[英]Export class as Node.js module in TypeScript
我熟悉 TypeScript 中的export
關鍵字,以及兩種使用 TypeScript 從 Node 模塊導出事物的規范方法(當然,也可以使用 TypeScript 模塊,但它們與我正在尋找的更遠):
export class ClassName { }
和一系列
export function functionName () { }
然而,我通常編寫模塊的方式是:
var ClassName = function () { };
ClassName.prototype.functionName = function () { };
module.exports = ClassName;
有沒有辦法使用 TypeScript 導出語法來做到這一點?
您可以在 TypeScript 0.9.0 中非常簡單地做到這一點:
class ClassName {
functionName () { }
}
export = ClassName;
以下是我使用 TypeScript 導出 CommonJS (Node.js) 模塊的方法:
src/ts/user/User.ts
export default class User {
constructor(private name: string = 'John Doe',
private age: number = 99) {
}
}
src/ts/index.ts
import User from "./user/User";
export = {
user: {
User: User,
}
}
配置文件
{
"compilerOptions": {
"declaration": true,
"lib": ["ES6"],
"module": "CommonJS",
"moduleResolution": "node",
"noEmitOnError": true,
"noImplicitAny": true,
"noImplicitReturns": true,
"outDir": "dist/commonjs",
"removeComments": true,
"rootDir": "src/ts",
"sourceMap": true,
"target": "ES6"
},
"exclude": [
"bower_components",
"dist/commonjs",
"node_modules"
]
}
dist/commonjs/index.js(編譯模塊入口點)
"use strict";
const User_1 = require("./user/User");
module.exports = {
user: {
User: User_1.default,
}
};
//# sourceMappingURL=index.js.map
dist/commonjs/user/User.js(編譯的用戶類)
"use strict";
Object.defineProperty(exports, "__esModule", { value: true });
class User {
constructor(name = 'John Doe', age = 72) {
this.name = name;
this.age = age;
}
}
exports.default = User;
//# sourceMappingURL=User.js.map
測試代碼(test.js)
const MyModule = require('./dist/commonjs/index');
const homer = new MyModule.user.User('Homer Simpson', 61);
console.log(`${homer.name} is ${homer.age} years old.`); // "Homer Simpson is 61 years old."
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