[javascript] How to determine whether an object has a given property in JavaScript

How can I determine whether an object x has a defined property y, regardless of the value of x.y?

I'm currently using

if (typeof(x.y) !== 'undefined')

but that seems a bit clunky. Is there a better way?

const data = [{"b":1,"c":100},{"a":1,"b":1,"c":150},{"a":1,"b":2,"c":100},{"a":2,"b":1,"c":13}]

const result = data.reduce((r, e)  => {
  r['a'] += (e['a'] ? e['a'] : 0)
    r['d'] += (e['b'] ? e['b'] : 0)
  r['c'] += (e['c'] ? e['c'] : 0)

  return r
}, {'a':0, 'd':0, 'c':0})

console.log(result)
`result` { a: 4, d: 5, c: 363 }

Since question was regarding clunkiness of property checking, and one regular usecase for that being validation of function argument options objects, thought I'd mention a library-free short way of testing existence of multiple properties. Disclaimer: It does require ECMAScript 5 (but IMO anyone still using IE8 deserves a broken web).

function f(opts) {
  if(!["req1","req2"].every(opts.hasOwnProperty, opts)) {
      throw new Error("IllegalArgumentException");
  }
  alert("ok");
}
f({req1: 123});  // error
f({req1: 123, req2: 456});  // ok

Underscore.js or Lodash

if (_.has(x, "y")) ...

:)

If you want to know if the object physically contains the property @gnarf's answer using hasOwnProperty will do the work.

If you're want to know if the property exists anywhere, either on the object itself or up in the prototype chain, you can use the in operator.

if ('prop' in obj) {
  // ...
}

Eg.:

var obj = {};

'toString' in obj == true; // inherited from Object.prototype
obj.hasOwnProperty('toString') == false; // doesn't contains it physically

includes

Object.keys(x).includes('y');

The Array.prototype.includes() method determines whether an array includes a certain value among its entries, returning true or false as appropriate.

and

Object.keys() returns an array of strings that represent all the enumerable properties of the given object.

.hasOwnProperty() and the ES6+ .? -optional-chaining like: if (x?.y) are very good 2020+ options as well.

ES6+:

There is a new feature on ES6+ that you can check it like below:

if (x?.y)

Actually, the interpretor checks the existence of x and then call the y and because of putting inside if parentheses the coercion happens and x?.y converted to boolean.

Why not simply:

if (typeof myObject.myProperty == "undefined") alert("myProperty is not defined!");

Or if you expect a specific type:

if (typeof myObject.myProperty != "string") alert("myProperty has wrong type or does not exist!");

One feature of my original code

if ( typeof(x.y) != 'undefined' ) ...

that might be useful in some situations is that it is safe to use whether x exists or not. With either of the methods in gnarf's answer, one should first test for x if there is any doubt if it exists.

So perhaps all three methods have a place in one's bag of tricks.

You can trim that up a bit like this:

if ( x.y !== undefined ) ...