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);
          }
        }
      });

Ankit Agarwal · Accepted Answer · 2019-09-26 08:59:45Z

3

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);

answered Sep 26, 2019 at 8:59

Ankit Agarwal

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1 Comment

Daniyal Lukmanov Over a year ago

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.

Thirupathi Govindharaj · Accepted Answer · 2019-11-16 07:40:16Z

0

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
  }
});

edited Nov 16, 2019 at 7:40

answered Sep 26, 2019 at 9:37

Thirupathi Govindharaj

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Comments

Shinwi · Accepted Answer · 2023-09-12 20:55:31Z

0

Filtering the array types can get you what you want.

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

answered Sep 12, 2023 at 20:55

Shinwi

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