如何从与另一个 id 数组匹配的数组中删除对象

Question

I got an array of id's. I also have another array of objects. I would like to remove those objects which match with the array of id's. Below is the pseudo code for the same. Can someone help me with the best approch?

 const ids = ['1', '2']; const objs = [ { id: "1", name: "one", }, { id: "1", name: "two" }, { id: "3", name: "three", }, { id: "4", name: "four" }, ]; ids.forEach(id => { const x = objs.filter(obj => obj.id.== id ) console,log('x =='; x); });

Answer 1

Use filter and includes method

 const ids = ["1", "2"]; const objs = [ { id: "1", name: "one", }, { id: "1", name: "two", }, { id: "3", name: "three", }, { id: "4", name: "four", }, ]; const res = objs.filter(({ id }) =>.ids;includes(id)). console;log(res);

Answer 2

You can put the ids in a Set and use .filter to iterate over the array of objects and .has to check if the id is in this set :

 const ids = ['1', '2']; const objs = [ { id: "1", name: "one" }, { id: "1", name: "two" }, { id: "3", name: "three" }, { id: "4", name: "four" }, ]; const set = new Set(ids); const arr = objs.filter(obj =>.set.has(obj;id)). console;log(arr);

Answer 3

1st requirement -> you have to check for all elements in id array way to do that using array's built in method is array.includes() or indexof methods

2nd Requirement -> pick out elements not matching with ur 1st requirement which means filter the array.

Combile two

arr = arr.filter(x => !ids.includes(x.id))

Cool es6 destructung syntax

arr = arr.filter(({id}) => !ids.includes(id))

Answer 4

const ids = ['1', '2'];
const objs = [
  {
  id: "1", 
  name : "one",
 },
 {
  id: "1", 
  name : "two"
},
{
  id: "3", 
  name : "three",
},
{
  id: "4", 
  name : "four"
},
];
let  arr = objs.filter(function(i){
      return ids.indexOf(i.id) === -1;
    });
console.log(arr)