.card.exp_month;

var cardYear = new Date(year, month, 0); // returns date according to card and sets day to last day of month becuase credit cards are set to expire on the last day of a month
var nextYearFromNow = new Date(new Date().setFullYear(new Date().getFullYear() + 1));

var expiry = compareTime(cardYear, nextYearFromNow); // if false card is too short

function compareTime(time1, time2) 
    return new Date(time1) > new Date(time2); // true if time1 is later