javascript - 当使用所有文件都需要的模块时，如何将代码拆分到多个文件中？

Question

在本例中，我使用的是 puppeteer chrome headless 浏览器和 MongoDB。这是我的代码:

var browser = await puppeteer.launch()
var page = await browser.newPage()
var db = await MongoClient.connect("mongodb://localhost:27017/bot")
var scraped_users = db.collection("scraped_users")

我想将我的代码拆分成相关文件，如 chrome.js、twitter.js、database,js、utilities .js 等。问题是我几乎在每个文件中都需要上面声明的那 4 个变量，在每个函数中传递它们似乎很愚蠢。我也不能在每个文件中重新声明它们，因为那样会打开一个新的 chrome 浏览器并建立一个新的数据库连接。

Answer 1

假设您有一个文件 init.js:

/*let isInitialized, browser, page, db, scraped_users

async function init() {
    if (!isInitialized) {
    browser = await puppeteer.launch()
    page = await browser.newPage()
    db = await MongoClient.connect("mongodb://localhost:27017/bot")
    scraped_users = db.collection("scraped_users")
    isInitialized = true
    }
    return { browser, page, db, scraped_users }
}

module.exports = init*/

let common = {}; //store some global's.

//this should only be called once, eg. when App starts
common.init = async function () {
    this.browser = await puppeteer.launch()
    this.page = await this.browser.newPage()
    this.db = await MongoClient.connect("mongodb://localhost:27017/bot")
    this.scraped_users = this.db.collection("scraped_users")
}

module.exports = common;

然后像这样使用它:

/*const init = require('./init.js')

init().then(({browser, page, db, scraped_users}) => {
    console.log(browser, page, db, scraped_users)
})*/

const common = require('./common');
const doSomething = require('./do-something');

async function run() {
    await common.init(); //this only needs to be done once
    console.log(common.browser); //etc
    await doSomething();
}

//now inside other units, eg. do-something.js
const common = require('./common');
async function doSomething() {
    console.log(common.browser); 
}