这个问题直接类似于TypeScript中的类类型检查
我需要在运行时找出任何类型的变量是否实现了接口。这是我的代码:
interface A{
member:string;
}
var a:any={member:"foobar"};
if(a instanceof A) alert(a.member);
如果您在typescript游乐场中输入这段代码，最后一行将被标记为错误，“名称A不存在于当前作用域”。但事实并非如此，该名称确实存在于当前作用域中。我甚至可以更改变量声明为var a: a ={成员:"foobar"};没有编辑的抱怨。在浏览网页并找到其他问题后，我将接口更改为类，但我不能使用对象字面量来创建实例。
我想知道A类型是如何消失的，但看看生成的javascript就能解释这个问题:
var a = {
member: "foobar"
};
if(a instanceof A) {
alert(a.member);
}
没有将A表示为接口，因此不可能进行运行时类型检查。
我知道javascript作为一种动态语言没有接口的概念。是否有方法对接口进行类型检查?
typescript游乐场的自动完成显示typescript甚至提供了一个方法实现。我怎么使用它?
export interface ConfSteps {
group: string;
key: string;
steps: string[];
}
private verify(): void {
const obj = `{
"group": "group",
"key": "key",
"steps": [],
"stepsPlus": []
} `;
if (this.implementsObject<ConfSteps>(obj, ['group', 'key', 'steps'])) {
console.log(`Implements ConfSteps: ${obj}`);
}
}
private objProperties: Array<string> = [];
private implementsObject<T>(obj: any, keys: (keyof T)[]): boolean {
JSON.parse(JSON.stringify(obj), (key, value) => {
this.objProperties.push(key);
});
for (const key of keys) {
if (!this.objProperties.includes(key.toString())) {
return false;
}
}
this.objProperties = null;
return true;
}
下面是我使用类和lodash想出的解决方案:(它有效!)
// TypeChecks.ts
import _ from 'lodash';
export class BakedChecker {
private map: Map<string, string>;
public constructor(keys: string[], types: string[]) {
this.map = new Map<string, string>(keys.map((k, i) => {
return [k, types[i]];
}));
if (this.map.has('__optional'))
this.map.delete('__optional');
}
getBakedKeys() : string[] {
return Array.from(this.map.keys());
}
getBakedType(key: string) : string {
return this.map.has(key) ? this.map.get(key) : "notfound";
}
}
export interface ICheckerTemplate {
__optional?: any;
[propName: string]: any;
}
export function bakeChecker(template : ICheckerTemplate) : BakedChecker {
let keys = _.keysIn(template);
if ('__optional' in template) {
keys = keys.concat(_.keysIn(template.__optional).map(k => '?' + k));
}
return new BakedChecker(keys, keys.map(k => {
const path = k.startsWith('?') ? '__optional.' + k.substr(1) : k;
const val = _.get(template, path);
if (typeof val === 'object') return val;
return typeof val;
}));
}
export default function checkType<T>(obj: any, template: BakedChecker) : obj is T {
const o_keys = _.keysIn(obj);
const t_keys = _.difference(template.getBakedKeys(), ['__optional']);
return t_keys.every(tk => {
if (tk.startsWith('?')) {
const ak = tk.substr(1);
if (o_keys.includes(ak)) {
const tt = template.getBakedType(tk);
if (typeof tt === 'string')
return typeof _.get(obj, ak) === tt;
else {
return checkType<any>(_.get(obj, ak), tt);
}
}
return true;
}
else {
if (o_keys.includes(tk)) {
const tt = template.getBakedType(tk);
if (typeof tt === 'string')
return typeof _.get(obj, tk) === tt;
else {
return checkType<any>(_.get(obj, tk), tt);
}
}
return false;
}
});
}
自定义类:
// MyClasses.ts
import checkType, { bakeChecker } from './TypeChecks';
class Foo {
a?: string;
b: boolean;
c: number;
public static _checker = bakeChecker({
__optional: {
a: ""
},
b: false,
c: 0
});
}
class Bar {
my_string?: string;
another_string: string;
foo?: Foo;
public static _checker = bakeChecker({
__optional: {
my_string: "",
foo: Foo._checker
},
another_string: ""
});
}
在运行时检查类型:
if (checkType<Bar>(foreign_object, Bar._checker)) { ... }
我知道我偶然发现了一个github包，它正确地解决了这个问题，在我的搜索历史中，我终于找到了它。检查typescript-is -尽管它要求你的代码使用ttypescript编译(我目前正在强迫它与create-react-app一起工作，稍后将在成功/失败时更新)，你可以用它做各种疯狂的事情。与ts-validate-type不同，这个包也是主动维护的。
你可以检查某个东西是否是字符串或数字，并将其作为字符串或数字使用，而编译器不会抱怨:
import { is } from 'typescript-is';
const wildString: any = 'a string, but nobody knows at compile time, because it is cast to `any`';
if (is<string>(wildString)) { // returns true
// wildString can be used as string!
} else {
// never gets to this branch
}
if (is<number>(wildString)) { // returns false
// never gets to this branch
} else {
// Now you know that wildString is not a number!
}
你也可以检查你自己的接口:
import { is } from 'typescript-is';
interface MyInterface {
someObject: string;
without: string;
}
const foreignObject: any = { someObject: 'obtained from the wild', without: 'type safety' };
if (is<MyInterface>(foreignObject)) { // returns true
const someObject = foreignObject.someObject; // type: string
const without = foreignObject.without; // type: string
}
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。
使用字符串文字是很困难的，因为如果你想重构你的方法或接口名称，那么你的IDE可能不重构这些字符串文字。
我为您提供我的解决方案，如果在接口中至少有一个方法
export class SomeObject implements interfaceA {
public methodFromA() {}
}
export interface interfaceA {
methodFromA();
}
检查object是否为interface类型:
const obj = new SomeObject();
const objAsAny = obj as any;
const objAsInterfaceA = objAsAny as interfaceA;
const isObjOfTypeInterfaceA = objAsInterfaceA.methodFromA != null;
console.log(isObjOfTypeInterfaceA)
注意:即使我们删除了'implements interfaceA'，我们也会得到true，因为SomeObject类中仍然存在该方法
