假设我有这个:
const color = {
red: null,
green: null,
blue: null
};
const newColor = ['red', 'green', 'blue'].filter(e => color[e]);
底部附近的误差为color[e]，其中:
元素隐式具有"any"类型，因为表达式的类型为
我试着在TypeScript文档上到处找，但我怎么能把这个值设为interface，这样它就可以接受color[e]呢？
你遇到的问题不是color是错误的类型，而是TypeScript推断['red', 'green', 'blue']的类型是string[].通常，这种类型的推理是可取的，因为(据编译器所知)您可能希望将'purple'推到它上面.但在本例中，您希望编译器知道，唯一的成员是三个字符串文本'red'、'green'和'blue'.也就是说，您需要一个至少与Array<'red'|'green'|'blue'>一样具体的类型.
Assuming you're using TS3.4 or later, the easiest way to get this kind of type inference from the compiler is to use a const assertion:
const constAssertionTest = ["red", "green", "blue"] as const;
// const constAssertionTest: readonly ["red", "green", "blue"];
The as const causes the compiler to infer a tuple composed of exactly the three string literals in the array, in the exact order you've set. (It's even a read-only tuple). That is good enough to fix your error:
const newColor = (['red', 'green', 'blue'] as const).filter(e => color[e]); // okay
好吧，希望有帮助.祝你好运