How to build minified and uncompressed bundle with webpack?

Here's my webpack.config.js

var webpack = require("webpack");

module.exports = {

  entry: "./entry.js",
  devtool: "source-map",
  output: {
    path: "./dist",
    filename: "bundle.min.js"
  },
  plugins: [
    new webpack.optimize.UglifyJsPlugin({minimize: true})
  ]
};

I'm building with

$ webpack

In my dist folder, I'm only getting

• bundle.min.js
• bundle.min.js.map

I'd also like to see the uncompressed bundle.js

webpack.config.js:

const webpack = require("webpack");

module.exports = {
  entry: {
    "bundle": "./entry.js",
    "bundle.min": "./entry.js",
  },
  devtool: "source-map",
  output: {
    path: "./dist",
    filename: "[name].js"
  },
  plugins: [
    new webpack.optimize.UglifyJsPlugin({
      include: /\.min\.js$/,
      minimize: true
    })
  ]
};

Since Webpack 4, webpack.optimize.UglifyJsPlugin has been deprecated and its use results in error:

webpack.optimize.UglifyJsPlugin has been removed, please use config.optimization.minimize instead

As the manual explains, the plugin can be replaced with minimize option. Custom configuration can be provided to the plugin by specifying UglifyJsPlugin instance:

const webpack = require("webpack");
const UglifyJsPlugin = require('uglifyjs-webpack-plugin');

module.exports = {
  // ...
  optimization: {
    minimize: true,
    minimizer: [new UglifyJsPlugin({
      include: /\.min\.js$/
    })]
  }
};

This does the job for a simple setup. A more effective solution is to use Gulp together with Webpack and do the same thing in one pass.

You can use a single config file, and include the UglifyJS plugin conditionally using an environment variable:

const webpack = require('webpack');
const TerserPlugin = require('terser-webpack-plugin');

const PROD = JSON.parse(process.env.PROD_ENV || '0');

module.exports = {

  entry: './entry.js',
  devtool: 'source-map',
  output: {
    path: './dist',
    filename: PROD ? 'bundle.min.js' : 'bundle.js'
  },
  optimization: {
    minimize: PROD,
    minimizer: [
      new TerserPlugin({ parallel: true })
  ]
};

and then just set this variable when you want to minify it:

$ PROD_ENV=1 webpack

Edit:

As mentioned in the comments, NODE_ENV is generally used (by convention) to state whether a particular environment is a production or a development environment. To check it, you can also set const PROD = (process.env.NODE_ENV === 'production'), and continue normally.

You can run webpack twice with different arguments:

$ webpack --minimize

then check command line arguments in webpack.config.js:

var path = require('path'),
  webpack = require('webpack'),
  minimize = process.argv.indexOf('--minimize') !== -1,
  plugins = [];

if (minimize) {
  plugins.push(new webpack.optimize.UglifyJsPlugin());
}

...

example webpack.config.js

To add another answer, the flag -p (short for --optimize-minimize) will enable the UglifyJS with default arguments.

You won't get a minified and raw bundle out of a single run or generate differently named bundles so the -p flag may not meet your use case.

Conversely the -d option is short for --debug --devtool sourcemap --output-pathinfo

My webpack.config.js omits devtool, debug, pathinfo, and the minmize plugin in favor of these two flags.

Maybe i am late here, but i have the same issue, so i wrote a unminified-webpack-plugin for this purpose.

Installation

npm install --save-dev unminified-webpack-plugin

Usage

var path = require('path');
var webpack = require('webpack');
var UnminifiedWebpackPlugin = require('unminified-webpack-plugin');

module.exports = {
    entry: {
        index: './src/index.js'
    },
    output: {
        path: path.resolve(__dirname, 'dist'),
        filename: 'library.min.js'
    },
    plugins: [
        new webpack.optimize.UglifyJsPlugin({
            compress: {
                warnings: false
            }
        }),
        new UnminifiedWebpackPlugin()
    ]
};

By doing as above, you will get two files library.min.js and library.js. No need execute webpack twice, it just works!^^