Defining array with multiple types in TypeScript

My TS lint was complaining about other solutions, so the solution that was working for me was:

item: Array<Type1 | Type2>

if there's only one type, it's fine to use:

item: Type1[]

Defining array with multiple types in TypeScript

Use a union type (string|number)[] demo:

const foo: (string|number)[] = [ 1, "message" ];

I have an array of the form: [ 1, "message" ].

If you are sure that there are always only two elements [number, string] then you can declare it as a tuple:

const foo: [number, string] = [ 1, "message" ];

IMPORTANT NOTE

This won't work with complex types with different properties, when you want to access a property available on only one of the types.

See this newer answer.

TypeScript 3.9+ update (May 12, 2020)

Now, TypeScript also supports named tuples. This greatly increases the understandability and maintainability of the code. Check the official TS playground.

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So, now instead of unnamed:

const a: [number, string] = [ 1, "message" ];

We can add names:

const b: [id: number, message: string] = [ 1, "message" ];

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Note: you need to add all names at once, you can not omit some names, e.g:

type tIncorrect = [id: number, string]; // INCORRECT, 2nd element has no name, compile-time error.
type tCorrect = [id: number, msg: string]; // CORRECT, all have a names.

Tip: if you are not sure in the count of the last elements, you can write it like this:

type t = [msg: string, ...indexes: number];// means first element is a message and there are unknown number of indexes.

If you're treating it as a tuple (see section 3.3.3 of the language spec), then:

var t:[number, string] = [1, "message"]

or

interface NumberStringTuple extends Array<string|number>{0:number; 1:string}
var t:NumberStringTuple = [1, "message"];