如何使TS编译器知道未知变量具有某些属性（不使用“ any”类型）

Question

function f(): unknown {
    return {abc: "ABC"};
}

const a = f();

if (a && a instanceof Object && a.hasOwnProperty("abc")) {
    console.log(a.abc);
}

I have a variable a which is the unknown type at first (>= TypeScript 3.0). I want to use the abc property of a inside an if statement when a has the property.

However, as the above VS code screenshot shows, the TypeScript compiler gives the TS2339 error even though I checked a.hasOwnProperty("abc") in the if condition.

Property 'abc' does not exist on type 'object'. ts(2339)

I know that I can bypass this error by casting a to any type, but it will make the compiler ignore the typos of members' names, and it will be bad if there are many properties in a and many things to do with the variable a .

Is there a way for the TypeScript compiler to know that a has certain properties?

Answer 1

You can use { [index: string]: unknown } instead of unknown to tell TypeScript your object is a dictionary of unknown properties.

declare const unfamiliar: { [index: string]: unknown };

if ('foo' in unfamiliar) {
  unfamiliar.foo;
}

Answer 2

The error comes from the assertion that a is an Object . The Object type does not contain abc so the following assertion does not make a difference.
If you only keep the a.hasOwnProperty() assertion it should work.