Typescript string type that must start with specific characters

user_1979665 0
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I know in TypeScript we can enforce specific strings like so:

const myStrings = ['foo', 'bar', 'baz'] as const;
type MyStringTypes = typeof myStrings[number];

However, what I would need is to enforce only the first characters of my final strings.

For instance, I would like to create a type (MyPrefixTypes) with something like 'prefix1', 'prefix2', ..., 'prefixN', followed by any other character.

Using this, I would be able to check if the string is correct or not.

Example:

const foo: MyPrefixTypes = 'prefix1blablabla'; // OK
const bar: MyPrefixTypes = 'incorrectprefixblablabla'; // NOT OK
typescript typechecking types
1 Answers

As of TypeScript 4.1, you can use a template literal type for this.

type StartsWithPrefix = `prefix${string}`;

This works for type unions as well:

// `abc${string}` | `def${string}`
type UnionExample = `${'abc' | 'def'}${string}`;

// For your example:
type MyPrefixTypes = `${MyStringTypes}${string}`;

const ok: UnionExample = 'abc123';
const alsoOk: UnionExample = 'def123';
const notOk: UnionExample = 'abdxyz';

// Note that a string consisting of just the prefix is allowed
// (because '' is assignable to string so
// `${SomePrefix}${''}` == SomePrefix is valid)
const ok1: MyPrefixTypes = 'foo'
const ok2: MyPrefixTypes = 'barxyz'
const ok3: MyPrefixTypes = 'bazabc'
const notOk1: MyPrefixTypes = 'quxfoo'

You can even define a helper type:

type WithPrefix<T extends string> = `${T}${string}`;

type StartsWithPrefix = WithPrefix<'prefix'>;
type UnionExample = WithPrefix<'abc' | 'def'>;
type MyPrefixTypes = WithPrefix<MyStringTypes>;

Playground link

Related:

user_82899180
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