带有扩展接口的参数的 Typescript 通用回调，TS2345

Question

我如何创建一个接受参数的回调，其中参数可以是任何扩展某些接口的东西，但除此之外我不知道确切的类型？

例如

const fnct = function( data: DataAny, callback: GenericCallback<DataAny> ): void {
    callback( data );
};
// DataAny here should be like "argument type that can be anything but has to extend DataAny"

例子

更具体地说，我怎样才能避免这里最后两行的打字错误
（从下往上阅读可能最容易）:


// -- unchangeable interface, comes from external library --
interface DataAny {
    type: string;
    [more: string]: any;
}

// -- my interfaces / types --
interface DataBook extends DataAny { pages: number }
interface DataWater extends DataAny { liters: number }

type GetQuantity<D> = ( data: D ) => void; // callback function type

// -- generic function --
const genericQuantityInfo = function( data: DataAny, f: GetQuantity<DataAny> ): void {
    console.log('quantity: ', f( data ));
};

// -- data --
const book: DataBook  = { type: 'book',  pages: 303 };
const water: DataWater = { type: 'water', liters: 12 };

// -- callbacks --
const getQuantityBook: GetQuantity<DataBook>  = function( data: DataBook ){ return data.pages + ' pages'; };
const getQuantityWater: GetQuantity<DataWater> = function( data: DataWater ){ return data.liters + ' liters'; };

// -- calls --
genericQuantityInfo( book,  getQuantityBook ); // <-- error TS2345: 'GetQuantity<DataBook>' is not assignable to 'GetQuantity<DataAny>'. 'pages' is missing in 'DataAny' but required in 'DataBook'
genericQuantityInfo( water, getQuantityWater ); // <-- error (similar to above)

不想要

请注意，我可以不传递特定类型，例如 genericQuantityInfo<DataBook> 此处（有效）：

const genericQuantityInfo = function<T>( data: T, f: GetQuantity<T> ): void { console.log('quantity: ', f( data )); };
genericQuantityInfo<DataBook>( book,  getQuantityBook );
genericQuantityInfo<DataWater>( water, getQuantityWater );

我不能这样做的原因是我的用例实际上更复杂，使用创建者函数，我不想传递所有可能的类型<DataBook | DataWater | ...（我也无法让它工作）。喜欢：

const creator = function<T>(){
    return function( data: T, f: GetQuantity<T> ): void {
        console.log('quantity: ', f( data ));
    };
};
creator<DataBook | DataWater /* | ... */ >()(book, getQuantityBook); // <-- still error anyway

但是由于我在这里进行了很多简化，所以也欢迎任何关于我误解的地方的想法。谢谢。

Answer 1

我想我明白了：

与其键入 DataAny，不如尝试键入 T extends DataAny，这意味着任何具有 DataAny 道具的东西。

interface DataAny {
    type: string;
    [more: string]: any;
}

// -- my interfaces / types --
interface DataBook extends DataAny { pages: number }
interface DataWater extends DataAny { liters: number }

type GetQuantity<D extends DataAny> = ( data: D ) => void; // callback function type

// -- generic function --
const genericQuantityInfo = function<T extends DataAny>( data: T, f: GetQuantity<T> ): void {
    console.log('quantity: ', f( data ));
};

// -- data --
const book: DataBook  = { type: 'book',  pages: 303 };
const water: DataWater = { type: 'water', liters: 12 };

// -- callbacks --
const getQuantityBook: GetQuantity<DataBook>  = function( data: DataBook ){ return data.pages + ' pages'; };
const getQuantityWater: GetQuantity<DataWater> = function( data: DataWater ){ return data.liters + ' liters'; };

// -- calls --
genericQuantityInfo( book,  getQuantityBook ); // OK
genericQuantityInfo( water, getQuantityWater ); // OK

playground

带有扩展接口的参数的 Typescript 通用回调，TS2345

Typescript generic callback with arguments that extend an interface, TS2345

arguments

interface

callback

extend

typescript-generics

例子

不想要