Object.assign的安全輸入

Question

我們正在尋找一種使用Object.assign的類型安全方法。但是，我們似乎無法使其工作。

爲了表示我們的問題，我將使用copyFields方法從Generics文檔

function copyFields<T extends U, U>(target: T, source: U): T { 
    for (let id in source) { 
     target[id] = source[id]; 
    } 
    return target; 
} 

function makesrc(): Source { return {b: 1, c: "a"}} 

interface Source { 
    a?: "a"|"b", 
    b: number, 
    c: "a" | "b" 
} 

我想要的發動機，以防止我從創建未申報的性質

/*1*/copyFields(makesrc(), {d: "d"}); //gives an error 
/*2*/copyFields(makesrc(), {a: "d"}); //gives an error 
/*3*/copyFields(makesrc(), {c: "d"}); //should give an error, but doesn't because "a"|"b" is a valid subtype of string. 

//I don't want to specify all the source properties 
/*4*/copyFields(makesrc(), {b: 2}); //will not give me an error 
/*5*/copyFields(makesrc(), {a: "b"}); //should not give an error, but does because string? is not a valid subtype of string 

我們已經試圖解決這個明確提供複製字段的類型請致電 ，但我們無法找到將使所有示例都能正常工作的調用。

例如： 使5個工作你可以稱之爲copyFields這樣的：

/*5'*/copyFields<Source,{a?:"a"|"b"}>(makesrc(), {a: "b"}); 

但後續的源類型的改變（例如去掉了「B」選項）現在將不再導致鍵入錯誤

有沒有人知道一種方法使這項工作？

Answer 1

打字稿2.1.4至救援！

Playground link

interface Data { 
    a?: "a"|"b", 
    b: number, 
    c: "a" | "b" 
} 

function copyFields<T>(target: T, source: Readonly<Partial<T>>): T { 
    for (let id in source) { 
     target[id] = source[id]; 
    } 
    return target; 
} 

function makesrc(): Data { return {b: 1, c: "a"}} 

/*1*/copyFields(makesrc(), {d: "d"}); //gives an error 
/*2*/copyFields(makesrc(), {a: "d"}); //gives an error 
/*3*/copyFields(makesrc(), {c: "d"}); //gives an error 

//I don't want to specify all the source properties 
/*4*/copyFields(makesrc(), {b: 2}); //will not give me an error 
/*5*/copyFields(makesrc(), {a: "b"}); //will not give me an error 

Answer 2

我能想到的最佳解決方法是定義第二個接口（我稱之爲SourceParts），它與Source完全相同，但所有成員都是可選的。

function copyFields<T extends U, U>(target: T, source: U): T { 
    for (let id in source) { 
     target[id] = source[id]; 
    } 
    return target; 
} 

function makesrc(): Source { return {b: 1, c: "a"}} 

interface Source { 
    a?: "a"|"b", 
    b: number, 
    c: "a" | "b" 
} 

interface SourceParts { 
    a?: "a"|"b", 
    b?: number, 
    c?: "a" | "b" 
} 

/*1*/copyFields<Source, SourceParts>(makesrc(), {d: "d"}); //gives an error 
/*2*/copyFields<Source, SourceParts>(makesrc(), {a: "d"}); //gives an error 
/*3*/copyFields<Source, SourceParts>(makesrc(), {c: "d"}); //gives an error 

//I don't want to specify all the source properties 
/*4*/copyFields<Source, SourceParts>(makesrc(), {b: 2}); //will not give me an error 
/*5*/copyFields<Source, SourceParts>(makesrc(), {a: "b"}); //will not give me an error 

這裏是在Typescript Playground。

Answer 3

我有這樣的功能：

/** 
    * Take every field of fields and put them override them in the complete object 
    * NOTE: this API is a bit reverse of extend because of the way generic constraints work in TypeScript 
    */ 
    const updateFields = <T>(fields: T) => <U extends T>(complete: U): U => { 
     let result = <U>{}; 
     for (let id in complete) { 
      result[id] = complete[id]; 
     } 
     for (let id in fields) { 
      result[id] = fields[id]; 
     } 
     return result; 
    } 

用法：

updateFields({a:456})({a:123,b:123}) // okay 
updateFields({a:456})({b:123}) // Error 

。

我在不同的上下文中提到過此功能

更多：https://stackoverflow.com/a/32490644/390330

PS：事情會變得更好的JavaScript一次獲得此上演3：https://github.com/Microsoft/TypeScript/issues/2103