Conditional build based on environment using Webpack

Dominic picture Dominic · Feb 17, 2015 · Viewed 72.1k times · Source

I have some things for development - e.g mocks which I would like to not bloat my distributed build file with.

In RequireJS you can pass a config in a plugin file and conditonally require things in based on that.

For webpack there doesn't seem to be a way of doing this. Firstly to create a runtime config for an environment I have used resolve.alias to repoint a require depending on the environment, e.g:

// All settings.
var all = {
    fish: 'salmon'
};

// `envsettings` is an alias resolved at build time.
module.exports = Object.assign(all, require('envsettings'));

Then when creating the webpack config I can dynamically assign which file envsettings points to (i.e. webpackConfig.resolve.alias.envsettings = './' + env).

However I would like to do something like:

if (settings.mock) {
    // Short-circuit ajax calls.
    // Require in all the mock modules.
}

But obviously I don't want to build in those mock files if the environment isn't mock.

I could possibly manually repoint all those requires to a stub file using resolve.alias again - but is there a way that feels less hacky?

Any ideas how I can do that? Thanks.

Answer

Matt Derrick picture Matt Derrick · Apr 24, 2015

You can use the define plugin.

I use it by doing something as simple as this in your webpack build file where env is the path to a file that exports an object of settings:

// Webpack build config
plugins: [
    new webpack.DefinePlugin({
        ENV: require(path.join(__dirname, './path-to-env-files/', env))
    })
]

// Settings file located at `path-to-env-files/dev.js`
module.exports = { debug: true };

and then this in your code

if (ENV.debug) {
    console.log('Yo!');
}

It will strip this code out of your build file if the condition is false. You can see a working Webpack build example here.