如何基于 2 个属性删除数组中的重复对象? [英] How to remove duplicates objects in array based on 2 properties?

问题描述

我有一个房间对象数组，我目前正在根据它们的 room_rate_type_id 属性从数组中删除重复的对象:

I have an array of room objects and I am currently removing duplicates objects from the array based on their room_rate_type_id property:

const rooms = [{
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

const newRooms = rooms.filter((room, index, array) => {
  const roomRateTypeIds = rooms.map(room => room.room_rate_type_id);
  // Returns the first index found.
  return roomRateTypeIds.indexOf(room.room_rate_type_id) === index;
});

console.log(newRooms);

但是，我还需要确保仅当对象的 room_rate_type_id 和价格都匹配时才会移除对象.

However I also need to make sure that objects only get removed if not only their room_rate_type_id matches but also their price.

我可以理解过滤器功能在我给定的示例中是如何工作的，但我不确定如何干净地检查价格，最好是在 ES6 中.

I can understand how the filter functionality works in my given example but I am unsure how to cleanly do a check for the price as well, preferably in ES6.

推荐答案

你可以做的

const rooms = [
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

let result = rooms.filter((e, i) => {
    return rooms.findIndex((x) => {
    return x.room_rate_type_id == e.room_rate_type_id && x.price == e.price;}) == i;

});

console.log(result);

这将过滤除第一次出现的任何对象之外的所有重复项

This would filter all duplicates except the first occurrence of any object

这篇关于如何基于 2 个属性删除数组中的重复对象?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！