Here is the code I am working with:
type CustomType<T extends string = string> = `custom-${T}-type`;
const value: CustomType = 'custom-example-type';
The code above functions as expected, but it does not enforce the expected structure. The following code is also accepted:
const value: CustomType = 'custom--type';
How can I ensure that the T parameter is utilized?
In this scenario, the task at hand is to validate against an empty string. For more insights on this issue, you can explore the details here: How to define a string type without including an empty string in TypeScript
Building upon the previous stack trace, a possible solution is outlined below:
type Character = 'a' | 'b' | 'c' | ... ; // Including all single-letter strings
type NonEmptyString = `${Character}${string}`;
type Foo<T extends string = NonEmptyString> = `bar-${T}-foo`;
const v: Foo = 'bar-a-foo' // ✔
const x: Foo = 'bar--foo' // ❌ Type '"bar--foo"' is not matching the assigned type
Using a function that assigns a value of a generic type dynamically may not always be the optimal approach. However, depending on your specific use case, this method could still be considered valid.
type NonEmptyString<T extends string> = T extends "" ? never : T;
function fn<T extends string>(a: `bar-${NonEmptyString<T>}-foo`) {
// ...
}
fn("bar--foo"); // Error
fn("bar-g-foo"); // No error
fn("bar-gg-foo"); // No error
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