How to filter an Array with another Array

Question

I have an Array of Objects:

const array = [{id: 1, bar: "test" }, {id: 2, bar: "test2" }, {id: 3, bar: "test3" }]

I have a second array containing the ID's that I want to filter out of the first Array:

const ids = [1, 2]

How do I create a new Array of Objects without the ID's found in ids.

Answer 1

This is a fairly simple filter operation

const array = [{id: 1, bar: "test" }, {id: 2, bar: "test2" }, {id: 3, bar: "test3" }];

const ids = [1, 2];

var result = array.filter( x => !ids.includes(x.id));
console.log(result);

Answer 2

If you need to mutate the original array you can do like this:

const array = [{id: 1, bar: "test" }, {id: 2, bar: "test2" }, {id: 3, bar: "test3" }];

const ids = [1, 2];

ids.forEach(idToDelete => {
    const index = array.findIndex(({ id }) => id === idToDelete);
    array.splice(index, 1);
});

console.log(array);

If you need a new array you can do like this:

const array = [{id: 1, bar: "test" }, {id: 2, bar: "test2" }, {id: 3, bar: "test3" }];

const ids = [1, 2];

const result = array.filter(({ id }) => !ids.includes(id));

console.log(result);

You could also reassign a new array to the array variable:

let array = [{id: 1, bar: "test" }, {id: 2, bar: "test2" }, {id: 3, bar: "test3" }];

const ids = [1, 2];

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

console.log(array);