DRY:如何将这个重复的日期代码简化为功能

Question

• 是否可以将其填充到函数或数组中？
• 如何简化代码？

var yesterday = new Date();
    yesterday.setDate(yesterday.getDate() - 1);
    yesterday.setHours(03, 0, 0, 0);
    var today = new Date();
    today.setHours(03, 0, 0, 0);
    var tomorrow1 = new Date();
    tomorrow1.setDate(tomorrow1.getDate() + 1);
    tomorrow1.setHours(03, 0, 0, 0);
    var tomorrow2 = new Date();
    tomorrow2.setDate(tomorrow2.getDate() + 2);
    tomorrow2.setHours(03, 0, 0, 0);
    var tomorrow3 = new Date();
    tomorrow3.setDate(tomorrow3.getDate() + 3);
    tomorrow3.setHours(03, 0, 0, 0);

Answer 1

您可以将将小时和日期添加到函数的重复代码移动到函数中，然后将该函数用作map中的参数

const getNthDay = (n) => {
  const day = new Date();
  day.setDate(day.getDate() + n);
  day.setHours(03, 0, 0, 0);
  return day;
}
const arrayOfDates = [-1,0,1,2,3].map(getNthDay);
console.log(arrayOfDates)

Answer 2

您可以使用

map()方法：

const [yesterday, today, tommorow1, tommorow2, tommorow3] = [-1, 0, 1, 2, 3].map(x => {
  const date = new Date();
  date.setDate(date.getDate() + x);
  date.setHours(03, 0, 0, 0);
  return date;
});
console.log(yesterday, today, tommorow1, tommorow2, tommorow3);

Answer 3

映射和解构。

var [yesterday, today, tomorrow1, tomorrow2, tomorrow3] = [-1, 0, 1, 2, 3].map(offset => {
   var d = new Date();
   d.setDate(d.getDate() + offset);
   d.setHours(03, 0, 0, 0);
   return d;
});
console.log(yesterday);
console.log(today);
console.log(tomorrow1);
console.log(tomorrow2);
console.log(tomorrow3);