Check if any object in an array contains a property listed in another array

Question

I've tried a few solutions to this and have had a look round on here but can't seem to get this working.

I have 2 arrays, and want to check if any of the items in the array contain any of the strings in the other array.

const stepsShown = ["ref2", "ref7"];
const items = [
{ name: "item1", element: "ref1"},
{ name: "item2", element: "ref2"}
];
const refsExist = items.some((r) =>{stepsShown.indexOf(r.element);});

In this example I would expect refsExist to be true

I think I am a little muddled in my syntax! Can anyone help please? :)

You want includes, not indexOf - and you need to return the result from the function — Bergi
– Bergi, Commented Aug 28, 2017 at 18:50

apsillers · Accepted Answer · 2017-08-28 18:48:21Z

3

You have it almost right, except that you need to test for a non -1 value, and your arrow function shouldn't have brackets.

If your arrow function consists of a single expression, you don't need brackets:

items.some(r=>stepsShown.indexOf(r.element)!=-1);

If your arrow function includes statements or multiple expressions, you need curly brackets. When you use curly brackets, you must include an explicit return statement.

items.some(r=>{
    for(some;thing;) { ... }
    return otherThing;
});

Instead of relying on the less-readable .indexOf(..)!=-1 test, you can simply use includes:

items.some(r=>stepsShown.includes(r.element));

answered Aug 28, 2017 at 18:48

apsillers

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marvel308 · Accepted Answer · 2017-08-28 18:49:46Z

0

You can find an element and later check if it is undefined or not

const stepsShown = ["ref2", "ref7"];
const items = [
{ name: "item1", element: "ref1"},
{ name: "item2", element: "ref2"}
];
const refsExist = items.find((r) =>{    
    return stepsShown.indexOf(r.element)>-1;
});

console.log(refsExist !== undefined);

answered Aug 28, 2017 at 18:49

marvel308

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