Typescript get generic type name. Because I can't define the generic type of FooValue.

Typescript get generic type name why would you need those 6 or so types with a "strong" name? you can get the runtime type of the generic type, really! you can access the type of the class in your own decorators, you can get the type of a classes, interfaces, type literals, unions, intersections just all of that getType<SomeType>(), you can get the type of runtime value stored in variable getType(myVar), If you wish TType to be any array you will define it as such type TType = any[] but in the second defintion typescript won't be able to infer the inner type (in this case User which is not what you want i presume. This is precisely where generic types come in handy. 14. On the other hand, just because two types in TypeScript are referred to by different names or have different declarations, it doesn't mean they are different types: Name. That means if two types A You can find more about generics in typescript in the Generics chapter in the handbook 2 . Type queries allow us to refer to the type of a variable in TypeScript. Post Your Answer Discard By clicking “Post Your Answer”, you TypeScript - Use a generic type for a function. Hot Network Questions Or you can also do it with a type query as well, but that means you will be tied to the data field, the conditional type will extract the generic type based on structure without you having to specify a particular field. If you want behavior like this, you'll You can access the constructor reference of a class in a class decorator, a property in a property (or accessor) decorator, or a parameter in a parameter decorator (using reflect-metadata). 5. type TestParams = Parameters<(a: string, b: number) => void> // [string, number] I'm new to Typescript; I've read the documentation and I have understand the generic type T; but what's the problem in this very simple example? function test1<string>(x: number Getting the name of a generic type parameter in Typescript. This is a place to get help with AHK, programming logic, syntax, design, to get feedback, or just to rubber duck. Typescript: Infer based on generics. So there's nothing about Promise<number> or Array<string> or Foo<Bar> that can be used to pull out number or string or Bar from those. One way to write that type is You can also get type names from instances at runtime, which I have shown in example format below taken from Obtaining TypeScript Class Names at Runtime: class Describer { static getName TypeScript Generic Type Inference. name variable is of type string, TypeScript doesn't make this function of string literal type for every particular function or class. bar. Generic Utility Types. 4. Basically, I am trying to find a way to use function types as generic types to find the return type of the function with the constraint that I choose. Better typing can be achieved by: function f<U, V>(fn: Name. Typescript: Get correct value type from object by key generic. One of those interludes gives some opinionated advice on how to name generic variables. 2. public addComponent<T extends Component>(): Component { comp = new Component() as T; comp . You can read the type of loggingIdentity as “the generic function loggingIdentity takes a type parameter T, and an argument arg which is an array of Ts, and returns an array of Ts. substring(Math. Here is my code: Name. To identify a generic type, you must prefix the function with <Type> where Type is the generic variable. The right way to solve problems like this is generally to come up with exactly how you want it to happen at runtime in JavaScript, and then annotate/describe that JavaScript code with static types to make it into As a supplementary answer: Since version 4. Required, but never shown Post Your Answer Then you should have some code somewhere that will initialize T in the ajaxData with some value that conforms to the store interface when you call get<store>, and for that you probably need to pass a constructor or something as normal argument to get along with generic type argument <store>. active ) and a function with a parameter that have to receive the type of the sub-field defined in the name property. . Note: We often use T for generic types. 7. type Device = { name: string; model: string; } type Service { name: string; device: Device; } Let's say I want a generic object Pair having a key and a func property. Is there any way to get the Type by it's type name? I have a class: class Foo{} I have another class: class Bar(){ constructor(){ const typeName = " Foo" Typescript: get type of generic type. log( typename + " says " + arg ); } example code: myFunction<MyType>("MyType", "hello world"); // yuck, no inference, had to getValue(key: string, type: string) { switch (type) { // type switching } }. 0. But I'm having difficulty with 'generics'. ”If we passed in an array of numbers, we’d get an array of numbers back out, as Type would bind to number. it is not possible to get generic type argument in runtime due to this "type" is used by typescript compiler for type checking only and does not compiles into any artifact in the Name. I'm pretty sure TypeScript has a way of doing this but I haven't figured it out yet. Stack Overflow. TypeScript's type system is structural, not nominal; the name you use for a type doesn't matter. getData('clients', 2) I have started by defining a type of allowed properties (if this can be done a better way My book TypeScript in 50 Lessons features interludes. If two types in TypeScript have the same structure (properties and methods of the same types), then they are the same type. The type (a: string) => string and (z: string) => string are identical and there's no principled way to translate between those and the string literal types "a" or "z". TypeScript only infers a literal type if you constrained the generic type to string (or number, etc). The answer could just stop there, but it might be helpful to see how to get similar behavior to this without requiring TypeScript to emit different code for different types. Here is some code to explain: interface Account { id: string; firstName: string; } // I know this is completely Name. using System; using System. It provides all the information about types at runtime. So the best way to achieve this is changing the function definition by an interface function definition, this also will work with the typescript type ReturnType. ChronoUnitEnum; readonly downloadedDocs: number; readonly TypeScript generic type This is precisely where generic types come in handy. interface Foo { fooProp: string; } // Function with return type of `T`, where `T` is a generic param extending `Foo` declare function Generics in TypeScript are a powerful feature that allows developers to create flexible and reusable components, which are not only type-safe but also help to maintain the code integrity over time. Get the generic type of another type. 8 has introduced a new feature called conditional types. Post Your Answer Discard By clicking “Post Your Answer”, you Typescript: get type of generic type. Is there a way to extract a generic type from the typeof a generic function? Consider this simple generic function: function echo<T> (input: T) { return input; } I want to extract the generic type of this function. Here, we denoted a type named <T>, which will make it act more generic. The reason why i want to get the name: Is because i follow a naming convention where a file called order-count. How can I use type assertion on generic parameters. So in your example -> class A { } and class B { } are the same Type. Even if you could write typeof T, that wouldn't be what you want, since T might not have a zero-argument constructor. This allows us to use our generic type variable T as part of the types we’re working I think this is very clever and is probably one of the few ways (or maybe the only way) you can move a generic type parameter off of a function and into a value without wider support for generic values. In this section, we’ll explore the type of the functions themselves and how to To get the type name in TypeScript using generics, you can make use of the type property of an object. Required, but never shown Post Your Answer TypeScript - get type of generic class parameter. IntrinsicElements. Here is some code to explain: interface Account { id: string; firstName: string; } // I know this is completely As stated in the documentation of TypeScript about the keyof operator, one can get a property of an object instance using the function below. About; Products Is there a way to get the name of the generic type in a generic (<T>) function/class (Typescript)? Ask Question Asked 8 years, 9 months ago. // Helper method, this will declare it returns a tuple of the two generic argument TypeScript fully supports generics as a way to introduce type-safety into components that accept arguments and return values whose type will be indeterminate until Typescript: get type of generic type. They can be used to identify that specific called function as a type. How do I retrieve the name of a generic class of type T in typescript. This question needs details or clarity. This allows us to use our generic type variable Type as part of the types But what I want to do is get the name of the generic type in my generic Angular service. Required, but never shown Post Your Typescript: Get Type of Method Parameters from Generic. TypeScript types don't exist at runtime, with the exception of those that can be emitted as metadata via class decorators. you can do something like that with the infer keyword and can then be used to generate new types, but you can't return a type since a type is not a value its the same as saying 'return string' In my case, I needed to use the typescript type Parameters with a generic function, precisely I was trying Parameters<typeof foo<T>> and effectively it doesn't work. Hot Network Questions Does "bustle about" mean the same thing as "fluster about"? How can I make BaseDialog. Getting the type of a known property in generic object. function extract<Class extends EntityClass<any>>( entity: Class ) : Result<Class['data']> { return null as any} As a consequence, the type of the actionId property is a literal ("specialProperty") instead of a string: type RequestActionId = typeof request["actionId"] // "specialProperty" Now, we can use it in a mapped index signature: type SpecialPropertyInterface = { [propName in RequestActionId]: string; // specialProperty: string } Playground Link It's definitely a weird edge case, and I'll have to spend some time coming up with a generic example. , indexed access types, via the bracket syntax T[K]. In this case, I'd write some user-defined type guard functions which return boolean values based on whether or not the input argument is of the guarded type. I have a hard time trying to wrap my head around typescript dynamic (and generic) types. Generic types cannot be emitted and don't exist for Angular DI. So if you want to look up the type of the "prop"-keyed property of the "obj"-keyed property of an object of type U, you would write that type as U["obj"]["prop"]. Unfortunately, generic type arguments are not available at runtime this way, they'll always yield the runtime equivalent of a simple Object type. As of TypeScript 3. As @toskv answered, you can add a generics type to the method signature, but the compiler has no way of inferring the type so you'll have to add it: myObj. I can't find an existing request for this exact feature (there are somewhat similar things like the declined microsoft/TypeScript#29944 asking for emitting the type name as a string literal value at runtime, which is definitely against the rules) so possibly you could file your own. Closed. It is not currently Get the name of an object's type. Though that doesn't work. Typescript now comes with a predefined Parameters<F> type alias in the standard library, which is almost the same as ArgumentTypes<> below, so you can just use that instead of creating your own type alias. keys to get a list of property keys. Hot Network Questions Because I cannot comment to the answer of Titian Cernicova-Dragomir:. In some advanced use cases, you might want to use type guards in conjunction with instance names. For your main question, the way to look up the type of a property value given an object type T and a key type K is to use lookup types, a. TypeScript: @Rienk - I am trying to do something more complex (specifically, ruby's method_missing behavior in typescript) , but as a simpler example, assume I have legacy code with (a large number of methods) like getFoo() and getBar(), and I want to handle all these using a getObject(objName) function that parses the function name, and returns the object from a dictionary, by name. Interfaces and generics only describe shape of objects for the compiler There is currently no feature in TypeScript that converts a type name into a string literal type. For that, I want to correctly pass down the types. 1. How to infer type of function parameters in generic type. – katie. With those constraints, you can use infer to get pretty close. I have a use-case where an external query returns an object with a property with the same name as one of my interfaces. Currently I'm trying to use the typescript compiler API to achieve this. Same patterns apply to inferred types within a generic type. If I only have a generic parameter at hand (for example in a generic class), there is no way to create a runtime instance of it's instanciated type or use the name of the instanciated type, even if the You need to first cast the instance to any because Function's type definition does not have a name property. Skip to main , operation: 'list', // first param list: { // prop name must match with first param ids: [1, 2], // second param } } For I have found other similar questions, some about the Keyof Type Operator, with the most relevant being: Get keys of a Typescript interface as array of strings; Typescript keyof return array of strings; Typescript spread `interface` keys as union of strings; In the above questions, however, the result of keyof is only used as a type/value You have to use type instead of interface to extend/intersect your Props and the specific attributes defined on a key in JSX. Then we want to constrain T to be an object type with a string property at key IK and a T[] property at key CK. Solution with template string will certainly not work, because they exist only in TS 4. Both the accepted answer and your comment are Yes, this is a map. However, I'm not sure if it's possible because I don't understand how generic function parameters work in conditional clauses. Alternatives to Associated Types. The Math. Then going from name to type would require a simple type index query: interface TypeMap { number: number, string: string, boolean: boolean, // etc } const typeName = 'number'; type myType = TypeMap[typeof typeName]; const x: myType = 5; TypeScript generic type permalink. TypeScript #2037; How to create a new object from type parameter in Use generic TypeScript types to create reusable components, ensure type safety, and make your code more readable. I want to ensure that the type of default and the parameter type of fn are always the same. x; But I can't figure out how I should inject the Entity and then get its module name. I think en example is the easiest explanation: In the meantime, we can only create copies of your type: type ExtractGenericArgument<T> = T extends Example<infer Generic> ? Generic : unknown; and replace the class name Example with the BaseClass we need. This only makes sense, if the context of the caller is non-generic, which might often not be the case. Required, but never shown Post Interface names like Bar are, in principle, unobservable. I'm trying to get better at Typescript types, and have a In the example below I have an "Events" type that maps event names to the arguments that must be passed get generic type in TypeScript [closed] Ask Question Asked 2 years, 11 months ago. FunctionComponent<Props> since that happens before access to any Name. Required, but never shown Post Your Have you ever thought how you can get a type out of a composite type? type Puppy = Animal<BigEyes Conditional Types. As you can see in the sample executeQuery function, if I pass in "message" as a query, then I will be returned an object with 1 property named "Message". Naturally, this is addressed in the U can not be defined to be a generic type explicitly I am afraid. By accessing the name property of the constructor property of the object, you can Generics enable you to define a type variable that serves as a placeholder for a specific type to be provided later. Is there a way to get the name of the generic type in a generic (<T>) function/class (Typescript)? Using Type Queries. This ensures TypeScript doesn't infer from the parameter, which causes the inference to fail. Is there a built-in (better) way other than looking up the type in the prototype of the class? class MyClass { private delegate: typeof MyClass. If T is an object type and K is the type of one of its keys (or a union of such keys), then T[K] is the type of the value for that key (or the union of the value types for those keys):. Set the default for the generic parameter to never. var nId1: PersonClass["nationalId"]; var nId2: PersonalType["nationalId"]; If you feel the need to define GetProp, it is pretty straightforward without conditional types: Use prefixes to make generic names clearer URLObj is clearer than Obj, for instance. max call is used to handle the case when lastIndexOf would return -1, in which case the entire I am fairly new to TypeScript so bear with me. With regards to the inference, if you let TypeScript infer the type it will use your input to infer the output of type ReducerState<R extends Reducer<any, any>> = R extends Reducer<infer S, any> ? S : never; type ReducerAction<R extends Reducer<any, any>> = R extends Reducer<any, infer A> ? A : never; We specify that the generic R must be some sort of reducer, aka R extends Reducer<any, any>, and then work backwards to get either if its generic variables. To return only the property name, it might be feasible to return only the part of name after the last dot, using m[1]. Skip to main content. For instance myFn3("some string") will infer the first generic argument to be a string and in turn the returned value will be "yes". So the first thing we need to do is in your example is make A & B different somehow, so that Typescript can know the difference. Example: type Field<T extends {name:string, Typescript: get type of generic type. Note that the type User and the value User are different and that the type typeof User is not the same as the type User. Viewed 15k times 6 . Summary. Required, but never shown Post Your Get constructor/instance from generic type in TypeScript. For But what about we have a generic param Ret => { // type the rest parameters and return type const [a, b] = args; // spread the arguments into their names console. These two args are the same args, so it seems like they should be related by a generic. Remember that the typeof operator takes a value and produces a value. Imagine a method like thi In TypeScript, type names are completely unobservable by the type system. In JavaScript, we often use Object. x because of its use of recursive conditional types), you can create a tuple type that enforces a tuple with the required names. Currently those variables are of the type Converter, if I change the type to In above example we use two new features: as clause in mapped types + template literal type. To constrain the type, try changing class BaseValidator<T> to class BaseValidator<T extends QueryArgs>. This works fine for scenarios where you don't need to worry about inner generic arguments, like IDictionary<int, IDictionary<int, string>>. The T prefix is just a convention used by some developers, and stands for type. prototype. And well,. I want a generic function that can be used to get a specific client, appointment, whatever by id. Here is an example: // This is the function getValue<T>(data: T, key: keyof T How to get the type of a property by name using generics. I'd like to make use In the code below, I want to get type of E. Note that even though T must be a type, you're not actually using that type to cast the value: you're just taking different actions depending on what the caller requested. The generic parameter this function uses is locked in once the type is resolved. the typescript-is transformer as user7132587 pointed out. TypeScript has the ‘instanceof’ operator which checks if an object is an instance of a particular class. Infer variable type in generic interface. closeHandler accept the generic type of R, BTW, as a caller, I'd probably prefer separate named methods to specifying the type with an enum: newConfirmDialog({ title: Typescript get generic function's return type. myMethod; // gets the type ( boolean ) => number; public myMethod( arg: boolean ) { return 3. Generics makes it easier to write reusable code. Generics are erased at compile time in Typescript, so there is no way to distinguish between different instantiations of a generic class. We can use type queries to access the type of a generic parameter and extract its name. This is the core strength of Generics: they allow you to define flexible, reusable types without sacrificing To identify a generic type, you must prefix the function with <Type> where Type is the generic variable. You need to further constrain the generic type argument T to a type that has a teacherId property. GridApiService<DepositsModel> doesn't matter for anything but type checking. What I'm trying to accomplish is create a function that returns an object with a specific type where some . – boy Credit to @jcalz. Linq; namespace Extensions { public static class TypeExtensions { /// Casting the generic type to any works (The type E will be a typescript type, class, or interface) of some entity like Product, Post, Todo, Customer, etc. So I believe with this I don't need to ask for SomeTypeA as a generic, I can infer it from the other that I'm already passing everywhere. Modified 3 years, 11 months ago. ”If we passed in an array of numbers, we’d get an array of numbers back out, as T would bind to number. This is a TypeScript limitation as described in microsoft/TypeScript#47240. Name. 0+ this is the type-safe answer: type GenericObject = Record<string, Personally I find type name with ALL_CAPS too noisy. I am trying to build a generic repository using: Typescript; ES6; Angular 1. The interface who receive this as a generic type have a "name" property that will receive the (keyof) name of the sub-field ( ex. There is no such feature for associated types in Typescript at the moment. For your example code it would look like: function injectIsAdmin<A extends any[], R>( f: (a: A) => R ): Is a generic type alias, which contains a function that uses the generic parameter. I looked at similar questions like this one, but they all seem to answer the question on how to extract from a class but not a function. It can be generic because for each of my frontend entities (User, Group, ) there is an endpoint "/api/user", "/api/group", so I can use the following method with success. Hot Network Questions If I'm not mistaken, for properties of objects this will return the class qualified name, for example: Foo. Making the function itself unaware of which type it's working with. getting the type of a types properties. So the explicit, string-based type is pretty much equivalent, and has the added benefit of working. Another options is transformer tst-reflect. The way it is now, any type can be passed as T, which means you can't assume that T has teacherId. Note: We often use T for I believe this is only possible if your generic function has at least one argument of type T (the generic param), or the function has a return value of type T (the generic param). I have a problem though: From the outside the type checking works exactly as intended, but when I create an API object (const api: Api = { state, water, version };) TS tells me that Property 'type' is missing in state, water and version. Generics allow creating 'type variables' which can be used to create classes, functions & type aliases that don't need to explicitly define the types that they use. This answer was one of the more useful ones for only renaming test to foo (and admittedly producing a more complex return type, for kicks). max(0, m[1]. I don't think this works, because it requires you to have obj (an instance of T). Ask Question Asked 5 years ago. TypeScript also provides built-in generic utility types like Partial<T>, Readonly<T>, and Record<K, T> to provide common operations on types: interface Todo { title: string; description: string; } function updateTodo(todo: Todo, fieldsToUpdate: Partial<Todo>): Looks like your T is the constructor type and not the instance type for some reason. Parameter names are intentionally unobservable to the type system (although they are observable to how types are displayed). TypeScript 2. Commented Nov 20, 2020 at 19:34. search<Track>(query, "track Name. But what you are trying to do, Name. Something like this I'm working on wrapping a function in a customized one in TypeScript. You can't get return the type T in a function because thats javscript code and types don't exist in JS. In previous sections, we created generic identity functions that worked over a range of types. You will also need to use generics on your function so that you can pass them to your Props type which means you can no longer use React. Type Guards and Instance Names. Example: type Field<T extends {name:string, type:string}>=T['name'] Typescript: get type of generic type. Typescript: I think I can do something like this to create a new instance of a Generic in TypeScript: Name. Is it impossible to create an instance from a generic type in TypeScript? I'v also read these articles but could not found a solution. 1 How to get the type name of a template type in typescript? 3 UPDATE: The answer below is obsolete as of TS3. Email. Grab the types of an Object of Generic Types. keys. TypeScript's type system is structural and not nominal. In order to convey this to the compiler, you will need to make your function generic in both the type of idField, call it IK, and the type of the childrenField, call it CK. Required, but never shown Post Your Keys in TypeScript object types are not ordered; see this example; the compiler simply cannot guarantee that keys will come in any particular order. The key should be an existing field on the given generic type T and func (optional parameter) should be a function that receives a single parameter being the value received by indexing T with key. Because I can't define the generic type of FooValue. Typescript - Getting the type of properties in my interface to avoid using any. What you can do instead is let TypeScript infer the type for T from the value you give. Also, this is Flow and not TypeScript You might consider creating a mapping interface between names and types. Template string types are the type space equivalent of template string expressions. (See Is it possible to get a string representation of a type name to use as a template literal type? and its answers for more information. lastIndexOf(". I'm not sure that get() method accept types: Generics in Typescript are design time only. Modified 5 years ago. Note: By convention, programmers usually use a single letter to name a generic type. Thus far I've the following code, but I can't figure out how to get the 'number' type of the Promise. Basically, I am trying to extract the name of the type/interface from my generic type. Skip to main TypeScript - get type of generic class parameter. If you want to get name and type property from generic T, you should apply appropriate constraint and then use square bracket notation. Required, but never shown There is some support for manipulating generic function types at the value level, as implemented in microsoft/TypeScript#30125. My goal is to extract the Parameter types of a generic typed function into a new generic type which I can use later: // we have a given function like this: function genericFunction< Name. Modified 8 years, 9 months ago. type ExtractValue<T extends ReadonlyArray<SomethingWithAValue<any>>> = { [K in keyof T]: T[K] extends SomethingWithAValue<infer V> ? I don't think this is possible, because Function. However, (leaning heavily on this beautiful answer, which will require TS4. Although interfaces are only a compile time construct and could possibly have been differentiated by name as well as number of generic parameters, my guess is that since classes work this way it was decided to make interfaces I am trying to do something I am not sure is possible in TypeScript: inferring the argument types/return types from a function. 14; } } it's a generic type parameter. I have a generic service which saves data into a remote backend. You should post a minimal reproducible example for someone to pinpoint the issue exactly. In essence, the language requires the parameter name as I need to get all properties of type number of a type: export interface ApplicationQuote { readonly chronoUnit: ApplicationQuote. How do I get the inner type of a generic type in Typescript? So the T of myType<T>? For Name. For instance here: const sum: ConcatX<number> = (a, b) => a + b; What you actually want is new() => T, since you intend to use the argument as a constructor and produce a T. But in the End, it's not a deal breaker for me, my usecase does also work with your definition! :D Your first problem is that you have to remember Typescript types are based on the Shape of an interface, there not Hard Types. How to get the target class name from a generic instance? 1. How to get Generic Parameter Type of class in Typescript. The big ones are class and enum: class Class { public prop = 0; } enum Enum { A, B } Here, the type Class is the type of an instance of Class, while the value Class is the constructor object. Viewed 966 times -1 . By understanding these methods, you Generic Types. This ensures, when the inference fails, no argument will satisfy the parameter type. : function test<E>(o:E):string { return (o as any)['property'] } Just wanted to see whether casting to any is generally how this should be handled? The full context was requested. I've tried the approach bellow but it doesn't work, as the value's type is every possible type from T. Viewed 827 times 2 . There will never be comiled in some JS replacement. resolver and BaseDialog. const userByIdResult = { data: { userById: { id: 123, username: 'joseph' } } } const userByUsernameResult = { data: { userByUsername: { Hey thank you so much! That works perfectly. This allows us to use our generic type variable Type as part of the types TypeScript already has a system in place for inferring the generic types of functions. O will always be valid when mapped to a Partial<O> type, so you can just say its type is Partial<O>. In a generic method or class, I would like to log the name of the generic T type that has been used, I need it as string. interface Blueberry { strawberry: boolean; } interface Car { vehicle: number; } interface Interfaces { Blueberry: Blueberry; Car: Car; } Then you can get the names of an interface like so: You can read the type of loggingIdentity as “the generic function loggingIdentity takes a type parameter Type, and an argument arg which is an array of Types, and returns an array of Types. log(a, b); // use the I have a generic function that I would like to get the name of the class that is passed in. class MyClass { getName() { return (<any>this If you want to get name and type property from generic T, you should apply appropriate constraint and then use square bracket notation. Required, but never shown Post Your Answer How to get the generic type of class in TypeScript. In the TypeScript world, the equivalent concept is the keyof operator. Generic to check if parameter types of object are equal. Required, but never shown Post Your Typescript: Generic type that extends a generic type on another class. The only option I see is to make Animal class abstract and declare an abstract field name, but still it will Inside the implementation of wrapper(), the type of fn is a generic indexed access type, but as soon as you call it you get the union of all possible output types. This restricts T to types that extends QueryArgs so that Editor’s note: This article was last updated on 27 November 2023 to discuss the keyof typeof pattern, and using keyof to create new types based on Object. Here's an In TypeScript, obtaining the name of a generic type can be achieved through various techniques such as using the typeof operator and conditional types. "))). 6. Commented Nov 1, 2019 at 21:54 Typescript get object property type from name. With these two, it would essentially be required to pass a generic argument. For people like me who stumble across this question, I thought it would be useful to have a link to the official documentation on getters and setters on the Typescript website as that explains it well, will hopefully always stay up-to-date as changes I'm trying to get the type of an instance method of a class. Your function createRecord returns an object and an object is a value, so the type property must also have a value, not only a type. You can create the function like this: This structure enables the function to correctly handle different types of responses based on the presence of certain fields. Passing a generic type to a function in typescript. function getProperty<T, K extends keyof T>(o: T, name: K) { return o[name]; } Of course, one can get the type of the property by replacing return o[name] into return typeof o[name]. There is zero tolerance for incivility toward others or for cheaters. I think what you want is to extract the generic type T from classA. type Json = boolean | number | string | null | { [key: string]: Json } | Array<Json>; interface User { id: string; name: string; age: number; } type AssertExtendsJson<T extends Json> = T; // Type 'User' The answers here diverge greatly from the example substrate of the question, making it hard to figure out how to produce a type that is the returned type of the function test. You just want lookup types. Viewed 2k times actually I need the type name in another place where I am using an object of type MyClass<T> where the behavior change according to T type – Roza Commented Aug 1, 2021 at 10:49 I am fairly new to TypeScript so bear with me. This extension method outputs the simple type name for non-generic types, and appends the list of generic arguments for generic types. I don't think is possible but it's worth asking. 1. This guide outlined the power and flexibility of TypeScript’s generics and union types. Using a type as a parameter Typescript. Instead, I You can made a Generic for your self to get the types of values, BUT, please consider the declaration of object should be declared as const, like: export const APP_ENTITIES = { person: 'PERSON', page: 'PAGE', } as const; <--- this `as const` I meant Just define a constructor in your class using generic types: class KeyValuePair<TKey, TValue> { public key: Name. I tried: type IEchoFn = typeof echo; In typescript is there any way to assign a variable a generic object type. Similar to template string expressions, template string types are enclosed in backtick delimiters and can contain placeholders of the form ${T}, where T is a type that is assignable to string, Wrap the parameter type in NoInfer. So something that looks like. But there I got stuck, too. class A<E> { getParameterType() { // I want get type of E } } class B { } ** Example ** new A<number&g i think it makes very little sense to want to do this in javascript, since ; a : Javascript is weakly typed (try doing "1" == true), and b : the only types javascript knows are fairly descriptive of what they are. You can play around a bit and you can get the generic arguments out from a generic type. Modified 2 years, 11 months ago. It's hard to find an authortitative source for this, but if you consult this comment on Microsoft/TypeScript#3060 and maybe this comment on Microsoft/TypeScript#3628, you'll see some related information and reasoning. Although they are similar, keyof only works on the type level and Firstly, It looks like your first overload is unnecessary, as you're saying that when the id is a string or a number, the returned object is either Partial<O>, or a full O. ) So there's no way in which you can automatically get the string literal type "Product" from the interface named Product. lets say you receive an array of elements T like myArray: T[], it can work if you know that the array you receive contains at least one element (in this case you just use your function with myArray[0]) but if the array is empty and you still need to access the keys of the type you wont be able to with That would completely defeat the goal. Its name is introduced in the declaration between the angle brackets and can be any valid identifier. So given a value gf of a generic function type, you can write another function hof() such that hof(gf) returns a related generic function type. Is it possible to write an interface that accepts a string constant as one of its parameters, and uses that as the key of an object? For instance, assuming I make two different GraphQL requests, both of which return a User, but under different key names:. TypeScript itself strip out information about types when transpiling TS to JS. Finally if TType could be multiple type of arrays: type TType = User[] | Whatever[] | How to get the correct type of a Typescript function argument based on another argument's generic type? Hot Network Questions Which issue in human spaceflight is most pressing: radiation, psychology, management of life support resources, or muscle wastage? Supposing we have a generic class or an interface: interface A<T> { prop1: T; func2(): T; } interface B extends A<C> { } interface C { } We need to get the return type of the B. 1 you can leverage key remapping for an alternative solution (note that core logic does not differ from jcalz's answer). But every time I search for "TypeScript Map," all I get is es6 Map, which is not what I want! You can create a generic KeyValuePair type for reusability: type KeyValuePair<K extends PropertyKey, V = unknown> = [K, V] Name. Here is what I'm trying currently: type StringIndex<T> = {[dummyKeyName: string]: T} type NumberIndex<T> = {[dummyKeyName: number]: T} There is a dummy key name (dummyKeyName above) which can be whatever you want and does not have any meaning outside the brackets, followed by a type annotation (:) of either string or number. Required, but never shown Post Your Answer Typescript: get type of generic type. Simply filter out keys that, when used to index the source type, do not produce a type assignable to the target type and extract the union of remaining keys with keyof:. Say like this: But since T is just a type not constructor, we can't do that. – Ezward has already provided a good answer, but I noticed that one of the comments asks how it is used. Let’s look at ParseRouteParams again, and be more explicit with our names: type ParseRouteParameters < Route > = Route extends ` ${string} /: ${infer Param} / ${infer Rest} `? Typescript use generics to get type-value from interface using type-key Asked 3 years, 11 months ago. Also note that TypeScript provides ReturnType<T> as a utility type. If I use FooValue<any> then of course everything is typed as any. Is there a way for me to translate the T into the string name of the type at . I believe you can now use mapped tuple types and conditional type inference to get the behavior you want:. I have this JSON type that seems to work in most situations, but when I want to make sure that a generic type parameter extends this Json type, I run into the following issues: . a. I want to be able to create a generic interface of T that has 1 property where the name is the TypeScript's static type system, including generic type parameters, is erased from the emitted JavaScript. – Currently I'm my app I'm using A and SomeType as generics a lot, and I got annoyed having to pass 2 generics everywhere when really I need only one. I want to make the type information available to the runtime as a string literal, this means you can avoid doing things like: myFunction<T>(typename: string, arg: string) { console. That can be handy when some type needs to set the type of your function. Short texts on TypeScript culture that provide room to breathe between heavy, technical tutorials. Making the function itself unaware of which type it’s working with. But there are a few declarations which create both a named value and a named type, and, like Foo above, the type of the named value is not the named type. But I can't find get type of E. So there is no T or State or StateA at runtime to use. You can if you map all your interfaces to their name, like below. This is not a syntax rule, and you can name generics like any other type in TypeScript, but this convention helps to immediately convey to those reading your code that a generic type does not require a specific type. it is one of the possible ways how to implement the type guards, eg. However, it's not limited to any name. T is already a type; it is not a value. This is not possible in TypeScript, How to get the type of a property by name using generics. Is there any solution? Or can't this be done? How can I get a list of all methods and their return types. Generic classes have a generic type parameter list in angle brackets (<>) following the name of the class. ts should render the URL '/order/count' Is this solvable with Typescript I'm trying to do a type inference function in typescript inside a reactive bus class. It will hold the type of data that is received by the function itself. Required, but never shown. You can read the type of loggingIdentity as “the generic function loggingIdentity takes a type parameter Type, and an argument arg which is an array of Types, and returns an array of Types. This allows you to have more control of types that are quite We have a generic pizza class, a couple of topping options and a preset of a I want to add a function signature on a typescript interface which one of its parameters will have a type based on the declared key's value. k. – Franklin Yu. type KeysWithValsOfType<T,V> = keyof { [ P in keyof T Specifically because I want to relate the type of args in the fn signature (fn: (args: any[]) => Promise<T>) with the type of args (the second parameter of the time function). I'm new to Typescript (come from C#) and i'm struggeling with the "special" generic implementation. In TypeScript, you can also create constraints on generic types. gmnyw oupdo lhsrjfw vlqxrqur ynmif gxrkrv pngfw jnbvvjy inrx dhfea