Filter an array<objects> by two other array<strings>

Question

i have a short question:

I want to filter an array of objects by two arrays of strings.

My Array looks like this:

[
  {
    "id": 12345,
    "title": "Some title",
    "contains": [
      {
        "slug": "fish",
        "name": "Fish"
      }, {
        "slug": "soy", // search term, like a id
        "name": "Soy"
      } 
    ], "complexity": [
      {
        "slug": 4, // search term, like a id
        "name": "hard"
      }, {
}],..
      },

{...}

and that are my two arrays:

// recipes should not contain this ingredients
let excludedIngredientsArray = Array<string> = ["soy", "fish"]; 

// recipes should not contain this complexities
let excludedComplexityArray = Array<string> = [1, 2];

Now i want to filter the recipes by these two arrays and want to remove all recipes which contain the excluded terms

Whats the best way to do this?

Thanks a lot!

edit:

recipeArray looks like this:

interface recipeArray {
    reciepe: Array<{
        name: string,
        contains: Array<{slug: string, name: string}> //Ingredients array
        complexity: Array<{slug: string, name: string}> //complexity array
    }>
}

What is the type definition for the first array?

Nitzan Tomer
– Nitzan Tomer

2017-01-26 14:41:38 +00:00
Commented Jan 26, 2017 at 14:41 — Nitzan Tomer
– Nitzan Tomer, Commented Jan 26, 2017 at 14:41
At the moments its <any>, but i will add this later.

Fargho
– Fargho

2017-01-26 14:43:30 +00:00
Commented Jan 26, 2017 at 14:43 — Fargho
– Fargho, Commented Jan 26, 2017 at 14:43

Nitzan Tomer · Accepted Answer · 2017-01-26 14:50:15Z

2

If your first array is like this:

interface Item {
    id: number;
    title: string;
    contains: { slug: string; name: string }[],
    complexity: { slug: number; name: string }
}

let data: Item[];

Then you can have your desired filtered array like so:

let excluded = data.filter(item => {
    return item.contains.every(obj => excludedIngredientsArray.indexOf(obj.slug) < 0)
        && excludedComplexityArray.indexOf(item.complexity.slug) < 0;
});

answered Jan 26, 2017 at 14:50

Nitzan Tomer

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1 Comment

Fargho Over a year ago

Oh nice, that was easy! Thanks!

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