Literal String Union Autocomplete

[MrTarantula]

Autocomplete works for literal string unions, but adding a union of string negates autocomplete entirely. This has been brought up before but I believe there is enough value in this feature to be reconsidered.

My use case is to have a union of string literals for several colors, but also allow hex codes without having to add 16.7 million string literals.

TypeScript Version: 3.4.0-dev.20190202

Search Terms: Literal string union autocomplete

Code

interface Options {
  borderColor: 'black' | 'red' | 'green' | 'yellow' | 'blue' | string
};

const opts: Options = {borderColor: 'red'};

Expected behavior:

Actual behavior:

Playground Link: https://stackblitz.com/edit/typescript-bwyyab

Related Issues: #12687 #13614

[RyanCavanaugh]

'black' | 'red' | 'green' | 'yellow' | 'blue' | string

From the compiler's point of view, this is just a very fancy way of writing string. By the time we're looking for suggestions at borderColor, all the literals are lost.

You could write something like this:

Naturally this doesn't stop you from writing "bluck". You might want to track #6579

[sindresorhus]

By the time we're looking for suggestions at borderColor, all the literals are lost.

Why not improve the compiler to keep more metadata around?

[MrTarantula]

You could write something like this:

It may accomplish the same behavior, but that's not intuitive to me at all. I doubt I could have gotten there on my own, and I know I would have trouble explaining it to someone newly coming to TS from JS.

[spcfran]

It would be great to have this built in, although understand it may be difficult to implement on the compiler.

In the meantime, a generic workaround based on @RyanCavanaugh's solution might help:

type LiteralUnion<T extends U, U = string> = T | (U & { zz_IGNORE_ME?: never })

type Color = LiteralUnion<'red' | 'black'>

var c: Color = 'red'                    // Has intellisense
var d: Color = 'any-string'             // Any string is OK
var d: Color = { zz_IGNORE_ME: '' }     // { zz_IGNORE_ME } placeholder is at the bottom of intellisense list and errors because of never 

type N = LiteralUnion<1 | 2, number> // Works with numbers too

See it in action here

[BendingBender]

Would be great if this could be implemented. It has the potential to improve even core Node.js APIs. The hash.digest encondig param, for example should accept plain strings but there are values available on all platforms that could be enumerated for ease of use.

[neurolag]

Hey Guys
Though the issue still isn't solved but I found out that in TS 3.5.1 you don't even have to create some weird property in order to get the workaround done:

type LiteralUnion<T extends U, U = string> = T | (U & {});
let x: LiteralUnion<"hello" | "world">;
x = "hey";

While this code is perfectly valid, "hello", and "world" are still showing up in the autocompletion.
Thank you for this great fix, @RyanCavanaugh and @spcfran!

[frenic]

I noticed a problem with type guards using this hack:

type LiteralUnion<T extends U, U = string> = T | (U & {});

function something(arg: LiteralUnion<'a' | 'b'>): 'a' {
  if (arg === 'a') {
    return arg; // Type '(string & {}) | "a"' is not assignable to type '"a"'
  }
}

Is there a way around this?

[neurolag]

I think I might have found a solution:

Use a type like my UnpackedLiteralUnion type to unpack the actual type of the LiteralUnion:

type LiteralUnion<T extends U, U = string> = T | (U & {});
type UnpackedLiteralUnion<T> = T extends LiteralUnion<any, infer U> ? U : never

function something(arg: LiteralUnion<'a' | 'b'>): 'a' {
    let unpackedArg = arg as UnpackedLiteralUnion<typeof arg>;

    if (unpackedArg === "a") {
        return unpackedArg;
    }
    else {
        return "a";
    }
}

[AhmedElywa]

@manuth Your solution not work for me can you know why ?

[neurolag]

@AhmedElywa it's because you're using (U & never).
Edit: After a year or something I finally noticed that my very own example was incorrect... sorry, pal 😅😂

Here's the longer explanation:

let x: (string & never); // x has type `never`
let y: number | (string & never); // y has type `number`
let z: ("Hello" | "world") | (string & never); // z has type `"Hello" | "world"`

In order to get the solution to work you have to use U & {}.
Following might give you an idea why this solution works.

How the solution works

let a: ("hello" | "world") | string;

In this snippet a will have type string because both "hello" and "world" inherit string.

let a: ("hello" | "world") | (string & {});

In this code-snippet a will be of type "hello" | "world" | (string & {}) because though both "hello" and "world" inherit string, they do not inherit string & {}, which means "hello" | "world" and string & {} are treated as distinguishable types.

Hope this helped you understanding.

[papb]

For those interested, there is a workaround for this called LiteralUnion in type-fest.

[kotarella1110]

Is there a way to use this hack with the object key?

TypeScript Playground

[sidwebworks]

is this https://www.typescriptlang.org/play/?#code/FAFwngDgpgBAsgUQCoAkDyARAyjAvDAIgHFkCYAfQgBTSyTIHoGYA6N4USWNAIwCsoAYxAAmPDADewGDJgQATgHsIALhgADACQTEqTFgC+MbQQYBDCAEsGANwCMZSqYvWbIx4XNXbAZg8AKAGcQeUsAOwBzGAAySQMASgN1AG5gAw5BRTDgmDM1XgFhMXwpWTklVUJGZgA5RVyAVxBFTIBbCAAbKBAoNKA related to this issue?

[mfulton26]

is this https://www.typescriptlang.org/play/?#code/FAFwngDgpgBAsgUQCoAkDyARAyjAvDAIgHFkCYAfQgBTSyTIHoGYA6N4USWNAIwCsoAYxAAmPDADewGDJgQATgHsIALhgADACQTEqTFgC+MbQQYBDCAEsGANwCMZSqYvWbIx4XNXbAZg8AKAGcQeUsAOwBzGAAySQMASgN1AG5gAw5BRTDgmDM1XgFhMXwpWTklVUJGZgA5RVyAVxBFTIBbCAAbKBAoNKA related to this issue?

yes, as a workaround you can use type-fest's LiteralUnion

[sidwebworks]

@mfulton26 it does not seem to work even with this helper. Adding both unions into a single string literal causes the whole thing to become just string.

[mfulton26]

@mfulton26 it does not seem to work even with this helper. Adding both unions into a single string literal causes the whole thing to become just string.

it works if you move the template literal type into LiteralUnion<> (instead of nesting LiteralUnion inside the template literal type)

https://www.typescriptlang.org/play/?#code/PTDECcFMHMEsHsB2ACAYueBbZAXAngA6QC0AZpAM44BQkAHgfODroZMgAriyaw6wA3dgF5qASAA+yRAFcANnPFSZiACaRSsRJFVLkVbomh7ZmAEaRwes-HhzIAQ0R6Kec3euw4iHAG5aDEws+ETIADJ8lg5yAKqICIgAPOIROFFyACpsADTiAEIOFJBZofRpahSc3Lz8QrkAfMjC4ZHg0SXsUgAUBUUdyABkyABKkADGTKqJ2kLg2dKQs-UAlP4goJBqUHBI1NQh7ACyAKIZABIA8gAiAMpNyABEAOKnD8hSDxwXNxlvIMgAOiBewOyAuZgAVuMcAAme4Ab2oyGRyAIGAIAC4Wmk2rF4khEgADAAk8JO52uNwAvshSQ9gA4CLBgAIAIxvD4MpksmEcx5c5kCADMDyphPmJLJp0utxppIMWmgYvq-ipewmiCoyAcWPBULGsIRSJRaPgmMeLwyj2Q-wcMhw8AmmAI9jSyAA7kwANaVRDwd3UKlAA

[iartemiev]

If I'm understanding correctly, the reason the suggested solutions work is that a union of a primitive type and some variation of an empty object produces the wrapper/boxed object type corresponding to that primitive.

For example, string & Record<never, never> appears to be equivalent to String, the return type of new String('').

This prevents the union from getting reduced to the wider type, as the compiler will not reduce a literal to an object type. However, assignability is preserved, so we're still able to assign an arbitrary string value in this case.

type Str1 = 'foo' | string;
//    ^? string

type Str2 = 'bar' | String;
//    ^? String | 'bar' 

let a: Str2;
a = 'foo'; // 'foo' suggested via autocomplete
a = 'bar'; // arbitrary string does not error

Is this the correct interpretation or am I making some incorrect assumptions along the way?