工具生产构建Node js

Question

我们在nodeJs应用程序中实现了swagger。到目前为止，我们是创建生产建设使用webpack，并删除控制器和服务文件。

bin/www.js

const YAML = require('yamljs');
const swaggerTools = require('swagger-tools');
const swaggerDoc = YAML.safeLoad('./swagger.yaml');

// swaggerRouter configuration
const swaggerOptions = {
  controllers: path.join(__dirname, '../public/javascripts/controllers'),
  useStubs: true, // Conditionally turn on stubs (mock mode)
};


// Initialize the Swagger middleware
swaggerTools.initializeMiddleware(swaggerDoc, (middleware) => {
  // Interpret Swagger resources and attach metadata to request - must be first in swagger-tools middleware chain
  app.use(middleware.swaggerMetadata());

  // validate the security using JWT token
  app.use(middleware.swaggerSecurity({
      Bearer: auth.verifyToken
  }));
  // Validate Swagger requests
  app.use(middleware.swaggerValidator({
    validateResponse: true
  }));

  // Route validated requests to appropriate controller
  app.use(middleware.swaggerRouter(swaggerOptions));

  // Serve the Swagger documents and Swagger UI
  app.use(middleware.swaggerUi());

});

如果我们在生产过程中做同样的事情，那么我们就会看到同样的方式来解决这个问题。生成后，我们删除公用文件夹。

Webpack码

const path = require('path');
const nodeExternals = require('webpack-node-externals');

module.exports = {
  entry: {
    server: './bin/www',
  },
  output: {
    path: path.join(__dirname, 'dist'),
    publicPath: '/',
    filename: 'server.build.js',
  },
  target: 'node',
  node: {
    // Need this when working with express, otherwise the build fails
    __dirname: false, // if you don't put this is, __dirname
    __filename: false, // and __filename return blank or /
  },
  externals: [nodeExternals()],
  module: {
    rules: [
      {
        // Transpiles ES6-8 into ES5
        test: /\.js$/,
        exclude: /node_modules/,
        use: {
          loader: 'babel-loader',
        },
      },
    ],
  },
};

请求帮助我们使用swagger中间件创建一个构建。

提前感谢

Answer 1

装腔作势的工具不是像webpack那样的包包。因此，您仍然需要向它提供控制器文件。因为您要从prod中删除/public，所以swagger工具中间件无法获得它需要的文件。在这种情况下，Webpack基本上是从代码中构建一个dist，这就是为什么删除控制器和服务是可以的。