NPM
NPM - 应用程序根目录的路径(NPM - path to root of the application)我正在开发一个NPM包，它需要从应用程序根目录中获取一个模块。 将其视为需要属性模块的包，该模块由用户放置在应用程序根目录中。
如何获取应用程序根目录的路径？
更新：
总结一下我要做的事情，express允许你做这样的事情：
app.use(express.router(myNPMModule.router));路由器功能将获取app作为参数。 这个想法是允许用户在单独的文件中指定路由，他们只需要遵循某些约定，例如将他们的控制器放在/controllers文件夹中，我的模块将能够动态地解析路径，然后调用正确的正确模块中的方法。 我已经有了它的工作，但只是意识到，如果我将它打包在NPM中，我就无法访问应用程序的路径，所以我无法动态调用控制器模块的方法。
I am developing a NPM package, which needs to require a module from the application root. Think of it as the package requiring a properties module, which is placed in the application root by the user.
How do I get the path to root of the application?
UPDATE:
To summarize what I am trying to do, express allows you to do something like this:
app.use(express.router(myNPMModule.router));The router function will get app as the parameter. The idea is to allow the users to specify the routes in a separate file, and they just need to follow certain conventions like putting their controllers in /controllers folder, and my module will be able to dynamically parse the path, and then invoke the correct method in the correct module. I have already got it working, but just realized that if I package it in NPM, I no longer have access to the path of the app, so I can't invoke a controller module's method dynamically.
最满意答案让用户调用一个设置变量的函数：
myModule.setProps(json) OR myModule = new MyModule(json)然后，用户可以要求他们的根配置并使用它来启动您的模块。
更新回复编辑过的问题：
Express允许您定义如下路线：
var routes = require('./routes')app.get('/info', routes.info);//Routes Moduleexports.info = function(req, res) { res.render('info', {title: 'Express'})};我相信这正是你想要做的。 不是吗？
更新2：
您可以使用__dirname找到当前正在执行的脚本所在的目录
console.log(__dirname);Make the user call a function that sets the variables:
myModule.setProps(json) OR myModule = new MyModule(json)Then the user can require their root config and init your module with it.
Update In response to edited question:
Express allows you to define routes like this:
var routes = require('./routes')app.get('/info', routes.info);//Routes Moduleexports.info = function(req, res) { res.render('info', {title: 'Express'})};I believe this is exactly what you are trying to do. Is it not?
Update 2:
You can find the directory the currently executing script is in using __dirname
console.log(__dirname);