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
.
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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