How to creates an object with only the listed keys?

Question

Creates an object composed of the picked source properties.

Parameters

• source - Any JavaScript Object
• keys - An array of JavaScript Strings

Return Value

A new Object containing all of the properties of source listed in keys. If a key is listed in keys, but is not defined in source, then that property is not added to the new Object.

Examples

pick({ foo: 1, bar: 2, baz: 3 }, ['foo', 'baz']) // -> { foo: 1, baz: 3 }
pick({ qux: 4, corge: 5 }, ['bar', 'grault'])    // -> {}
pick({ bar: 2 }, ['foo', 'bar', 'baz'])          // -> { bar: 2 }

I have

function pick(source, keys) {
  let result = {};
  for (key in source) {
    if (key === keys) {
      result[key];
    }
  }
  return result;
}

so far

Answer 1

You're not assigning anything to result[key], that should be result[key] = source[key].

You're not testing whether key is in keys correctly. === does exact comparison, you want to use keys.includes(key) to test inclusion.

function pick(source, keys) {
  let result = {};
  for (key in source) {
    if (keys.includes(key)) {
      result[key] = source[key];
    }
  }
  return result;
}

console.log(pick({ foo: 1, bar: 2, baz: 3 }, ['foo', 'baz'])) // -> { foo: 1, baz: 3 }
console.log(pick({ qux: 4, corge: 5 }, ['bar', 'grault']))    // -> {}
console.log(pick({ bar: 2 }, ['foo', 'bar', 'baz']))          // -> { bar: 2 }