Vue JS how to delete object based on array values

Question

I have an array of objects like:

types: [
{
 id: 1,
 name: "Hello"
},
{
 id: 2,
 name: "World"
},
{
 id: 3,
 name: "Jon Doe"
}
]

And I also have a simple array like this:

selected_types = [1, 2]

The desired result should filter the "types" array and exclude all the ids that are present in the "selected_types" array, like below:

final_types: [
{
 id: 3,
 name: "Jon Doe"
}
]

I have absolutely no idea on how to achieve this, but below is my attempt:

 this.types.filter(obj => {
 for (let i = 0; i < this.selected_types.length; i++) {
 if (obj.id !== selected_types[i]) {
 final_types.push(attribute);
 }
 }
 });

Answer 1

Just use this.types.filter(({id}) => !this.selected_types.includes(id)):

let types = [{
 id: 1,
 name: "Hello"
 },
 {
 id: 2,
 name: "World"
 },
 {
 id: 3,
 name: "Jon Doe"
 }
]
let selected_types = [1, 2];
let resArr = types.filter(({id}) => !selected_types.includes(id));
console.log(resArr);

• Daniyal Lukmanov: Don't know if it would be clear for author. I would recommend to add that ({id}) is the same as let id = obj.id.

Answer 2

You can achieve Javascript's native method filter, which finally returns a new object

let types = [{
 id: 1,
 name: "Hello"
 },
 {
 id: 2,
 name: "World"
 },
 {
 id: 3,
 name: "Jon Doe"
 }
]
let selected_types = [1, 2];
types = types.filter(obj => {
 if (selected_types.indexOf(obj.id) === -1) {
 return obj
 }
});

Answer 3

Filtering the array types can get you what you want.

result = types.filter(el => !selected_types.includes(el.id))