Javascript 是否有一种方法可以从全局级别的网页包中排除节点_模块？

我有一个用于演示的简单配置，可以从js和jsx文件中排除节点模块，但找不到全局排除节点模块的解决方案，所以我不需要在每个加载程序中始终指定此排除

const path = require("path");
const HtmlWebpackPlugin = require("html-webpack-plugin");
const CleanWebpackPlugin = require("clean-webpack-plugin");
const ExtractTextPlugin = require("extract-text-webpack-plugin");

module.exports = {
  entry: {
    app: path.resolve(__dirname, "../src/index.js")
  },
  output: {
    path: path.resolve(__dirname, "../dist"),
    filename: "[name].[chunkhash].js",
    publicPath: "/",
    chunkFilename: "[name].[chunkhash].js"
  },
  module: {
    rules: [
      { test: /\.js$/, loaders: ["babel-loader"], exclude: /node_modules/ },
      { test: /\.jsx$/, loaders: ["babel-loader"], exclude: /node_modules/ },
      {
        test: /\.css$/,
        use: ExtractTextPlugin.extract({
          fallback: "style-loader",
          use: "css-loader"
        })
      }
    ]
  },
  plugins: [
    new CleanWebpackPlugin([path.resolve(__dirname, "../dist")]),
    new HtmlWebpackPlugin({ template: path.resolve(__dirname, "../public/index.html") }),
    new ExtractTextPlugin("styles.css")
  ]
};

有没有办法做到这一点？我在网页3上，你可以使用

{ test: /\.jsx?$/, loaders: ["babel-loader"], exclude: /node_modules/ },