Getting values of all id as type from an object

I have an object like

var myObj = [{ id: 'a', value: 1}, { id: 'b', value: 2 }, { id: 'c', value: 3}]

and I have a function to which I want to pass just possible value of ids

function abc(id: 'a' | 'b' | 'c') { /* */ }

I tried like this

var myObj = [{ id: 'a', value: 1}, { id: 'b', value: 2 }, { id: 'c', value: 3}]

type OJ = typeof myObj;
type id = keyof OJ[number]['id'];

but its showing id as
type id = number | typeof Symbol.iterator | "toString" | "charAt" | "charCodeAt" | "concat" | "indexOf" | "lastIndexOf" | "localeCompare" | "match" | "replace" | "search" | "slice" | ... 30 more ... | "padEnd"

Your error will be fixed if you remove keyof from type id = keyof OJ[number]['id'];. This will make id match string. If you want more specific type, ie, you want id to be "a" | "b" | "c", then you can give as const after the object.

function abc(id: 'a' | 'b' | 'c') { /* */ }

var myObj = [{ id: 'a', value: 1}, { id: 'b', value: 2 }, { id: 'c', value: 3}] as const

type OJ = typeof myObj;
type id = OJ[number]['id'];
//    ^?

playground

You don't need the keyof in your type id:

const myObj = [{ id: 'a', value: 1}, { id: 'b', value: 2 }, { id: 'c', value: 3}];

type OJ = typeof myObj;
type id = OJ[number]['id'];

With this approach, however, the type of id will be string, because TypeScript considers myObj to be open-ended and you could add additional elements to the myObj later on.

To prevent this, use a const assertion:

const myObj = [{ id: 'a', value: 1}, { id: 'b', value: 2 }, { id: 'c', value: 3}] as const;

type OJ = typeof myObj;
type id = OJ[number]['id'];