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Typescript: Why destructuring an object doesn't hold type?

In the following, I would expect this to fail as I've stated afternoon should be of type number, but it doesn't fail.

My guess is I'm destructuring incorrectly as hovering over the word afternoon when it returns shows the type as any (in VS Code).

Alternatively, it could all be wrong, this is literally my first play with Typescript, not getting it, yet!

const greetings =  {
    morning: "Good morning.",
    evening: "Good evening.",
    afternoon: "Good afternoon.",
};

interface AboutGreetingTypes {
    morning: string,
    evening: string,
    afternoon: number,
};

const aboutGreeting = ({copy}): AboutGreetingTypes => {
    const d = new Date();
    const timeInHours = d.getHours();

    const { morning, evening, afternoon } = copy;

    if (timeInHours < 12) {
        return morning;
    }
    if (timeInHours > 18) {
        return evening;
    }
    if (timeInHours > 12) {
        return afternoon;
    }
};

console.log(aboutGreeting({copy: greetings}));

about 4 years ago · Juan Pablo Isaza
2 answers
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0

Your function parameter type should be {copy: AboutGreetingTypes} not just AboutGreetingTypes

about 4 years ago · Juan Pablo Isaza Report

0

const aboutGreeting = ({copy}): AboutGreetingTypes => {

This line declares a function that accepts a single untyped object and returns an AboutGreetingTypes. Since the argument has no declared type, it is assumed to be any, which lets you do anything without without a type error.


What I think you want is a function that accepts an object that has a copy property that has value of an AboutGreetingTypes.

That would look like this:

const aboutGreeting = ({ copy }: { copy: AboutGreetingTypes }) => {

See playground which still has some errors, but those resolving those are answers to different questions.

about 4 years ago · Juan Pablo Isaza Report
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