Using the keyof operator to determine the data type of a property within a TypeScript class

Question

Using the keyof operator to determine the data type of a property within a TypeScript class

According to TypeScript's documentation on the keyof operator, you can access a property of an object instance using this function below.

function getProperty<T, K extends keyof T>(o: T, name: K) {
    return o[name];
}

If you want to obtain the type of the property without passing any object instance, is there a possible way to do so?

function getPropertyType<T>(name: keyof T) {
    // Is there a method similar to T[name]?
}

typescript

Answer 1

Answer №1

Absolutely, you can use lookup types without any issues:

type BarType = FooType['bar'];

In this scenario, it assumes that FooType is structured like an object:

type FooType = {
    bar: string;
}

The assignment of BarType mirrors the type of FooType['bar'], which in this case is a string.

Additionally, keep in mind that FooType could also represent an interface or class.

Answer 2

Absolutely, you can use lookup types without any issues:

type BarType = FooType['bar'];

In this scenario, it assumes that FooType is structured like an object:

type FooType = {
    bar: string;
}

The assignment of BarType mirrors the type of FooType['bar'], which in this case is a string.

Additionally, keep in mind that FooType could also represent an interface or class.

Answer 3

Answer №2

Is this the solution you seek?

type PropType<TObject, TProperty extends keyof TObject> = TObject[TProperty];

Retrieve the type of an object property with the following:

type MyPropType = PropType<ObjectType, '<property>'>;

This method mirrors the use of Pick in typescript, providing compile-time error reporting for invalid property keys.

Updates:

Thanks to @astoilkov's suggestion, a simpler alternative would be PropType['property'].

Answer 4

Is this the solution you seek?

type PropType<TObject, TProperty extends keyof TObject> = TObject[TProperty];

Retrieve the type of an object property with the following:

type MyPropType = PropType<ObjectType, '<property>'>;

This method mirrors the use of Pick in typescript, providing compile-time error reporting for invalid property keys.

Updates:

Thanks to @astoilkov's suggestion, a simpler alternative would be PropType['property'].

Answer 5

Answer №3

If you were to replace return o[name] with return typeof o[name], you would actually receive a return value of
"string" | "number" | "boolean" | "symbol" | "undefined" | "object" | "function"
. This is because the JavaScript typeof operator used at runtime does not provide the compile-time type information seen by TypeScript.

In TypeScript, the keyword typeof serves as the compile-time type query operator, distinct from the JavaScript typeof operator used for runtime evaluation. For example, in TypeScript, Array<typeof a> evaluates to Array<{foo: number}> where a is an object defined as { foo: 0 }.

TypeScript does not facilitate retrieving compile-time type information at runtime through a typeof operator. However, it does support attaining type details about properties at compile time using lookup types, as achieved in functions like getProperty().

To obtain compile-time type information about object properties, one can use lookup types such as T[K], representing "the type of property with a key of type K on an object of type T." This allows declaring values of specific property types without necessitating values of the parent type objects.

This response aims to address your inquiry based on the available context. If further details are provided, a more tailored explanation can be offered. Best of luck!

Answer 6

If you were to replace return o[name] with return typeof o[name], you would actually receive a return value of
"string" | "number" | "boolean" | "symbol" | "undefined" | "object" | "function"
. This is because the JavaScript typeof operator used at runtime does not provide the compile-time type information seen by TypeScript.

In TypeScript, the keyword typeof serves as the compile-time type query operator, distinct from the JavaScript typeof operator used for runtime evaluation. For example, in TypeScript, Array<typeof a> evaluates to Array<{foo: number}> where a is an object defined as { foo: 0 }.

TypeScript does not facilitate retrieving compile-time type information at runtime through a typeof operator. However, it does support attaining type details about properties at compile time using lookup types, as achieved in functions like getProperty().

To obtain compile-time type information about object properties, one can use lookup types such as T[K], representing "the type of property with a key of type K on an object of type T." This allows declaring values of specific property types without necessitating values of the parent type objects.

This response aims to address your inquiry based on the available context. If further details are provided, a more tailored explanation can be offered. Best of luck!

Answer 7

Answer №4

Discover various methods to determine the type of a property within an object. While some have been mentioned, let's delve deeper into this topic.

A helpful tip: incorporating Conditional Types can assist in avoiding type errors, depending on the specific requirements at hand.

1. Utilizing Conditinal Types with infer

Imagine having the following object:

const student = {
    name: 'John Doe',
    age: 22,
    dormitory: true
}

The objective is to ascertain the type of age (for instance)

type AgeTypeUsingInfer<T> = T extends {age: infer K} ? K : never;
type AgeType = AgeTypeUsingInfer<typeof student>

The resultant AgeType will correctly indicate a number type.

If an object lacking the age property is used, the return type will be never.

const musician = {
    art: 'Pop',
    city: 'Gotham'
}

type MusicianAgeType = AgeTypeUsingInfer<typeof musician>

In such instances, MusicianAgeType will resolve as never sans any type errors.

2. Implementing Conditional Types with Indexed Access

type AgeTypeUsingConditionalTypesAndIndexedAccess<T> = T extends {age: any} ? T['age'] : never;

type AgeType2 = AgeTypeUsingConditionalTypesAndIndexedAccess<typeof student>; // number
type MusicianAgeType2 = AgeTypeUsingConditionalTypesAndIndexedAccess<typeof musician>; // never

3. Employing Indexed Access

type AgeTypeUsingIndexedAccess<T extends {age: any}> = T['age'];

type AgeType3 = AgeTypeUsingIndexedAccess<typeof student>; // number
type MusicianAgeType3 = AgeTypeUsingIndexedAccess<typeof musician>; // type error

Based on your needs, you can designate the type as never, or permit a type error to promptly emerge if an incompatible object is provided.

This explanation aims to aid in your programming pursuits... happy coding :)

Answer 8

Discover various methods to determine the type of a property within an object. While some have been mentioned, let's delve deeper into this topic.

A helpful tip: incorporating Conditional Types can assist in avoiding type errors, depending on the specific requirements at hand.

1. Utilizing Conditinal Types with infer

Imagine having the following object:

const student = {
    name: 'John Doe',
    age: 22,
    dormitory: true
}

The objective is to ascertain the type of age (for instance)

type AgeTypeUsingInfer<T> = T extends {age: infer K} ? K : never;
type AgeType = AgeTypeUsingInfer<typeof student>

The resultant AgeType will correctly indicate a number type.

If an object lacking the age property is used, the return type will be never.

const musician = {
    art: 'Pop',
    city: 'Gotham'
}

type MusicianAgeType = AgeTypeUsingInfer<typeof musician>

In such instances, MusicianAgeType will resolve as never sans any type errors.

2. Implementing Conditional Types with Indexed Access

type AgeTypeUsingConditionalTypesAndIndexedAccess<T> = T extends {age: any} ? T['age'] : never;

type AgeType2 = AgeTypeUsingConditionalTypesAndIndexedAccess<typeof student>; // number
type MusicianAgeType2 = AgeTypeUsingConditionalTypesAndIndexedAccess<typeof musician>; // never

3. Employing Indexed Access

type AgeTypeUsingIndexedAccess<T extends {age: any}> = T['age'];

type AgeType3 = AgeTypeUsingIndexedAccess<typeof student>; // number
type MusicianAgeType3 = AgeTypeUsingIndexedAccess<typeof musician>; // type error

Based on your needs, you can designate the type as never, or permit a type error to promptly emerge if an incompatible object is provided.

This explanation aims to aid in your programming pursuits... happy coding :)

Using the keyof operator to determine the data type of a property within a TypeScript class

Answer №1

Answer №2

Answer №3

Answer №4

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