TypeScript: Object.keys return string[]

Question

When using Object.keys(obj), the return value is a string[], whereas I want a (keyof obj)[].

const v = {
    a: 1,
    b: 2
}

Object.keys(v).reduce((accumulator, current) => {
    accumulator.push(v[current]);
    return accumulator;
}, []);

I have the error:

Element implicitly has an 'any' type because type '{ a: number; b: number; }' has no index signature.

TypeScript 3.1 with strict: true. Playground: here, please check all checkboxes in Options to activate strict: true.

Answer 1

1. npm install ts-extras (written by sindresorhus)

2. Use it:

import { objectKeys } from 'ts-extras'

objectKeys(yourObject)

That's it.

---

Here's another pkg I made before I knew about ts-extras:

npm install object-typed --save

import { ObjectTyped } from 'object-typed'

ObjectTyped.keys({ a: 'b' })

This will return an array of type ['a']

Answer 2

Here's a more accurate utility function:

const keys = Object.keys as <T>(obj: T) =>
    (keyof T extends infer U ? U extends string ? U : U extends number ? `${U}` : never : never)[];

Explanation: keyof T extends string | number | symbol, however Object.keys omits the symbol keys and returns number keys as strings. We can convert number keys to string with a template literal `${U}`.

Using this keys utility:

const o = {
    x: 5,
    4: 6,
    [Symbol('y')]: 7,
};
for(const key of keys(o)) {
    // key has type 'x' | '4'
}

Answer 3

Here is a pattern I use for copying objects in a typesafe way. It uses string narrowing so the compiler can infer the keys are actually types. This was demonstrated with a class, but would work with/between interfaces or anonymous types of the same shape.

It is a bit verbose, but arguably more straightforward than the accepted answer. If you have to do the copying operation in multiple places, it does save typing.

Note this will throw an error if the types don't match, which you'd want, but doesn't throw an error if there are missing fields in thingNum. So this is maybe a disadvantage over Object.keys.

    class thing {
    a: number = 1;
    b: number = 2;
    }
    type thingNum = 'a' | 'b';
    const thingNums: thingNum[] = ['a', 'b'];
    
    const thing1: thing = new thing();
    const thing2: thing = new thing(); 
...
    
    thingNums.forEach((param) => {
        thing2[param] = thing1[param];
    });

playground link

Answer 4

Object.keys returns a string[]. This is by design as described in this issue

This is intentional. Types in TS are open ended. So keysof will likely be less than all properties you would get at runtime.

There are several solution, the simplest one is to just use a type assertion:

const v = {
    a: 1,
    b: 2
};

var values = (Object.keys(v) as Array<keyof typeof v>).reduce((accumulator, current) => {
    accumulator.push(v[current]);
    return accumulator;
}, [] as (typeof v[keyof typeof v])[]);

You can also create an alias for keys in Object that will return the type you want:

export const v = {
    a: 1,
    b: 2
};

declare global {
    interface ObjectConstructor {
        typedKeys<T>(obj: T): Array<keyof T>
    }
}
Object.typedKeys = Object.keys as any

var values = Object.typedKeys(v).reduce((accumulator, current) => {
    accumulator.push(v[current]);
    return accumulator;
}, [] as (typeof v[keyof typeof v])[]);

Answer 5

As a possible solution, you can iterate using for..in over your object:

for (const key in myObject) {
  console.log(myObject[key].abc); // works, but `key` is still just `string`
}

While this, as you said, would not work:

for (const key of Object.keys(myObject)) {
  console.log(myObject[key].abc); // doesn't!
}

Answer 6

I completely disagree with Typescript's team's decision...
Following their logic, Object.values should always return any, as we could add more properties at run-time...

I think the proper way to go is to create interfaces with optional properties and set (or not) those properties as you go...

So I simply overwrote locally the ObjectConstructor interface, by adding a declaration file (aka: whatever.d.ts) to my project with the following content:

declare interface ObjectConstructor extends Omit<ObjectConstructor, 'keys' | 'entries'> {
    /**
     * Returns the names of the enumerable string properties and methods of an object.
     * @param obj Object that contains the properties and methods. This can be an object that you created or an existing Document Object Model (DOM) object.
     */
    keys<O extends any[]>(obj: O): Array<keyof O>;
    keys<O extends Record<Readonly<string>, any>>(obj: O): Array<keyof O>;
    keys(obj: object): string[];

    /**
     * Returns an array of key/values of the enumerable properties of an object
     * @param obj Object that contains the properties and methods. This can be an object that you created or an existing Document Object Model (DOM) object.
     */
    entries<T extends { [K: Readonly<string>]: any }>(obj: T): Array<[keyof T, T[keyof T]]>
    entries<T extends object>(obj: { [s: string]: T } | ArrayLike<T>): [string, T[keyof T]][];
    entries<T>(obj: { [s: string]: T } | ArrayLike<T>): [string, T][];
    entries(obj: {}): [string, any][];
}

declare var Object: ObjectConstructor;

Note:

Object.keys/Object.entries of primitive types (object) will return never[] and [never, never][] instead of the normal string[] and [string, any][]. If anyone knows a solutions, please, feel free to tell me in the comments and I will edit my answer

const a: {} = {};
const b: object = {};
const c: {x:string, y:number} = { x: '', y: 2 };

// before
Object.keys(a) // string[]
Object.keys(b) // string[]
Object.keys(c) // string[]
Object.entries(a) // [string, unknown][]
Object.entries(b) // [string, any][]
Object.entries(c) // [string, string|number][]

// after
Object.keys(a) // never[]
Object.keys(b) // never[]
Object.keys(c) // ('x'|'y')[]
Object.entries(a) // [never, never][]
Object.entries(b) // [never, never][]
Object.entries(c) // ['x'|'y', string|number][]

So, use this with caution...

Answer 7

You can use the Extract utility type to conform your param to only the keys of obj which are strings (thus, ignoring any numbers/symbols when you are coding).

const obj = {
  a: 'hello',
  b: 'world',
  1: 123 // 100% valid
} // if this was the literal code, you should add ` as const` assertion here

// util
type StringKeys<objType extends {}> = Array<Extract<keyof objType, string>>

// typedObjKeys will be ['a', 'b', '1'] at runtime
// ...but it's type will be Array<'a' | 'b'>
const typedObjKeys = Object.keys(obj) as StringKeys<typeof obj>

typedObjKeys.forEach((key) => {
  // key's type: 'a' | 'b'
  // runtime: 'a', 'b', AND '1'
  const value = obj[key]
  // value will be typed as just `string` when it's really `string | number`
})

All that said, most developers would probably consider having numbers as keys a poor design decision/bug to be fixed.

Answer 8

Based on Titian Cernicova-Dragomir answer and comment

Use type assertion only if you know that your object doesn't have extra properties (such is the case for an object literal but not an object parameter).

Explicit assertion

Object.keys(obj) as Array<keyof typeof obj>

Hidden assertion

const getKeys = Object.keys as <T extends object>(obj: T) => Array<keyof T>

Use getKeys instead of Object.keys. getKeys is a ref to Object.keys, but the return is typed literally.

Discussions

One of TypeScript’s core principles is that type checking focuses on the shape that values have. (reference)

interface SimpleObject {
   a: string 
   b: string 
}

const x = {
   a: "article", 
   b: "bridge",
   c: "Camel" 
}

x qualifies as a SimpleObject because it has it's shape. This means that when we see a SimpleObject, we know that it has properties a and b, but it might have additional properties as well.

const someFunction = (obj: SimpleObject) => {
    Object.keys(obj).forEach((k)=>{
        ....
    })
}

someFunction(x)

Let's see what would happen if by default we would type Object.keys as desired by the OP "literally":

We would get that typeof k is "a"|"b". When iterating the actual values would be a, b, c. Typescript protects us from such an error by typing k as a string.

Type assertion is exactly for such cases - when the programmer has additional knowledge. if you know that obj doesn't have extra properties you can use literal type assertion.

Answer 9

See https://github.com/microsoft/TypeScript/issues/20503.

declare const BetterObject: {
  keys<T extends {}>(object: T): (keyof T)[]
}

const icons: IconName[] = BetterObject.keys(IconMap)

Will retain type of keys instead of string[]