How to filter array of objects and only leave certain fields?

Question

For example, my array filter on user status works correctly, but I only want the users' name and id with no other fields in the response.

let users = [{
    id: 1,
    name: 'Zen',
    age: 21,
    status: true
}, {
    id: 2,
    name: 'Roy',
    age: 29,
    status: false
}, {
    id: 3,
    name: 'Zohn',
    age: 41,
    status: true
}]

let result = users.filter(({status, age}) => status === true);
console.log('result', result);

Log of result:

[{
    id: 1,
    name: 'Zen',
    age: 21,
    status: true
}, {
    id: 3,
    name: 'Zohn',
    age: 41,
    status: true
}]

But what I want is:

[{
    id: 1,
    name: 'Zen',
}, {
    id: 3,
    name: 'Zohn',
}]

SrThompson · Accepted Answer · 2018-10-19 13:01:11Z

4

The operation you're describing is a map (transformation of inputs), not a filter

users.filter(({status}) => status===true)
  .map(({id, name}) => ({id, name}));

answered Oct 19, 2018 at 13:01

SrThompson

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Comments

Christian Kellner · Accepted Answer · 2018-10-19 13:00:58Z

1

This should work:

array.filer(val => val.status)
  .map(val => {
      return {
       id: val.id,
       name: val.name
      }
   })

answered Oct 19, 2018 at 13:00

Christian Kellner

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Comments

tokland · Accepted Answer · 2018-10-19 13:03:31Z

1

Just add an extra map:

const users2 = users
  .filter(user => user.status)
  .map(user => ({id: user.id, name: user.name}));

edited Oct 19, 2018 at 13:03

answered Oct 19, 2018 at 13:00

tokland

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1 Comment

DontVoteMeDown Over a year ago

Check syntax: }); => }));

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