[英]Typescript: type guard for whether an object property is defined, when the key is a wide type?
I have a function that returns whether an object property is undefined
.我有一个返回对象属性是否为undefined
的函数。 I need this function instead of just doing obj[key] === undefined
because otherwise I'd sometimes get Property 'foo' does not exist on type 'Bar'.
我需要这个函数而不是仅仅做obj[key] === undefined
因为否则我有时会得到Property 'foo' does not exist on type 'Bar'.
. . It's straightforward to write the type when the property key is a literal.当属性键是文字时,编写类型很简单。 Ie: IE:
function hasDefinedProp<
Obj extends Partial<Record<string, any>>,
Prop extends string,
>(
obj: Obj,
prop: Prop,
): obj is Obj & Record<Prop, Prop extends keyof Obj ? Exclude<Obj[Prop], undefined> : unknown> {
return obj[prop] !== undefined;
}
const obj: Partial<Record<string, number>> = {};
if (hasDefinedProp(obj, 'foo')) {
obj.foo + 1; // No error.
obj.bar + 1; // "Object is possibly 'undefined'."
}
However, this doesn't work when the key's type is a wide type, ie:但是,当键的类型是宽类型时,这不起作用,即:
const obj: Partial<Record<string, number>> = {};
const key: string = '';
if (hasDefinedProp(obj, key)) {
obj[key] + 1; // No error.
obj.bar + 1; // No error. Should be "Object is possibly 'undefined'."
}
Is it possible to make the type guard work for wide types?是否可以使类型保护适用于宽类型?
AFAIK, it is not possible. AFAIK,这是不可能的。 Once you have added explicit string<\/code> type to
const key: string = '';<\/code>
将显式
string<\/code>类型添加到
const key: string = '';<\/code>
- TS is unable to narrow literal type of
key<\/code> .
-TS 无法缩小
key<\/code>的文字类型。
As a result, you are allowed to use any string you want to access a property.因此,您可以使用任何想要访问属性的字符串。 TS is unable to distinguish two types with type
string<\/code> in this example:
在这个例子中,TS 无法区分两种类型为
string<\/code>的类型:
const obj: Partial<Record<string, number>> = {};
const key: string = '';
if (hasDefinedProp(obj, key)) {
obj[key] + 1; // No error.
obj.bar + 1; // No error. Should be "Object is possibly 'undefined'."
}
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