Accessing object properties inside array through object method

Question

This could be a simple solution but I can't figure out if I can use this method or not. This question is asked before in different ways but I am looking through object methods not functions so bear with me please.

Is this possible to access objects properties and return them as an array through Object.getOwnPropertyNames(object)?

For example , this method working fine with below variable without array.

var y = {
  firstName: "John",
  lastName: "Doe",
  eyeColor: "Black",
  age: 50
};
console.log(
  Object.getOwnPropertyNames(y)
);

Returns 
(4) ["firstName", "lastName", "eyeColor", "age"]

can we use same method to access object properties inside array ?

var x = [{
  firstName: "John",
  lastName: "Doe",
  eyeColor: "Black",
  age: 50
}, {
  firstName: "Marry",
  lastName: "Smith",
  eyeColor: "Brown",
  age: 45
}];

Will all the objects in the array have the same structure? also are all properties mandatory? this is quite important and some of the answers make assumptions as to the answer of this question without asking it first. — Damo
– Damo, Commented Jan 6, 2019 at 14:16

mplungjan · Accepted Answer · 2019-01-06 14:24:50Z

1

You can use spread operator ... for this:

var x = [{
  firstName: "John",
  lastName: "Doe",
  eyeColor: "Black",
  age: 50
}, {
  firstName: "Marry",
  lastName: "Smith",
  eyeColor: "Brown",
  age: 45
}];

console.log(
  Object.getOwnPropertyNames(...x)
);

edited Jan 6, 2019 at 14:24

mplungjan

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answered Jan 6, 2019 at 14:06

Behrouz Pooladrak

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1 Comment

ibrahim mahrir Over a year ago

Why would you do destructuring (which is not what being used here by the way, it's the spread syntax), where only x[0] will suffice?

digitil · Accepted Answer · 2019-01-06 14:07:50Z

0

It wouldn’t work, but depending on your desired output structure, you could use map to transform each array element

x.map(Object.getOwnPropertyNames)

would return an array of property name arrays.

answered Jan 6, 2019 at 14:07

digitil

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Comments

westdabestdb · Accepted Answer · 2019-01-06 14:07:53Z

0

var x = [{firstName: "John", lastName : "Doe", eyeColor : "Black", age : 50}, {firstName: "Marry", lastName : "Smith", eyeColor : "Brown", age : 45}];

x.map(y => console.log(Object.getOwnPropertyNames(y)))

You can access the objects in array with map().

answered Jan 6, 2019 at 14:07

westdabestdb

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Comments

ibrahim mahrir · Accepted Answer · 2019-01-06 14:22:20Z

0

var propNames = Object.getOwnPropertyNames(x[0]);

x[0] is the first object in the array and since all objects seem to have the same keys, calling the function on the first object will get you the result you want.

Note: This solution assumes there is at least one object in the array, so you might want to check that first before doing x[0] or else you'll get an error.

Example:

var x = [{firstName: "John", lastName : "Doe", eyeColor : "Black", age : 50}, {firstName: "Marry", lastName : "Smith", eyeColor : "Brown", age : 45}];

var propNames = Object.getOwnPropertyNames(x[0]);

console.log(propNames);

edited Jan 6, 2019 at 14:22

answered Jan 6, 2019 at 14:11

ibrahim mahrir

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2 Comments

Damo Over a year ago

This assumes there will always be at least one element in the array

ibrahim mahrir Over a year ago

@Damo yes, but that's OP's concern. OP is asking how to get property names of objects inside an array, no need to complicate things. Note added anyway.

Damo · Accepted Answer · 2019-01-06 19:09:58Z

0

If the array of objects contains objects with different properties you can get a list of unique property names thus:

var x = [{
  firstName: "John",
  lastName: "Doe",
  eyeColor: "Black"
}, {
  firstName: "Marry",
  lastName: "Smith",
  age: 45
}];

var arrayOfArrays = x.filter(obj => obj).map(Object.getOwnPropertyNames)
var array = [].concat.apply([], arrayOfArrays);
var uniqueProperties = [...new Set(array)]

console.log(uniqueProperties);

This will not crash if the array is empty, or if the array contains null or undefined. But it does assume that x is an array, empty or otherwise.

answered Jan 6, 2019 at 19:09

Damo

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