使用 Webpack 基于环境条件构建 [英] Conditional build based on environment using Webpack

问题描述

我有一些东西需要开发 - 例如，我不想用它来膨胀我的分布式构建文件的模拟.

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

在 RequireJS 中，你可以在插件文件中传递一个配置，并根据它有条件地要求内容.

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

对于 webpack，似乎没有办法做到这一点.首先为我使用的环境创建运行时配置 resolve.alias 重新指向一个 require 依赖关于环境，例如:

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'));

然后在创建 webpack 配置时，我可以动态分配 envsettings 指向的文件(即 webpackConfig.resolve.alias.envsettings = './' + env).

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

但是我想做类似的事情:

However I would like to do something like:

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

但显然，如果环境不是模拟的，我不想构建那些模拟文件.

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

我可以再次使用 resolve.alias 手动将所有这些要求重新指向存根文件 - 但有没有一种感觉不那么讨厌的方法?

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?

任何想法我怎么能做到这一点?谢谢.

Any ideas how I can do that? Thanks.

推荐答案

您可以使用 define插件.

我通过在你的 webpack 构建文件中做一些像这样简单的事情来使用它，其中 env 是导出设置对象的文件的路径:

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 };

然后在你的代码中

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

如果条件为假，它将从您的构建文件中删除此代码.您可以在 此处查看 Webpack 构建示例.

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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