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()

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.

我也遇到了同样的问题，所以我编写了一个名为include的包。

包含通过定位包来确定项目根文件夹的句柄。Json文件，然后传递路径参数，你给它的本地require()没有所有的相对路径混乱。我认为这不是require()的替代品，而是需要处理非打包/非第三方文件或库的工具。类似的

var async = require('async'),
    foo   = include('lib/path/to/foo')

我希望这对你有用。

这是我六个多月来的实际做法。我在项目中使用一个名为node_modules的文件夹作为我的根文件夹，这样它将始终从我调用绝对require的任何地方查找该文件夹:

node_modules myProject index.js我可以require("myProject/someFolder/hey.js")而不是require("./someFolder/hey.js") 包含hey.js的someFolder

当你被嵌套到文件夹中时，这更有用，如果以绝对方式设置，更改文件位置的工作要少得多。我在整个应用程序中只使用了2个相对要求。

虽然这些答案工作，但他们没有解决npm测试的问题

例如，如果我在server.js中创建一个全局变量，它将不会为我的测试套件执行设置。

设置全局apot变量，避免../../..在npm start和npm test中都可以使用，参见:

Mocha使用额外的选项或参数进行测试

请注意，这是新的官方摩卡解决方案。

大局

这看起来“真的很糟糕”，但要给它时间。事实上，它真的很好。显式的require()提供了完全的透明性和易于理解，就像项目生命周期中的一股新鲜空气。

可以这样想:你正在阅读一个例子，尝试Node.js，你认为它“在我看来真的很糟糕”。你在质疑Node.js社区的领导者，他们比任何人都花了更多的时间来编写和维护Node.js应用程序。作者犯这样一个新手错误的可能性有多大?(我同意，从我的Ruby和Python背景来看，乍一看这像是一场灾难。)

围绕Node.js有很多炒作和反炒作。但是当尘埃落定后，我们将承认显式模块和“本地优先”包是采用的主要驱动力。

一般情况

当然，首先搜索当前目录中的node_modules，然后是父目录、祖父目录、曾祖父目录等等。所以您已经安装的包已经以这种方式工作了。通常你可以从项目中的任何地方要求(“express”)，它工作得很好。

如果您发现自己正在从项目的根目录加载公共文件(可能是因为它们是公共实用程序函数)，那么这就是一个很大的线索，说明是时候创建包了。包非常简单:将文件移动到node_modules/，并放入package.json 在那里。瞧!整个项目都可以访问该名称空间中的所有内容。包是将代码放入全局名称空间的正确方法。

其他解决方法

我个人不使用这些技巧，但它们确实回答了你的问题，当然你比我更了解自己的情况。

您可以将$NODE_PATH设置为项目根目录。当您需要()时，将搜索该目录。

接下来，您可以折衷一下，从所有示例中要求一个通用的本地文件。该通用文件只是重新导出祖父目录中的真实文件。

Examples /downloads/app.js(以及其他类似的文件)

var express = require('./express')

/下载/ express.js例子

module.exports = require('../../')

现在当你重新定位这些文件时，最坏的情况是修复一个垫片模块。