I in electron am doing:
path.dirname('C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron\\main')
That path is the actual value of my __dirname. How come it is not giving me C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron? I want that main part chopped off.
Assuming main is a directory inside electron. Also assuming that you have some file called index.js inside main folder where you want to have the path of electron directory.
So, you can do path.join this way:
var mainFolderParentPath = path.join(__dirname, '../');
Your original file location:
C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron\\main\\index.js
__dirname will return
C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron\\main
and then inside path.join '../', will chop off the main folder from path. So, you will be left off with:
C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron
Well you obviously didn't read the docs for dirname. It states that it works like the Unix command dirname which "strips non-directory suffix from file name", thus you get the C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron.
What you are looking for is basename.
path.basename('C:\\Users\\Blagoh\\Documents\\GitHub\\Screeenshoter\\dist\\electron\\main') will give you main.
Related
So I have developed a Node Api and on local the path works fine, but I want to move it now to a server online, and don't know how to address the path.
This is the path in local: const path = "/Users/username/Code/projectname/api/invoices"
And what I want to do is make it work also for online, also to make sure that the folder api is what is going to be uploaded, not the one with projectname as it contains the client folder and other folders, so only the api folder.
use __dirname to get the path of the folder.
The __dirname in a node script returns the path of the folder where the current JavaScript file resides.
const path = `${__dirname}/api/invoices`
You can use the __dirname variable.
For this, you first need to import path.
const nodePath = require("path");
Then, to get the path, you need to join the directory name and your custom path together.
const path = nodePath.join(__dirname, "/api/invoices");
This should correctly join the path together.
Note: Naming conventions changed to avoid naming conflicts.
In conclusion, you can use the __dirname variable!
Normally in node one would use a path similar to this:
../js/hereIsMyJs.js
However in mac for example (pc is different)
the path can be ~/Desktop/Ohms/js/hereIsMyJs.js
Is there any module or way to use the computer path I just presented in node?
I'm using a module that requires the path to where the file should be placed.
It has to work dynamic so the optimal solution would be for me just to feed it the "computer" path.
const fileName = '~/Desktop/Ohms/somewhere/here.jpg'
QRCode.toFile(fileName, 'https://example.com')
To achieve what you want, it's normal to use node's native path.resolve https://nodejs.org/api/path.html#path_path_resolve_paths
// Start from current directory, and gives absolute path valid to the current OS.
console.log( path.resolve("../js/hereIsMyJs.js") ); // ie: C:\\projects\\js\\hereIsMyJs.js
// It also accepts multiple arguments, so you can feed it partial paths
path.resolve( "..", "js" ,"hereIsMyJs.js" ); // same result as above.
Other things worth to mention:
The tilde character ~ is short for the home directory in the *nix world. And works most places not windows.
In node you can use require('os').homedir() to get the home directory
There is also __dirname (gives absolute path of the directory containing the currently executing file) and process.cwd() which gives the directory from where you executed your file
And finally there is path.join() which is similar to resolve, but works for joining relative paths, and doesn't care about the current directory.
I'm new in nodejs. I try to use some file in my filesystem. I would like to get them through some realtive path. Here is the structure:
-app
-resources
-tmp
-file1
-file2
-common
-common.js
-etc
So, I would like to reach the file1 and file2 in my resources/tmp folder from the common.js file in int common folder. I used two relative path methodic:
// First , wrong one
var tmpfolder = '../resources/tmp';
// Second, good one
var tmpfolder = './resources/tmp';
The file absolute path is something like this:
C:\Users\xy\Documents\work\nodejs\projects\project_wrapper_folder\project_folder\resources\tmp
If I log the first relative path I got this result:
C:\Users\xy\Documents\work\nodejs\projects\project_folder\resources\tmp
which is wrong, because it does not contains the wrapper folder.
But the second works fine.
Can somebody explain me this behaviour?
UPDATE
I see the meaning if '../', thanks your explanations!
I have tried #Lissy answer: "So baring that in mind ./ will resolve to the value of your project root..." that sounds great, but the result is not.
I have this logic in my app:
var tmpfolder = require('./otherFolder/orhetFile');
where otherFolder is the subfolder of my project_folder. So, when I used this here, I got an error called Cannot find module ...
But if I use this './' in fs module, here: /project_folder/otherFolder_2/otherFile_2 like:
var path = `./resources/tmp`;
fs.mkdirsSync(path);
It works!
these is strange for me. If './' means the current folder, than the example above sould not work (but it's works).
But if the './' means path of the project root, the example with require should work (but does not work).
Strange for me, is there some meaning of require or fs??
In Summary
./my-file means look at the root, whereas ../my-file, means come out of the current directory, and look in the parent folder.
Explanation of Relative and Absolute Paths in Node.js
The default document root is set with the NODE_PATH environment variable (however it would be terrible practice to modify that). So baring that in mind ./ will resolve to the value of your project root.
And so let tmpfolder = require('./resources/tmp'); would resolve to
C:\Users\......\project_wrapper_folder\project_folder\resources\tmp as you have seen
A relative path is denoted by not starting with ./, so for example let tmpfolder = require('my-scripts/some-script'); would look in the current directory for another folder called my-scripts.
Following on from that ../ is relative to the current directory, but goes out one (to it's parent) and looks from there. You can go out two directories by doing ../../my-dir and so on (this is not great practice though, as it can get hard to manage if you have ../../../../../../)
Better method
You can use the __dirname constant in Node to locate the directory name of the current module.
e.g.
fs.readFile(__dirname + '/../foo.bar');
Useful Links
I think you'll find this article on Better local require() paths for Node.js very useful.
Also, for a bit of further reading, here is the official Node.js documentation on paths
The .. means the previous dir. So you are getting out from project_wrapper_folder and then add to the script dir resources\tmp and that way the project_wrapper_folder is no longer in the relative path.
Lets assume you are inside a folder called app.
when you use ../resources/tmp, it will come out of the app folder and will search for tmp inside resources folder.
when you use ./resources/tmp, it will look within the app folder for resources folder and tmp inside it.
../ points to the parent directory while ./ points to current working directory.
A dot basically means "go one folder back in the path"
So:
'./' points at the folder you are in.(in your case the folder common/)
'../' points at the folder that contains the folder you are in. (in your case the folder app/)
Here's my folder structure :
Root Folder
...
...
...
app.js
escape_iframe.html
...
from within app.js I'm doing :
res.sendFile('escape_iframe.html' , {root : __dirname});
and the error I'm getting is :
Error: ENOENT, stat '/app/escape_iframe.html'
which I guess is a horrible way of saying that the file can't be found in the path provided ?
Anyway, as you can tell I simply want to serve a file that is a sibling of app.js (which is the file form which the res.sendFile(... call is being made)
I also tried : res.sendFile('escape_iframe.html'); and I got : path must be absolute or specify root to res.sendFile
What am I doing wrong ?
Paths specified with a . are relative to the current working directory, not relative to the script file. So the file might be found if you run node app.js but not if you run node folder/app.js.
To make a path relative to the script, you must use the __dirname variable.
res.sendFile(__dirname + '/path/to/escape_iframe.html');
Path specified with . are relative to the path and any number of forward slashes are truncated as single slash. You might get this error when the process node filename.js is not able to locate the filename.
https://nodejs.org/api/errors.html#errors_enoent_no_such_file_or_directory
My suggestion would be ::
var path = require('path')
//process.cwd() returns the current directory of the project
res.sendFile(process.cwd(),'/path/to/escape_iframe.html');
Let's say I have this directory structure
/Project
/node_modules
/SomeModule
bar.js
/config
/file.json
foo.js
-
foo.js:
require('bar');
-
bar.js:
fs.readdir('./config'); // returns ['file.json']
var file = require('../../../config/file.json');
Is it right that the readdir works from the file is being included (foo.js) and require works from the file it's been called (bar.js)?
Or am I missing something?
Thank you
As Dan D. expressed, fs.readdir uses process.cwd() as start point, while require() uses __dirname. If you want, you can always resolve from one path to another, getting an absolute path both would interpret the same way, like so:
var path = require('path');
route = path.resolve(process.cwd(), route);
That way, if using __dirname as start point it will ignore process.cwd(), else it will use it to generate the full path.
For example, assume process.cwd() is /home/usr/node/:
if route is ./directory, it will become /home/usr/node/directory
if route is /home/usr/node/directory, it will be left as is
I hope it works for you :D