JavaScript: How to filter array by values nested in object 2 level deep

Question

I have an array of objects.

Within each object of the array, there can be multiple "Book" objects, all with dynamic keys. I want the objects with at least one "Book" object that is new.

For example:

const arr = [
  {
    id: '123',
    book1242: {isNew: true},
    book9023: {isNew: false},
  },
  {
    id: '123',
    book0374: {isNew: false},
    book9423: {isNew: false},
  },
  {
    id: '123',
    book8423: {isNew: false},
    book9023: {isNew: false},
  },
  {
    id: '123',
    book6534: {isNew: true},
    book9313: {isNew: false},
  },
]

So my filtered array will consist of the first and last element of the original array

Expected filtered array

const arr = [
  {
    id: '123',
    book1242: {isNew: true},
    book9023: {isNew: false},
  },
  {
    id: '123',
    book6534: {isNew: true},
    book9313: {isNew: false},
  },
]

I have tried using filter and map, but I get to the point where I have to loop through and check which book is new and I'm not sure how to return that object within the filter.

david · Accepted Answer · 2020-08-03 23:01:34Z

4

It sounds like you want something like this:

arr.filter((o) => Object.values(o).some((b) => b.isNew))

Will your array only ever have keys that are id and bookwxyz? If not you may need to do some checking on the values to make sure they aren't undefined

You could also explicitly check the key using Object.entries and a regular expression:

arr.filter((o) => Object.entries(o).some(([key, value]) => /book\d{4}/.test(key) && value.isNew))

edited Aug 3, 2020 at 23:01

answered Aug 3, 2020 at 22:44

david

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3 Comments

david Over a year ago

@RokoC.Buljan Good point, I was trying to find a way to deal with the fact that there might be other things in the array that could contain isNew but that's so unlikely I've changed the answer.

Roko C. Buljan Over a year ago

Beautifully simple. Terrific job!

Roko C. Buljan Over a year ago

PS, to provide more longevity, instead or relying to 4 - I would do /book\d+$/ so that in N years from now I don't need to care about bugs arising just because a library got more books.

Unmitigated · Accepted Answer · 2020-08-03 22:44:59Z

2

You can use Array#some with Object.values.

const arr = [ { id: '123', book1242: {isNew: true}, book9023: {isNew: false}, }, { id: '123', book0374: {isNew: false}, book9423: {isNew: false}, }, { id: '123', book8423: {isNew: false}, book9023: {isNew: false}, }, { id: '123', book6534: {isNew: true}, book9313: {isNew: false}, }, ];
const res = arr.filter(obj=>Object.values(obj).some(({isNew})=>isNew));
console.log(res);

answered Aug 3, 2020 at 22:44

Unmitigated

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3 Comments

w35l3y Over a year ago

It won't work whenever there is any non-object values, like primitive boolean or float.

Unmitigated Over a year ago

@w35l3y Have you tried it? It does work fine. In fact, the id is a primitive in the example.

w35l3y Over a year ago

You are right. It won't work with null or undefined only.

symlink · Accepted Answer · 2020-08-03 22:51:06Z

1

const arr=[{id:"123",book1242:{isNew:!0},book9023:{isNew:!1}},{id:"123",book0374:{isNew:!1},book9423:{isNew:!1}},{id:"123",book8423:{isNew:!1},book9023:{isNew:!1}},{id:"123",book6534:{isNew:!0},book9313:{isNew:!1}}];

const res = arr.filter(obj => {
  for(const [key, val] of Object.entries(obj)){
    if(key.substring(0,4) === 'book' && val.isNew === true){
      return obj
    }
   }
})

console.log(res)

answered Aug 3, 2020 at 22:51

symlink

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1 Comment

david Over a year ago

This is certainly the most explicit and safe one, actually checking the key has book at the front. You could replace the for loop with some, and filter only requires you to return a boolean so you could just return key.substring(0,4) === 'book' && val.isNew directly

Fritzdultimate · Accepted Answer · 2020-08-04 10:27:26Z

0

 let sorted =[]
  arr.forEach((el, i) => {
         let obj_values = Object.values(el);
         obj_values = obj_values.filter((elem) => typeof elem == 'object' && elem.isNew == true);
if(obj_values.length) sorted.push(arr[i])
     })

edited Aug 4, 2020 at 10:27

answered Aug 4, 2020 at 10:02

Fritzdultimate

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