需要帮助切片日期值

Question

result_database = [
  { id: 1, name: "Tom Riddle", date: "2022-05-16T22:00:00.000Z" },
  { id: 2, name: "Hank Some", date: "2022-05-19T22:00:00.000Z" },
  { id: 3, name: "Family Man", date: "2022-05-17T22:00:00.000Z" },
];
var holiday_date = new Date().toJSON().slice(0, 10);
let holiday_date_remove = result_database.filter(
  (item1) =>
    !result_database.find(
      (item2) =>
        item1.name == item2.name && item2.date.slice(0, 10) == holiday_date
    )
);
const holiday_result = [
  ...new Map(
    holiday_date_remove.map((item) => [JSON.stringify(item.name), item])
  ).values(),
];
console.log(holiday_result);

我正在尝试.slice(0,10)日期。我把它显示成这样 "2022-05-16T22:00:00.000Z"，但我需要它作为 "2022-05-16"，所以我尝试 date.slice(0, 10) 但它不起作用。

Answer 1

const dataDate = '2022-05-17T22:00:00.000Z' 

const date = new Date(dataDate)

date.getDate()  + "-" + (date.getMonth()+1) + "-" + date.getFullYear()

console.log(date)

有关日期格式的更多信息，请阅读here

Answer 2

您可以先使用 Array.prototype.map and update the date property and then filter the array based on holiday_date using Array.prototype.filter 在 result_database 上映射。

const 
  result_database = [
    { id: 1, name: "Tom Riddle", date: "2022-05-16T22:00:00.000Z" },
    { id: 2, name: "Hank Some", date: "2022-05-19T22:00:00.000Z" },
    { id: 3, name: "Family Man", date: "2022-05-17T22:00:00.000Z" },
  ],
  holiday_date = new Date().toJSON().slice(0, 10),
  holiday_date_remove = result_database
    .map((item) => ({ ...item, date: item.date.slice(0, 10) }))
    .filter(({ date }) => date !== holiday_date);

console.log(holiday_date_remove);

Answer 3

当它是 ISOString 格式时，你应该用“T”分割。

const yourTypeDate = new Date().toISOString().split('T')[0];

下面的这个比较是有效的，它将检查数据库日期是否等于今天的日期，如果是，将删除索引，过滤器正在工作。

result_database = [
  { id: 1, name: "Tom Riddle", date: "2022-05-16T22:00:00.000Z" },
  { id: 2, name: "Hank Some", date: "2022-05-19T22:00:00.000Z" },
  { id: 3, name: "Family Man", date: "2022-05-17T22:00:00.000Z" },
];
var holiday_date = new Date().toISOString().split('T')[0];
let holiday_date_remove = result_database.filter(
  (item1) =>
    !result_database.find(
      (item2) => {
        item2.date = item2.date.split('T')[0];
        return ((item1.name == item2.name) && (item2.date.split('T')[0] == holiday_date));
        }
    )
);
const holiday_result = [
  ...new Map(
    holiday_date_remove.map((item) => [JSON.stringify(item.name), item])
  ).values(),
];
console.log(holiday_result);

需要帮助切片日期值

Need help slicing date value

javascript

node.js