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

我实现这一点的方法是创建“本地链接模块”。

的文件夹结构为例

db ¬ 
    models ¬
        index.js
    migrations
    seed
    config.json

routes ¬
    index.js
    user ¬
        index.js

如果从。/routes/user/index.js我想访问/db/models/index.js我会写

require('../../db/models/index.js')

为了使/db/models/index.js可以从任何地方访问,我在db文件夹中创建了一个名为_module_的文件夹,其中包含一个包。Json和一个main.js文件。

# package.json
{
    "name": "db", <-- change this to what you want your require name to be
    "version": "1.0.0",
    "description": "",
    "author": "",
    "repository": {},
    "main": "main.js"
}
// main.js
module.exports = require('../../db/models/index');

main.js中的路径必须是相对的,就像文件在node_modules中一样

node_modules ¬
    db ¬
        main.js

然后你可以运行npm install ./db/_module_,这将把。/db/_module_中的文件复制到。/node_modules/db中,在应用程序包的依赖项下创建一个条目。json之类的

"db": "file:db/_module_"

您现在可以在任何地方使用此包

const db = require('db');

当你运行npm install时,它会自动安装你的其他模块,工作跨平台(没有符号链接),并且不需要第三方包。

其他回答

如果你的应用程序的入口点js文件(即你实际运行“node”的文件)在你的项目根目录下,你可以很容易地用rootpath npm模块做到这一点。只需通过

npm install --save rootpath

...然后在入口点js文件的最上面,添加:

require('rootpath')();

从那时起,所有的require调用现在都相对于项目根——例如require('../../../config/debugging/log');就要求(“配置/调试/日志”);(其中配置文件夹在项目根目录下)。

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');

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.

我认为你不需要用你描述的方式来解决这个问题。如果您想在大量文件中更改相同的字符串,请使用sed。在你的例子中,

find . -name "*.js" -exec sed -i 's/\.\.\/\.\.\//\.\.\//g' {} +

/../变成了../

或者,您可以要求配置文件存储包含库路径的变量。如果您将以下文件存储为config.js在示例目录中

var config = {};
config.path = '../../';

在你的例子文件中

myConfiguration = require('./config');
express = require(config.path);

您将能够从一个文件控制每个示例的配置。

这只是个人喜好。

这里已经有很多好答案了。这只是说明这是一个普遍的问题,没有明确的最佳解决方案。当然最好是在Node.js中提供本地支持。以下是我目前使用的:

const r  = p => require (process.cwd() + p);
let see  = r ('/Subs/SubA/someFile.js' );
let see2 = r ('/Subs/SubB/someFile2.js');
...

我喜欢这个解决方案,因为require部分变得更短,不需要键入'require'很多次。绝对路径的主要好处是,您可以将它们从一个文件复制到另一个文件,而不必像使用相对路径那样对它们进行调整。因此,复制额外的一行箭头函数'r()'也不会有太多额外的工作。为了完成这个非常简单的任务,不需要导入额外的npm依赖项。