Typescript：有没有办法检查变量是否是具有嵌套属性的接口定义的对象？

Question

For example:

interface TEST_OBJECT {
    "test1" : TEST_TYPE;
    "test2" : TEST_INTERFACE
}

type TEST_TYPE = string|number;

interface TEST_INTERFACE {
    "test3" : boolean;
}

Is there a way to check if a variable is a TEST_OBJECT?

Answer 1

If you're looking for TypeScript to automatically provide runtime type guards for you, that's not possible. In practice though, the compiler is pretty good at narrowing types for you, depending on how much you know about the object you're inspecting. For example, say you have something you know is either a TEST_OBJECT or a string :

function hasKey<K extends string>(k: K, x: any): x is Record<K, {}> {
  return k in x;
}

declare const obj: TEST_OBJECT | string;

if (hasKey("test2", obj)) {
  // it knows that obj is a TEST_OBJECT now
  console.log(obj.test1);
  console.log(obj.test2.test3 === false); 
} else {
  // it knows that obj is a string
  console.log(obj.charAt(0));
}

If you have no idea what the object you're inspecting might be, then, as @JoshCrozier said, User-Defined Type Guards are probably the way to go. It's possible to build these up the same way you've built up your interfaces and types:

function hasKey<K extends string>(k: K, x: any): x is Record<K, {}> {
  return k in x;
}

function isTEST_INTERFACE(x: any): x is TEST_INTERFACE {
  return hasKey("test3",x) && (typeof x.test3 === 'boolean');
}

function isTEST_TYPE(x: any): x is TEST_TYPE {
  return (typeof x === 'string') || (typeof x === 'number');
}

function isTEST_OBJECT(x: any): x is TEST_OBJECT {
  return hasKey("test1", x) && isTEST_TYPE(x.test1) &&
    hasKey("test2", x) && isTEST_INTERFACE(x.test2);
}

// use it:
declare const obj: TEST_OBJECT | string
if (isTEST_OBJECT(obj)) {
  console.log(obj.test1);
  console.log(obj.test2.test3 === false); 
} else {
  console.log(obj.charAt(0));
}

Hope that helps.