node.js 中的 __dirname 和 ./有什么区别? [英] What is the difference between __dirname and ./ in node.js?

问题描述

在 Node.js 中编程并引用与当前目录相关的某个位置的文件时，是否有任何理由使用 __dirname 变量而不是常规的 ./?到目前为止，我一直在我的代码中使用 ./并且刚刚发现 __dirname 的存在，并且基本上想知道将我的 ./转换为它是否明智，如果是这样，为什么这是一个聪明的主意.

解决方案

要点

在 Node.js 中，__dirname 始终是当前正在执行的脚本所在的目录 (见这个).因此，如果您在 /d1/d2/myscript.js 中键入 __dirname，则该值将是 /d1/d2.

相比之下，当您使用像 这样的库时，. 为您提供在终端窗口(即您的工作目录)中运行 node 命令的目录路径 和 fs.从技术上讲，它最初是您的工作目录，但可以使用 process.chdir() 进行更改.

例外情况是当您将 . 与 require() 一起使用时.require 中的路径总是相对于包含对 require 调用的文件.

例如...

假设您的目录结构是

/dir1/目录2路径测试.js

和pathtest.js包含

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

你会

cd/dir1/dir2节点路径测试.js

你得到

<预><代码>.=/目录1/目录2__dirname =/dir1/dir2

您的工作目录是 /dir1/dir2 所以这就是 . 解析的内容.由于 pathtest.js 位于 /dir1/dir2，这也是 __dirname 解析的内容.

但是，如果您从 /dir1

运行脚本

cd/dir1节点 dir2/pathtest.js

你得到

<预><代码>.=/目录1__dirname =/dir1/dir2

在那种情况下，您的工作目录是 /dir1 所以这就是 . 解析为，但 __dirname 仍然解析为 /dir1/dir2.

在 require...

中使用 .

如果在 dir2/pathtest.js 中，你有一个 require 调用在 dir1 中包含一个文件，你会总是强>做

require('../thefile')

因为 require 中的路径总是相对于你调用它的文件.它与您的工作目录无关.

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.

解决方案

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) when you use libraries like path and fs. Technically, it starts out as your working directory but can be changed using process.chdir().

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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