typescript-types

问题 I'm writing a React app using TypeScript. I use material-ui for my components. I'm writing a custom wrapper for material-ui's Button component. It looks like this: import MUIButton, { ButtonProps } from "@material-ui/core/Button"; import withStyles, { WithStyles } from "@material-ui/core/styles/withStyles"; import classNames from "classnames"; import React, { PureComponent } from "react"; import styles from "./styles"; export interface OwnProps { color: "primary" | "danger" | "warning" |

问题 Is there a way to obtain proper type using approach described below. I have tried make makeInstance() generic but I haven't obtained Extended type. Code below. class Base { name = 'foo'; static makeInstance() { return new this(); } } class Extended extends Base { age = 10; } let base = Base.makeInstance() // is Base let extended = Extended.makeInstance(); //should be Extended console.log(base.name);//ok console.log(extended.age); //output ok; age doesn't exists console.log(extended.name);//

问题 Is there a way to obtain proper type using approach described below. I have tried make makeInstance() generic but I haven't obtained Extended type. Code below. class Base { name = 'foo'; static makeInstance() { return new this(); } } class Extended extends Base { age = 10; } let base = Base.makeInstance() // is Base let extended = Extended.makeInstance(); //should be Extended console.log(base.name);//ok console.log(extended.age); //output ok; age doesn't exists console.log(extended.name);//

问题 I'm trying to type in proper way function that applys name of function and argumens for this function. After that apply it and return the result. Here the code: const sum = (a: number, b: number) => a + b const concat = (a: string, b: string, c: string) => a + b + c const funs = { sum, concat } type Keys = 'sum' | 'concat' type Args<T> = T extends (...args: infer R) => any ? R : never type Sum = Args<typeof sum> type Concat = Args<typeof concat> function apply<K extends Keys>(funKey: K, ..

问题 I'm trying to type in proper way function that applys name of function and argumens for this function. After that apply it and return the result. Here the code: const sum = (a: number, b: number) => a + b const concat = (a: string, b: string, c: string) => a + b + c const funs = { sum, concat } type Keys = 'sum' | 'concat' type Args<T> = T extends (...args: infer R) => any ? R : never type Sum = Args<typeof sum> type Concat = Args<typeof concat> function apply<K extends Keys>(funKey: K, ..

问题 I'm building a React app using TypeScript. I'm writing a higher order component and I would like to overload it's arguments. How can I do that? Let me give you what I've tried so that you can understand it better: const myHOC = ((value: string, anotherValue: number) | completeProps: SomeProps) => (WrappedComponent: ComponentClass) => // ... HOC code So it should be either: const myHOC = (value: string, anotherValue: number) => (WrappedComponent: ComponentClass) => // ... or const myHOC =

问题 I want to initialize a generic class variable inside a generic class using a generic type, but I can't figure out if there is a way to do this. Initializing with types works fine, it doesn't seem like it work with generics though. type EventCallback<I, O> = (event: I) => O; type ListenerList<K extends string | symbol | number, I, O, V extends EventCallback<I, O>> = { [T in K]: V[]; }; const test: ListenerList<string, string, string, (event: any) => any> = {}; // Works fine export default

问题 Is it in TypeScript somehow possible to define a type so that it only includes objects but not functions? Example: type T = { [name: string]: any } // How to modify this to only accepts objects??? const t: T = () => {} // <- This should not work after modification Thanks for your help. 回答1: There's no perfect way to refer to a type like "an object which is not a function", since that would require true subtraction types, and that doesn't exist in TypeScript (as of 3.1 at least). One easy

问题 The type should detect if the array has duplicate items and throw error in typescript? type UniqueArray = [ // How to implement this? ] const a:UniqueArray = [1, 2, 3] // success const b:UniqueArray = [1, 2, 2] // error PS: I'am currently removing duplicate items using JS, but, curious if this error can be captured using typescript type before hand? 回答1: The only possible way this could work at compile time is if your arrays are tuples composed of literals. For example, here are some arrays

问题 I'm writing a React Native application using TypeScript. I have a component EmotionsRater that accepts one of two types: Emotion or Need. It should also either accept a function of type rateNeed or rateEmotion . I combined these types to one called rateBoth using the | operator. And it passes this combined type down to another component called EmotionsRaterItem . The problem is that EmotionsRaterItem then claims: Cannot invoke an expression whose type lacks a call signature. Type 'rateBoth'