狐狸.fox 2011-11-15 04:05 采纳率: 0%
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和 dirname 的区别是什么。 在 node.js 中?

When programming in Node.js and referencing files that are located somewhere in relation to your current directory, is there any reason to use the __dirname variable instead of just a regular ./? I've been using ./ thus far in my code and just discovered the existence of __dirname, and essentially want to know whether it would be smart to convert my ./'s to that, and if so, why that would be a smart idea.

转载于:https://stackoverflow.com/questions/8131344/what-is-the-difference-between-dirname-and-in-node-js

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  • 游.程 2013-08-16 23:28
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    The gist

    In Node.js, __dirname is always the directory in which the currently executing script resides (see this). So if you typed __dirname into /d1/d2/myscript.js, the value would be /d1/d2.

    By contrast, . gives you the directory from which you ran the node command in your terminal window (i.e. your working directory).

    The exception is when you use . with require(). The path inside require is always relative to the file containing the call to require.

    For example...

    Let's say your directory structure is

    /dir1
      /dir2
        pathtest.js
    

    and pathtest.js contains

    var path = require("path");
    console.log(". = %s", path.resolve("."));
    console.log("__dirname = %s", path.resolve(__dirname));
    

    and you do

    cd /dir1/dir2
    node pathtest.js
    

    you get

    . = /dir1/dir2
    __dirname = /dir1/dir2
    

    Your working directory is /dir1/dir2 so that's what . resolves to. Since pathtest.js is located in /dir1/dir2 that's what __dirname resolves to as well.

    However, if you run the script from /dir1

    cd /dir1
    node dir2/pathtest.js
    

    you get

    . = /dir1
    __dirname = /dir1/dir2
    

    In that case, your working directory was /dir1 so that's what . resolved to, but __dirname still resolves to /dir1/dir2.

    Using . inside require...

    If inside dir2/pathtest.js you have a require call into include a file inside dir1 you would always do

    require('../thefile')
    

    because the path inside require is always relative to the file in which you are calling it. It has nothing to do with your working directory.

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