我知道这个问题有点老了，但这只是我的五毛钱。这招对我很管用:

const container: Container = icc.controlComponent as unknown as Container;
if (container.getControlComponents) {
    this.allControlComponents.push(...container.getControlComponents());
}

容器是接口，而icc。controlComponent是我想检查的对象，getControlComponents是来自Container接口的一个方法。

你可以在运行时使用ts-validate-type验证TypeScript类型，就像这样(确实需要一个Babel插件):

const user = validateType<{ name: string }>(data);

因为在运行时类型是未知的，所以我写了如下代码来比较未知对象，不是与类型进行比较，而是与已知类型的对象进行比较:

创建正确类型的示例对象 指定它的哪些元素是可选的 将你的未知对象与这个样本对象进行深度比较

下面是我用于深度比较的(接口不可知)代码:

function assertTypeT<T>(loaded: any, wanted: T, optional?: Set<string>): T {
  // this is called recursively to compare each element
  function assertType(found: any, wanted: any, keyNames?: string): void {
    if (typeof wanted !== typeof found) {
      throw new Error(`assertType expected ${typeof wanted} but found ${typeof found}`);
    }
    switch (typeof wanted) {
      case "boolean":
      case "number":
      case "string":
        return; // primitive value type -- done checking
      case "object":
        break; // more to check
      case "undefined":
      case "symbol":
      case "function":
      default:
        throw new Error(`assertType does not support ${typeof wanted}`);
    }
    if (Array.isArray(wanted)) {
      if (!Array.isArray(found)) {
        throw new Error(`assertType expected an array but found ${found}`);
      }
      if (wanted.length === 1) {
        // assume we want a homogenous array with all elements the same type
        for (const element of found) {
          assertType(element, wanted[0]);
        }
      } else {
        // assume we want a tuple
        if (found.length !== wanted.length) {
          throw new Error(
            `assertType expected tuple length ${wanted.length} found ${found.length}`);
        }
        for (let i = 0; i < wanted.length; ++i) {
          assertType(found[i], wanted[i]);
        }
      }
      return;
    }
    for (const key in wanted) {
      const expectedKey = keyNames ? keyNames + "." + key : key;
      if (typeof found[key] === 'undefined') {
        if (!optional || !optional.has(expectedKey)) {
          throw new Error(`assertType expected key ${expectedKey}`);
        }
      } else {
        assertType(found[key], wanted[key], expectedKey);
      }
    }
  }

  assertType(loaded, wanted);
  return loaded as T;
}

下面是我如何使用它的一个例子。

在本例中，我期望JSON包含一个元组数组，其中第二个元素是一个名为User的接口实例(它有两个可选元素)。

TypeScript的类型检查将确保我的示例对象是正确的，然后assertTypeT函数检查未知(从JSON加载的)对象是否与示例对象匹配。

export function loadUsers(): Map<number, User> {
  const found = require("./users.json");
  const sample: [number, User] = [
    49942,
    {
      "name": "ChrisW",
      "email": "example@example.com",
      "gravatarHash": "75bfdecf63c3495489123fe9c0b833e1",
      "profile": {
        "location": "Normandy",
        "aboutMe": "I wrote this!\n\nFurther details are to be supplied ..."
      },
      "favourites": []
    }
  ];
  const optional: Set<string> = new Set<string>(["profile.aboutMe", "profile.location"]);
  const loaded: [number, User][] = assertTypeT(found, [sample], optional);
  return new Map<number, User>(loaded);
}

您可以在用户定义的类型保护的实现中调用这样的检查。

答案很简单。然而，这种解决方案至少在大约3/4的情况下是可能的(尽管并不总是理想的)。所以，换句话说，这可能与阅读这篇文章的人有关。

假设我有一个非常简单的函数，需要知道参数的接口类型:

const simpleFunction = (canBeTwoInterfaces: interfaceA | interface B) => { 
  // if interfaceA, then return canBeTwoInterfaces.A
  // if interfaceB, then return canBeTwoInterfaces.B
}

得到最多赞的答案往往是使用“功能检查”。也就是说,

const simpleFunction = (canBeTwoInterfaces: interfaceA | interface B) => { 
  if (canBeTwoInterfaces.onlyExistsOnInterfaceA) return canBeTwoInterfaces.A
  else return canBeTwoInterfaces.B
}

然而，在我正在使用的代码库中，我需要检查的接口主要包含可选参数。另外，我团队里的其他人可能会在我不知情的情况下突然改名字。如果这听起来像您正在使用的代码库，那么下面的函数要安全得多。

就像我之前说的，这对很多人来说可能是一件非常明显的事情。尽管如此，要知道何时何地应用给定的解决方案并不明显，不管它是否恰好像下面这样非常简单。

这就是我要做的:

const simpleFunction = (
  canBeTwoInterfaces: interfaceA | interface B,
  whichInterfaceIsIt: 'interfaceA' | 'interfaceB'
) => { 
  if (whichInterfaceIsIt === 'interfaceA') return canBeTwoInterface.A
  else return canBeTwoInterfaces.B
}

TypeGuards

interface MyInterfaced {
    x: number
}

function isMyInterfaced(arg: any): arg is MyInterfaced {
    return arg.x !== undefined;
}

if (isMyInterfaced(obj)) {
    (obj as MyInterfaced ).x;
}

与上面使用用户定义守卫的情况相同，但这次使用的是箭头函数谓词

interface A {
  member:string;
}

const check = (p: any): p is A => p.hasOwnProperty('member');

var foo: any = { member: "foobar" };
if (check(foo))
    alert(foo.member);