Getting the name of a generic type parameter in Typescript

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Is it possible to get the name of a generic type param in typescript.

having this method.

getName<T>(): string {
   .... use some operator or something 
}

using it like this

class MyClass{
}


getName<MyClass>(); ///=> should return 'MyClass'

I've tried using https://www.npmjs.com/package/ts-nameof but it does not work.

Doing 

const name = nameof<T>();

fails

Or maybe is there another way of achieving this ?

typescript
2 Answers

You can't, not with typescript alone. Typescript only compile typescript to javascript. Javascript does not have something like generic type, so:

getName<MyClass>();

is compiled to

getName();

Of course, you do not expect to have different results of the same getName() without parameter. You need something to generate more code in order to do it, or add a parameter to your function:

function getName<T extends new (...args: any[]) => any>(clazz: T): string {
  return clazz.name;
}

class MyClass{
}

getName(MyClass);

It will only work with Class, which exists in runtime.

user_110255820

You need to ensure that you've correctly setup the prerequisites for using TypeScript transform plugins.

TypeScript transform plugins like ts-nameof require some setup because they rely on the transform compiler,ttypescript.

What you need to do:

1. Install ttypescript and ts-nameof:

npm i ttypescript ts-nameof @types/ts-nameof -D

2. Add ts-nameof to your plugins array in tsconfig.json:

{
  "compilerOptions": {
    "noEmitOnError": true,
    "target": "ES2020",
    "plugins": [{ "transform": "ts-nameof", "type": "raw" }],
  }
}

3. Change the path to your tsdk to the location ofttypescript in your vscode users settings:

"typescript.tsdk": "node_modules/ttypescript/lib" // this is for intellisense

4. Run with npx

npx ttsc

Then:

 const name = nameof<MyClass>();

will generate:

const name = "MyClass";

Example: ts-nameof-example

Original documentation:

user_47164880
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