Obtaining the identifier of a generic type parameter in Typescript

Question

Obtaining the identifier of a generic type parameter in Typescript

Is there a method in TypeScript to retrieve the name of a generic type parameter?

Consider the following method:

getName<T>(): string {
   .... implement using some operator or technique 
}

You can use it like this:

class MyClass{
}


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

I attempted to utilize https://www.npmjs.com/package/ts-nameof, but it was unsuccessful.

Trying to do this:

const name = nameof<T>();

results in an error.

Are there any alternative methods for achieving this outcome?

typescript

Answer 1

Answer №1

Unfortunately, TypeScript alone does not have the capability to support generic types since it ultimately compiles down to JavaScript, which lacks that feature. As a result, using syntax like getName<MyClass>(); in TypeScript will be compiled to getName();, resulting in the same output without any parameter differentiation.

To work around this limitation, you can either use a code generator to produce additional code or modify your function by adding a parameter, as shown below:

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

class MyClass{
}

getName(MyClass);

It's important to note that this solution only applies to classes that exist at runtime and cannot be applied universally across all scenarios.

Answer 2

Unfortunately, TypeScript alone does not have the capability to support generic types since it ultimately compiles down to JavaScript, which lacks that feature. As a result, using syntax like getName<MyClass>(); in TypeScript will be compiled to getName();, resulting in the same output without any parameter differentiation.

To work around this limitation, you can either use a code generator to produce additional code or modify your function by adding a parameter, as shown below:

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

class MyClass{
}

getName(MyClass);

It's important to note that this solution only applies to classes that exist at runtime and cannot be applied universally across all scenarios.

Answer 3

Answer №2

To effectively utilize TypeScript transform plugins, it is crucial to ensure that the necessary prerequisites are properly set up.

Transform plugins such as ts-nameof require specific setup steps due to their reliance on the transform compiler, ttypescript.

Here's what needs to be done:

1. Installation of ttypescript and ts-nameof:

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

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

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

3. Updating the path for the tsdk to the location of ttypescript in your vscode user settings:

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

4. Executing with npx

npx ttsc

After completing these steps, utilizing the following code:

 const name = nameof<MyClass>();

will result in:

const name = "MyClass";

For reference: ts-nameof-example

Original documentation sources:

Answer 4

To effectively utilize TypeScript transform plugins, it is crucial to ensure that the necessary prerequisites are properly set up.

Transform plugins such as ts-nameof require specific setup steps due to their reliance on the transform compiler, ttypescript.

Here's what needs to be done:

1. Installation of ttypescript and ts-nameof:

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

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

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

3. Updating the path for the tsdk to the location of ttypescript in your vscode user settings:

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

4. Executing with npx

npx ttsc

After completing these steps, utilizing the following code:

 const name = nameof<MyClass>();

will result in:

const name = "MyClass";

For reference: ts-nameof-example

Original documentation sources:

Answer 5

Answer №3

One strategy involves creating a class type interface specifically for the function parameter:

export declare interface ClassType<T> {
    new (...args: any[]): T;
}
class NewClass {
}
function getClassName<TClass>(classType: ClassType<TClass>): string {
    return classType.name;
}

This approach requires passing the class name as an argument to retrieve it:

let myClassName = getClassName(NewClass);

The limitation is that you cannot apply this method to a generic that originates from a data type.

Answer 6

One strategy involves creating a class type interface specifically for the function parameter:

export declare interface ClassType<T> {
    new (...args: any[]): T;
}
class NewClass {
}
function getClassName<TClass>(classType: ClassType<TClass>): string {
    return classType.name;
}

This approach requires passing the class name as an argument to retrieve it:

let myClassName = getClassName(NewClass);

The limitation is that you cannot apply this method to a generic that originates from a data type.

Obtaining the identifier of a generic type parameter in Typescript

Answer №1

Answer №2

Answer №3

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