Eliminate text from template literals in TypeScript types

I have successfully created a function that eliminates the 'Bar' at the end of a string when using foo. However, is there a way to achieve the same result using the type statement? (refer to the code snippet below)

declare function foo<T extends string>(str: `${T}Bar`): T;
const justFoo = foo('fooBar');

// justFoo now has the type 'foo'

// How can we accomplish this?
type Foo<T extends string> = T;
type JustFoo = Foo<'fooBar'>;

Answer №1

To extract the type from a specific spot, you should utilize the infer keyword:

type Foo<T extends string> = T extends `${infer Prefix}Bar` ? Prefix : never;

This code checks if the string ends with "Bar". If it does, it captures the contents of the string before "Bar" as the type. If not, it sets the type to never, causing a type error upon usage.

Try it on Typescript Playground


You can also constrain the generic so that it cannot be never.

type Foo<T extends `${string}Bar`> = T extends `${infer Prefix}Bar` ? Prefix : never;
type JustFoo = Foo<'fooBar'>; // valid
type ExpectErrorHere = Foo<'nofoofoyoo'>; // expected error

In this scenario:

T extends `${infer Prefix}Bar`

Will always evaluate to true, preventing the use of the never clause. Nevertheless, infer must be part of a conditional type even in such cases.

Play around on Playground

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