Swagger Tools生产构建节点js

问题描述

我们在nodeJs应用程序中实现了大摇大摆。到目前为止，我们已经使用webpack创建了生产版本，并删除了控制器和服务文件。

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());

});

如果我们在生产构建中做过同样的事情，并且庞大的中间件期望解决的路径相同。构建完成后，我们将删除公用文件夹。

Webpack代码

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',],};

请帮助我们使用敏捷的中间件创建构建

预先感谢

解决方法

Swagger工具不是像webpack这样的打包程序。因此，您仍然需要为其提供控制器文件。由于您要从产品中删除/ public，因此摇摇欲坠的工具中间件无法获取所需的文件。在这种情况下，Webpack基本上是在与您的代码建立距离，这就是为什么可以删除控制器和服务的原因。

deployment deployment swagger-tools webpack-production