Detecting if a string is in sentence or title case with a typeguard

When setting the sameSite property of a cookie, it must be either Strict, Lax, or None. However, the package I'm using uses lowercase values for this attribute. Therefore, I need to adjust the first letter of the string:

let sentenceCaseSameSite: "None" | "Lax" | "Strict" | undefined
if (
    (sessionCookie.attributes.sameSite &&
        sessionCookie.attributes.sameSite === "strict") ||
        sessionCookie.attributes.sameSite === "lax" ||
        sessionCookie.attributes.sameSite === "none"
    ) {
        const firstLetter =
            sessionCookie.attributes.sameSite[0].toUpperCase()
        const sentenceCaseSameSite = (firstLetter +
            sessionCookie.attributes.sameSite.slice(1)) as
            | "None"
            | "Lax"
            | "Strict"
    }

This approach involves using

as | "None" | "Lax" | "Strict"
, but it doesn't feel like the optimal solution.

Is there a method to capitalize the initial letter of the string while maintaining type safety?

Answer №1

When working with the code below, it is important to note that Capitalize must be utilized within the cap function in order for TypeScript to recognize that x[0].toUpperCase() + x.slice(1) returns a capitalized type. The intention of TypeScript developers was for users to create a function like cap, which would then require a Captialize utility to define the return type.

A typeguard for L has been implemented on x.sameSite, enabling it to be passed safely to the cap function, resulting in the correct typing for sessionCookie.

TypeScriptPlayground

type L = "none" | "lax" | "strict";
function isL(s:any): s is L {
    return ["none","lax","strict"].includes(s);
}
function cap(x:L) {
    return x[0].toUpperCase() + x.slice(1) as Capitalize<L>;
}
declare const x: { sameSite: string | undefined };
if (isL(x.sameSite)){
    let sessionCookie = cap(x.sameSite); // type is "None" | "Lax" | "Strict"
}

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