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How to check if an array of objects contains every other element of an array of objects based on a property

February 17, 2022

given those arrays

arr1 = [
{name: "name1", id: 1},
{name: "name2", id: 2},
{name: "name3", id: 3}
]

arr2 = [
{name: "name1", id: 1},
{name: "name2", id: 4},
{name: "name3", id: 3}
]

The case above should return false, because arr2[1].id (4) doesn’t exist on arr1

I pretty much, must check that every element of arr2 exists on arr1 , based on its id

I was thinking something like

arr1.every(i => i.id === arr2.map(z =>z.id))

But im not really sure

If arr2 was

arr2 = [
    {name: "name1", id: 1},
    {name: "name3", id: 3}
    ]

It should return true, because every el of arr2, exists in arr1

>Solution :

You’re relaly close, but map is for mapping each element of the array to some new value, not for seeing if an element exists. You’d use some for that, and you’d want to reverse the arrays you’re calling every and some on:

const flag = arr2.every(element => arr1.some(({id}) => id === element.id));

That says: "Does every element of arr2 have at least one matching element in arr1 by id?"

Live Example:

const arr1 = [
    {name: "name1", id: 1},
    {name: "name2", id: 2},
    {name: "name3", id: 3}
];

const arr2 = [
    {name: "name1", id: 1},
    {name: "name2", id: 4},
    {name: "name3", id: 3}
];

const arr3 = [
    {name: "name1", id: 1},
    {name: "name3", id: 3}
];

const result1 = arr2.every(element => arr1.some(({id}) => id === element.id));
console.log("arr2 and arr1: " + result1);

const result2 = arr3.every(element => arr1.some(({id}) => id === element.id));
console.log("arr3 and arr1: " + result2);