Node.js:如何在模块中需要功能的地方获取文件名? [英] Node.js: How to get filename where function was required from within a module?

问题描述

我正在尝试从需要模块功能的位置获取原始文件名.我知道您可以使用__filename来获取当前文件，但是我想获取原始文件.

I'm trying to get the original filename from where a module's function has been required from. I know you can use __filename to get the current file, but I want to get the original file.

例如，我将拥有一个简单的模块

For example a simple module I have would be

module.js

module.exports(function() {
   return {
     print : function(message) {
        console.log(__filename + ' ' + message);
     };
   }
});

app.js

var module = require('./module')();
module.print('hello');

最终发生的事情是它将打印module.js hello，但是我真的很想看看app.js hello.

What ends up happening is it will print module.js hello but I really want to see app.js hello.

我正在探索获取它的方法，我知道您可以使用console.trace来查看调用堆栈，但是我无法解析它来执行我想要的操作.

I was exploring ways to get it and I know you can use console.trace to see the stack of calls but I can't parse it to do what I want.

现在，我已经通过使print函数采用另一个参数来解决此问题，您只需从app.js内部传递__filename，但是我有点想找到一种解决方案，我不必做到这一点.

Right now I've worked about it by making the print function take in another parameter and you simply pass __filename from within app.js but I kind of want to find a solution where I don't have to do this.

推荐答案

获取父模块

您可以使用 module.parent 来解决此问题，然后解析

Getting Parent Module

You can do this by using module.parent , then resolving the filename property like so:

module.exports(function() {
   return {
     print : function(message) {
        console.log(module.parent.filename + ' ' + message);
     };
   }
});

app.js

var module = require('./module')();
module.print('hello');

输出:

/path/to/app.js hello

您几乎要的是什么.

此解决方案为您提供了完全限定的路径，但是您可以通过用路径分隔符分割路径来获取文件名.

This solution provides you with a fully qualified path, but you get the filename by splitting the path by its delimiter.

var parentModFilename = module.parent.filename.split(/\\|\//).pop()

然后在parentModFilename中给您"/app.js".

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