Is it possible to skip over a parameter in TypeScript generics (e.g. function(…)) and only specify the 2nd parameter?
I have a function where TypeScript can infer the return type based on the type of the input parameters.
Use argument value as type key in declaration file
I’m working to create a Prisma extension. For reasons, I started with JavaScript, using JSDoc hints for typing. The extension actually works just fine; however, the type hints are terrible, which led me to believe I was doing something wrong. In fact, when I switched to a typescript transpiler, all sorts of errors appeared in my declaration files that were invisible when using esbuild
. My biggest current challenge in squashing these is, I need to use a value provided by the user when the initial extension is instantiated as a key in various other type declarations in my declaration files.
Generics Airthmetic in Typescript
I am a beginner in Typescript. I am trying to learn generics.
TypeScript subclass method override not assignable to same property in base type when returning super
Say I want to extend the Promise class with an overriden then()
method that does something before calling super.then()
:
Type assert that a string contains substrings matching keys in object
I’m working on some type guards for an API framework and want to connect the path parameters (a string) to a validation object.
Type ‘1’ is not assignable to type ‘One’, where T extends number
I have the following code (nb. this is a simplified version, there are more types allowed than just number or bigint, like time periods, etc.) :
Mapped types from arrays and generics
Consider the following code:
Is it possible to turn a utility type into a literal type?
I’m just playing around with TypeScript and wrote this:
How do I create a type that matches a Map’s value?
A JavaScript Map
object has the type Map<K, V>
. I would like to create a type that matches V
.
Is there any way to change my generic function to get warnings?
Here my generic function defc