我想要求我的文件总是通过我的项目的根,而不是相对于当前模块。

例如,如果查看https://github.com/visionmedia/express/blob/2820f2227de0229c5d7f28009aa432f9f3a7b5f9/examples/downloads/app.js第6行,您将看到

express = require('../../')

在我看来,这真的很糟糕。假设我想让我所有的例子都只靠近根结点一层。这是不可能的,因为我必须更新超过30个例子,并且在每个例子中更新很多次。:

express = require('../')

我的解决方案是有一个基于根的特殊情况:如果字符串以$开头,那么它相对于项目的根文件夹。

任何帮助都是感激的,谢谢

更新2

现在我使用require.js,它允许你以一种方式编写,在客户端和服务器上都可以工作。Require.js还允许你创建自定义路径。

更新3

现在我转移到webpack + gulp,我使用enhanced-require来处理服务器端模块。看这里的基本原理:http://hackhat.com/p/110/module-loader-webpack-vs-requirejs-vs-browserify/


当前回答

你可以用我做的一个模块,Undot。它没有什么高级的,只是一个助手,让你可以避免那些点地狱与简单。

例子:

var undot = require('undot');
var User = undot('models/user');
var config = undot('config');
var test = undot('test/api/user/auth');

其他回答

在Browserify手册中有一个非常有趣的章节:

avoiding ../../../../../../.. Not everything in an application properly belongs on the public npm and the overhead of setting up a private npm or git repo is still rather large in many cases. Here are some approaches for avoiding the ../../../../../../../ relative paths problem. node_modules People sometimes object to putting application-specific modules into node_modules because it is not obvious how to check in your internal modules without also checking in third-party modules from npm. The answer is quite simple! If you have a .gitignore file that ignores node_modules: node_modules You can just add an exception with ! for each of your internal application modules: node_modules/* !node_modules/foo !node_modules/bar Please note that you can't unignore a subdirectory, if the parent is already ignored. So instead of ignoring node_modules, you have to ignore every directory inside node_modules with the node_modules/* trick, and then you can add your exceptions. Now anywhere in your application you will be able to require('foo') or require('bar') without having a very large and fragile relative path. If you have a lot of modules and want to keep them more separate from the third-party modules installed by npm, you can just put them all under a directory in node_modules such as node_modules/app: node_modules/app/foo node_modules/app/bar Now you will be able to require('app/foo') or require('app/bar') from anywhere in your application. In your .gitignore, just add an exception for node_modules/app: node_modules/* !node_modules/app If your application had transforms configured in package.json, you'll need to create a separate package.json with its own transform field in your node_modules/foo or node_modules/app/foo component directory because transforms don't apply across module boundaries. This will make your modules more robust against configuration changes in your application and it will be easier to independently reuse the packages outside of your application. symlink Another handy trick if you are working on an application where you can make symlinks and don't need to support windows is to symlink a lib/ or app/ folder into node_modules. From the project root, do: ln -s ../lib node_modules/app and now from anywhere in your project you'll be able to require files in lib/ by doing require('app/foo.js') to get lib/foo.js. custom paths You might see some places talk about using the $NODE_PATH environment variable or opts.paths to add directories for node and browserify to look in to find modules. Unlike most other platforms, using a shell-style array of path directories with $NODE_PATH is not as favorable in node compared to making effective use of the node_modules directory. This is because your application is more tightly coupled to a runtime environment configuration so there are more moving parts and your application will only work when your environment is setup correctly. node and browserify both support but discourage the use of $NODE_PATH.

我编写了这个小包,它允许您通过项目根的相对路径来要求包,而不引入任何全局变量或覆盖节点默认值

https://github.com/Gaafar/pkg-require

它是这样工作的

// create an instance that will find the nearest parent dir containing package.json from your __dirname
const pkgRequire = require('pkg-require')(__dirname);

// require a file relative to the your package.json directory 
const foo = pkgRequire('foo/foo')

// get the absolute path for a file
const absolutePathToFoo = pkgRequire.resolve('foo/foo')

// get the absolute path to your root directory
const packageRootPath = pkgRequire.root()

一些答案说,最好的方法是将代码作为包添加到node_module,我同意,这可能是最好的方法,失去../../../ in require,但实际上没有一个给出这样做的方法。

从2.0.0版本开始,你可以从本地文件安装一个包,这意味着你可以在根目录下创建一个文件夹,里面有你想要的所有包,

-modules
 --foo
 --bar 
-app.js
-package.json

所以在包装上。Json,你可以添加模块(或foo和bar)作为一个包,而不需要发布或使用外部服务器,像这样:

{
  "name": "baz",
  "dependencies": {
    "bar": "file: ./modules/bar",
    "foo": "file: ./modules/foo"
  }
}

之后你执行npm install,你可以使用var foo = require("foo")访问代码,就像你对所有其他包所做的一样。

更多信息可以在这里找到:

https://docs.npmjs.com/files/package.json#local-paths

下面是如何创建一个包:

https://docs.npmjs.com/getting-started/creating-node-modules

我正在寻找完全相同的简单性,要求文件从任何级别,我发现模块-别名。

安装:

npm i --save module-alias

打开你的包裹。Json文件,在这里你可以为你的路径添加别名,例如。

"_moduleAliases": {
 "@root"      : ".", // Application's root
 "@deep"      : "src/some/very/deep/directory/or/file",
 "@my_module" : "lib/some-file.js",
 "something"  : "src/foo", // Or without @. Actually, it could be any string
}

使用你的别名只需:

require('module-alias/register')
const deep = require('@deep')
const module = require('something')

刚刚看到这篇文章提到了app-module-path。它允许你这样配置一个基础:

require('app-module-path').addPath(baseDir);