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Type guard on an object key


By : user2176105
Date : October 14 2020, 02:22 PM
To fix the issue you can do Reason why typescript allows you to index into the object has nothing to do with the union. Under default compiler settings typescript will consider any object as having an index property. The way to top this behavior is to enable the compiler option noImplicitAny. This will surface other erros in your code (I would argue good errors and implicitly typed symbols kill type safety and without this settign you might not even be aware of it).
code :


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Typescript Object is possibly undefined after type guard error TS2532


By : brzmath
Date : March 29 2020, 07:55 AM
To fix the issue you can do From the Feathers typings:
code :
interface Service<T> extends events.EventEmitter {

  create?(data: T | T[], params?: Params, callback?: any): Promise<T | T[]>;
let data = someService.create!({test: 'hello'});

Type Guard for empty object


By : Mohit Bisht
Date : March 29 2020, 07:55 AM
this will help In general, a TypeScript object type includes any object that has at least the specified properties. That means {} is effectively the type of any object. So if your custom type guard claims that param is not a {}, then it is not a {text: string} either, and you are left with never. (In fact, T | {} just simplifies to {}, though maybe it shouldn't quite.)
I don't believe there's a way to write a type guard for an empty object. If you say more about how this came up, I might have more suggestions.

How to type guard a get method on an object


By : user3150095
Date : March 29 2020, 07:55 AM
hope this fix your issue I have an object that contains values accessed from a get method. I want to type the get method such that it infers the return values from the key. , What you need is generics:
code :
interface ObjectWithGetter {
   get<K extends keyof Values>(key: K): Values[K];
}

Does 'key' in object work as a type guard in Typescript?


By : user3220252
Date : March 29 2020, 07:55 AM
it fixes the issue So, my understanding of the in operator was incorrect. 'key' in obj could be true even in { key: undefined } (as pointed out by @acdcjunior in the comments.
So, I opted to change my code to this to fix it:
code :
function test(options?: PaginationOptions) {
    const size = options && typeof options.size === 'number' ? options.size : 100
    const from = options && typeof options.page === 'number' ? size * (options.page - 1) : 0

    return {
        size,
        from,
    }
}

TypeScript: Type guard does not work with switch statement when putting object into variable


By : Akjo
Date : March 29 2020, 07:55 AM
This might help you switch statement serves as type guard in 'good' snippet and doesn't do that in 'bad' one.
code :
...
switch (name) {
    case 'A':
        const aaa = (<ParamA>param).aaa;
...
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