javascript - 无法使用传递给 "npm start"的参数配置 React Pre Hook

Question

我创建了一个全新的项目:

$ npx create-react-app test-react-app && cd test-react-app && npm start

这里是仓库:https://github.com/tlg-265/test-react-app

我创建了一个新脚本:/prepare-project.js，其中包含以下内容:

let project = 'ferrari' // replace this value with the one passed via: $ npm start [what to put here?]

console.log(`########################`);
console.log(`###### The current project is: ${project} ######`);
console.log(`########################`);

文件:/package.json的内容是:

{
  "name": "test-react-app",
  "version": "0.1.0",
  "private": true,
  "dependencies": {
    "@testing-library/jest-dom": "^4.2.4",
    "@testing-library/react": "^9.3.2",
    "@testing-library/user-event": "^7.1.2",
    "react": "^16.12.0",
    "react-dom": "^16.12.0",
    "react-scripts": "3.3.1"
  },
  "scripts": {
    "start": "react-scripts start",
    "build": "react-scripts build",
    "test": "react-scripts test",
    "eject": "react-scripts eject",
    "prestart": "node prepare-project.js"
  },
  "eslintConfig": {
    "extends": "react-app"
  },
  "browserslist": {
    "production": [
      ">0.2%",
      "not dead",
      "not op_mini all"
    ],
    "development": [
      "last 1 chrome version",
      "last 1 firefox version",
      "last 1 safari version"
    ]
  }
}

您可以在其中看到我配置了预 Hook :

"prestart": "node prepare-project.js"

我的目标是运行:

$ npm start [what to put here?]

所以在脚本中:/prepare-project.js我可以读取该值。

我尝试过:

$ node prepare-project.js --project=mclaren

尝试从脚本内部读取该值时没有运气:prepare-project.js。

知道如何实现这一目标吗？

谢谢!

Answer 1

您可以使用yargs模块来解析命令行参数，然后您可以使用child_process来执行您的命令。您的代码应如下所示，您可以添加自己的自定义逻辑。

const yargs = require('yargs');
const { execSync } = require('child_process');

const argv = yargs
    .option('project', {
        alias: 'p',
        description: 'name of the project',
        type: 'string',
    })
    .help()
    .alias('help', 'h')
    .argv;

console.log(argv.project);
console.log(`npm start --project=${argv.project}`);
const output = execSync(`npm start --project=${argv.project}`);
console.log(output)