使用 Webpack 基于环境的条件构建

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最佳答案

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.

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another way is using a JS file as a proxy, and let that file load the module of interest in commonjs, and export it as es2015 module, like this:

// file: myModule.dev.js
module.exports = "this is in dev"


// file: myModule.prod.js
module.exports = "this is in prod"


// file: myModule.js
let loadedModule
if(WEBPACK_IS_DEVELOPMENT){
loadedModule = require('./myModule.dev.js')
}else{
loadedModule = require('./myModule.prod.js')
}


export const myString = loadedModule

Then you can use ES2015 module in your app normally:

// myApp.js
import { myString } from './store/myModule.js'
myString // <- "this is in dev"

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I ended up using something similar to Matt Derrick' Answer, but was worried about two points:

The complete config is injected every time I use ENV (Which is bad for large configs).
I have to define multiple entry points because require(env) points to different files.

What I came up with is a simple composer which builds a config object and injects it to a config module.
Here is the file structure, Iam using for this:

config/
└── main.js
└── dev.js
└── production.js
src/
└── app.js
└── config.js
└── ...
webpack.config.js

The main.js holds all default config stuff:

// main.js
const mainConfig = {
apiEndPoint: 'https://api.example.com',
...
}


module.exports = mainConfig;

The dev.js and production.js only hold config stuff which overrides the main config:

// dev.js
const devConfig = {
apiEndPoint: 'http://localhost:4000'
}


module.exports = devConfig;

The important part is the webpack.config.js which composes the config and uses the DefinePlugin to generate a environment variable __APP_CONFIG__ which holds the composed config object:

const argv = require('yargs').argv;
const _ = require('lodash');
const webpack = require('webpack');


// Import all app configs
const appConfig = require('./config/main');
const appConfigDev = require('./config/dev');
const appConfigProduction = require('./config/production');


const ENV = argv.env || 'dev';


function composeConfig(env) {
if (env === 'dev') {
return _.merge({}, appConfig, appConfigDev);
}


if (env === 'production') {
return _.merge({}, appConfig, appConfigProduction);
}
}


// Webpack config object
module.exports = {
entry: './src/app.js',
...
plugins: [
new webpack.DefinePlugin({
__APP_CONFIG__: JSON.stringify(composeConfig(ENV))
})
]
};

The last step is now the config.js, it looks like this (Using es6 import export syntax here because its under webpack):

const config = __APP_CONFIG__;


export default config;

In your app.js you could now use import config from './config'; to get the config object.

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Use ifdef-loader. In your source files you can do stuff like

/// #if ENV === 'production'
console.log('production!');
/// #endif

The relevant webpack configuration is

const preprocessor = {
ENV: process.env.NODE_ENV || 'development',
};


const ifdef_query = require('querystring').encode({ json: JSON.stringify(preprocessor) });


const config = {
// ...
module: {
rules: [
// ...
{
test: /\.js$/,
exclude: /node_modules/,
use: {
loader: `ifdef-loader?${ifdef_query}`,
},
},
],
},
// ...
};

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Not sure why the "webpack.DefinePlugin" answer is the top one everywhere for defining Environment based imports/requires.

The problem with that approach is that you are still delivering all those modules to the client -> check with webpack-bundle-analyezer for instance. And not reducing your bundle.js's size at all :)

So what really works well and much more logical is: NormalModuleReplacementPlugin

So rather than do a on_client conditional require -> just not include not needed files to the bundle in the first place

Hope that helps

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While this is not the best solution, it may work for some of your needs. If you want to run different code in node and browser using this worked for me:

if (typeof window !== 'undefined')
return
}
//run node only code now

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Use envirnment variables to create dev and prod deployments:

https://webpack.js.org/guides/environment-variables/

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I've struggled with setting env in my webpack configs. What I usually want is to set env so that it can be reached inside webpack.config.js, postcss.config.js and inside the entry point application itself (index.js usually). I hope that my findings can help someone.

The solution that I've come up with is to pass in --env production or --env development, and then set mode inside webpack.config.js. However, that doesn't help me with making env accessible where I want it (see above), so I also need to set process.env.NODE_ENV explicitly, as recommended here. Most relevant part that I have in webpack.config.js follow below.

...
module.exports = mode => {
process.env.NODE_ENV = mode;


if (mode === "production") {
return merge(commonConfig, productionConfig, { mode });
}
return merge(commonConfig, developmentConfig, { mode });
};

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Faced with the same problem as the OP and required, because of licensing, not to include certain code in certain builds, I adopted the webpack-conditional-loader as follows:

In my build command I set an environment variable appropriately for my build. For example 'demo' in package.json:

...
"scripts": {
...
"buildDemo": "./node_modules/.bin/webpack --config webpack.config/demo.js --env.demo --progress --colors",
...

The confusing bit that is missing from the documentation I read is that I have to make this visible throughout the build processing by ensuring my env variable gets injected into the process global thus in my webpack.config/demo.js:

/* The demo includes project/reports action to access placeholder graphs.
This is achieved by using the webpack-conditional-loader process.env.demo === true
*/


const config = require('./production.js');
config.optimization = {...(config.optimization || {}), minimize: false};


module.exports = env => {
process.env = {...(process.env || {}), ...env};
return config};

With this in place, I can conditionally exclude anything, ensuring that any related code is properly shaken out of the resulting JavaScript. For example in my routes.js the demo content is kept out of other builds thus:

...
// #if process.env.demo
import Reports from 'components/model/project/reports';
// #endif
...
const routeMap = [
...
// #if process.env.demo
{path: "/project/reports/:id", component: Reports},
// #endif
...

This works with webpack 4.29.6.

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I use string-replace-loader to get rid of an unnecessary import from the production build, and it works as expected: the bundle size becomes less, and a module for development purposes (redux-logger) is completely removed from it. Here is the simplified code:

In the file webpack.config.js:

rules: [
// ... ,
!env.dev && {
test: /src\/store\/index\.js$/,
loader: 'string-replace-loader',
options: {
search: /import.+createLogger.+from.+redux-logger.+;/,
replace: 'const createLogger = null;',
}
}
].filter(Boolean)

In the file src/store/index.js:

// in prod this import declaration is substituted by `const createLogger = null`:
import { createLogger } from 'redux-logger';
// ...
export const store = configureStore({
reducer: persistedReducer,
middleware: createLogger ? [createLogger()] : [],
devTools: !!createLogger
});