The cwd is a concept that many CLI programs leverage to their advantage,
namely npm and node-gyp, which execute their actions aginst the cwd. That
said, it's a powerful tool, and not at all a problem, as stated pretty
well in this thread already.
There won't be any warning being added to node for
This is all working as intended.
process.cwd() returns the current working directory of the process.
Anything else would be surprising and wrong.
__dirname and __filename are local to the module (ie, like __FILE__ in
C programs). require() resolves relative links relative to the file
doing the
__dirname is local to any file. It works everywhere.
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Because it's local to the file, it doesn't work anywhere.
El lunes, 26 de agosto de 2013 14:46:55 UTC+2, mks escribió:
__dirname is local to any file. It works everywhere.
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Gagle, can you write down a concrete use case?
On Mon, Aug 26, 2013 at 9:59 AM, Gagle gagle...@gmail.com wrote:
Because it's local to the file, it doesn't work anywhere.
El lunes, 26 de agosto de 2013 14:46:55 UTC+2, mks escribió:
__dirname is local to any file. It works everywhere.
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Well then it's your problem to know where your local file is located in
relation to your main file
On Mon, Aug 26, 2013 at 9:59 AM, Gagle gagle...@gmail.com wrote:
Because it's local to the file, it doesn't work anywhere.
El lunes, 26 de agosto de 2013 14:46:55 UTC+2, mks escribió:
From http://nodejs.org/docs/latest/api/globals.html#globals_dirname it
states:
__dirname isn't actually a global but rather local to each module.
So it actually works in each module (not just the main file).
Best regards,
Jeroen Janssen
Op maandag 26 augustus 2013 14:25:26 UTC+2 schreef
$ pwd
/home/user1
$ mkdir dir
$ cat dir/app.js
console.log (process.cwd ());
$ node dir/app.js
/home/user1
$ cd dir node app.js
/home/user1/dir
Now suppose you have this code:
//app.js
var fs = require (fs);
if (fs.existsSync (settings.json)){
doSomethingUseful ();
}else{
//Warning!!
Ah! I see..
But this is not a problem for require, but for fs.
Ok, in my code I would use
if (fs.existsSync(path.join(__dirname, settings.json))
On Mon, Aug 26, 2013 at 10:27 AM, Gagle gagle...@gmail.com wrote:
$ pwd
/home/user1
$ mkdir dir
$ cat dir/app.js
console.log (process.cwd
But now suppose you have another file called b.js that is required by a.js
and is stored in a different directory and uses a relative path. You can't
use __dirname.
El lunes, 26 de agosto de 2013 15:32:17 UTC+2, ajlopez escribió:
Ah! I see..
But this is not a problem for require, but for
Ok, only to be aligned, can you write down a concrete use case? Now you
switched to a require issue, not fs.
On Mon, Aug 26, 2013 at 10:34 AM, Gagle gagle...@gmail.com wrote:
But now suppose you have another file called b.js that is required by a.js
and is stored in a different directory and
./a.js
require('./b_dir/b');
./b_dir/b.js
var fs = require('fs');
console.log(fs.readFileSync(__dirname + '/../a.js', 'utf-8'));
As I was saying, it's your problem to know where your required file is
located relative to the main file in your application. Or you could always
use some sort of
Node's require is always relative to the file that calls require. In fact,
the internal implementation of this is done by giving file a unique copy of
the require function that embeds that file's directory. If you wanted to
require relative to the cwd, then use process.cwd and path.resolve. If
+100
Relative requires combined with npm dependencies are the next big thing
since sliced bread ;-)
On Mon, Aug 26, 2013 at 12:05 PM, Tim Caswell t...@creationix.com wrote:
Node's require is always relative to the file that calls require. In
fact, the internal implementation of this is done
If node.js is better at requiring modules than other platforms why not
include a warning when you execute a file with a relative path different
than the directory of this file? The process.root solution is already
implemented:
process.root = path.dirname(process.mainModule.filename)
We could
On Mon, Aug 26, 2013 at 9:10 AM, Gagle gagle...@gmail.com wrote:
If node.js is better at requiring modules than other platforms why not
include a warning when you execute a file with a relative path different
than the directory of this file?
Because it's not a problem, it's a useful feature,
The point is that cwd is not problematic at all.
Think of a command line program that you want to do something with your
current working directory:
$ /usr/local/bin/myls
$ cd ~
$ myls
On Monday, August 26, 2013 6:10:07 PM UTC+2, Gagle wrote:
If node.js is better at requiring modules than
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