Question

150

array.indexOf in TypeScript/JavaScript

rated 0 times [ 156] [ 6] / answers: 1 / hits: 61003 / 8 Years ago, thu, february 4, 2016, 12:00:00

UPDATE: Although this question is marked as duplicated with this. But @ssube's way is neat and much smarter.

UPDATE2: Seems there is new way to do it in the comment by @Grungondola.

I am using Typescript.

This works well.

var array1 = [];

array1.push(5);

array1.push(6);

console.log(a, array2.indexOf(6));

But this does not work well. Because array2.indexOf returns -1 which means it does not find it.

var array2 = [];

array2.push({aa:5,bb:5});

array2.push({aa:6,bb:6});

console.log(array2.indexOf({aa:6,bb:6}));

Looks like indexOf does not support Object. Does TypeScript have its own ways to deal with this kind of problem? Thanks.

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haleeg

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2 Years ago

answered 8 Years ago ravenl · Accepted Answer

No. The problem is not with Object, but that you are creating two different objects.

The object literal syntax ({foo: 'bar'}) declares an object inline. When the script is executed, the object is created. Using that syntax multiple times creates multiple objects.

You can easily test that with {foo: 3} === {foo: 3}. This will evaluate to false, but they are not the same object (reference).

The indexOf method checks if the object, string, number, etc, is present in the array. You're passing a new object, which is not in the array.

If you have a reference to the object, you can use that and indexOf will work:

var foo = {aa:5,bb:5}, bar = {aa:6,bb:6};

var array2 = [];

array2.push(foo);

array2.push(bar);

console.log(array2.indexOf(foo));

Because you're referring to the same instance, this will print the index.

You can also use filter or find with a predicate to perform a deep search:

function deepIndexOf(arr, obj) {

  return arr.findIndex(function (cur) {

    return Object.keys(obj).every(function (key) {

      return obj[key] === cur[key];

    });

  });

}



var array2 = [];

array2.push(foo);

array2.push(bar);

console.log(deepIndexOf(array2, foo));

This won't recurse into nested objects, but will accomplish the comparison you're looking for (equivalence on two objects and their immediate fields).