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Question

我们正在寻找使用Object.assign的类型安全方式。但是，我们似乎无法使其发挥作用。

要显示我们的问题，我将使用Generics文档中的copyFields方法

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"
}

我希望引擎阻止我创建未声明的属性

/*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

我们试图通过明确地为copyfields调用提供类型来解决这个问题但是我们找不到能使所有例子都有效的电话。

例如：做5个工作你可以这样称为copyFields：

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

但是对源类型的后续更改（例如删除＆＃34; b＆＃34;选项）现在将不再导致类型错误

有没有人知道如何使这项工作？

Answer 1

我能想到的最好的解决方法是定义第二个接口（我称之为SourceParts）与Source完全相同，除了所有成员都是可选的。

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

此处位于Typescript Playground。

Answer 2

打字稿2.1.4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 3

我之前已经完成了以下解决方案：

   /**
     * assign with known properties from target.
     *
     * @param target
     * @param source
     */
    public static safeAssignment(target: any,  source: any) {
        if (isNullOrUndefined(target) || isNullOrUndefined(source)) {
            return;
        }

        for (const att of Object.keys(target)) {
            target[att] = source.hasOwnProperty(att) ? source[att] : target[att];
        }
    }

我希望有人能有所帮助。问候

Answer 4

我有这个功能：

   /**
     * 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

。

安全输入Object.assign

4 个答案:

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