CODEWITHSUNDEEP
javascript
Mads
Mads

Reputation: 127

Filter array object by array of Boolean values dynamically

I have my array object in the following format

menuItem = [{
    "name": "MOZZARELLA DI BUFALA & TOMATOES",
    "gluten": true,
    "eggs": false,
    "dairy": true,
     ...
},...

I want to filter hundreds of menuItems by the allergen values (Boolean).

While the following code does the trick:

menuItem.filter(el => !el.gluten && !el.fish && !el.soy && !el.dairy)"

The allergens are hardcoded and repetitive. I am struggling to make the code dynamic and elegant.

I tried the following solution

menuItem.filter(el => !el[allergens])"
var allergens = ['gluten', 'fish', 'soy', 'dairy']

However, it only works correctly with one allergen value. Multiple values as in the above example do not filter anything.

Upvotes: 0

Views: 586

Answers (2)

hgb123
hgb123

Reputation: 14891

You could use .some or .every, whichever match your case

const allergens = ['gluten', 'fish', 'soy', 'dairy']

const res = menuItem.filter(
  (el) => !allergens.every((allergen) => el[allergen])
);
// equivalent to `menuItem.filter(el => !el.gluten && !el.fish && !el.soy && !el.dairy)`


const menuItem = [
  {
    name: 'dish A',
    gluten: true,
    fish: true,
    soy: true,
    dairy: true,
  },
  {
    name: 'dish B',
    gluten: true,
    fish: true,
    soy: false,
    dairy: true,
  },
];

const allergens = ['gluten', 'fish', 'soy', 'dairy'];

const res = menuItem.filter(
  (el) => !allergens.every((allergen) => el[allergen])
);

console.log(res);

Upvotes: 2

CertainPerformance
CertainPerformance

Reputation: 371049

First extract the name property, then check that the rest of the properties in the object are falsey.

menuItem.filter(
  ({ name, ...rest }) => Object.values(rest).every(val => !val)
);

Ideally, I'd prefer to restructure the input array so the allergies are in a separate property, perhaps:

menuItem = [{
    "name": "MOZZARELLA DI BUFALA & TOMATOES",
    allergens: {
        "gluten": true,
        "eggs": false,
        "dairy": true,

to make accessing them easier.

menuItem.filter(
  ({ allergens }) => Object.values(allergens).every(val => !val)
);

Upvotes: 2

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