The file structure looks like this:
= (main-folder)
- package.json
- ...
= server
- index.js
- ...
= deploy-abc // new server
= src
- index.js
= src
- index.js
- ...
I am trying to install 'deploy-abc' as a module within the main-folder app. So, I ran : yarn add "/var/www/main-folder/deploy-abc". It installed correctly & I can see the 'deploy-abc' dependency listed in package.json.
However, when I try to access the deploy-abc's exported object, I get node error Error: Cannot find module 'deploy-abc'. E.g: In some file in the main folder:
const deployAbc = require("deploy-abc");
console.log(deployAbc); // error
Where am I going wrong?
As per the node docs: https://nodejs.org/api/modules.html#modules_loading_from_node_modules_folders
If the module identifier passed to require() is not a core module, and does not begin with '/', '../', or './', then Node.js starts at the parent directory of the current module, and adds /node_modules, and attempts to load the module from that location.
So, seeing as your module is now in the main folder, how you require it will depend on the relative location. If you are requiring it from say /src/index.js, then you will need:
const deployAbc = require('../deploy-abc')
You also don't need to specify the actual file in this case, as it defaults to index.js. If however the entry file was say, dabc.js or something else, you would need to also specify that in the location.
You might have to use the exact relative path. For example, const deployAbc = require("../deploy-abc")
Related
I have a website whose .js files handle imports by doing import {Foobar} from /path/to/foobar.js, where the import path starts with a slash.
I want to run this same JS on the server using node.js. Of course, the imports fail because the root path for the website does not correspond to the root directory for the server.
Do I have to change all the import statements? Or is there some setting (perhaps in package.json) where I can tell the module loader that / corresponds to a particular directory?
var absolute_path = __dirname;
So far, I've only seen the root directory be "/app"
I have my server.js file at let's say directory: /dir1
I start the server with node server.js.
I have my file routes.js in directory /dir1/app/.
I want to get the directory of the server.js file.
I am not too sure how to do this. I can place code that tells me the location of the current script. But what is the current script?
A web app makes use of multiple files. So how do i get the dir of server.js?
EDIT: Here is a code example:
// routes.js
const spawn = require('child_process').spawn;
const fs = require('fs');
module.exports = function(app)
{
//the file upload button does a POST the the '/' route
//i left the arrow function notation for reference
app.post('/', (req, res) => {
if (req.files)
{
var pathAndFileName = '/home/user1/Desktop/app/dataLocation/'+fileName;
As you can see, pathAndFileName is hardcoded.
It works on my system, but it does not for every system.
So i would like to be to get the location of the server.js file (in /dir1), inside the app.js file, so i can the append to the directory the '/dataLocation' + fileName.
If routes.js is in:
/home/user1/Desktop/app/dir1
And, server.js is in:
/home/user1/Desktop/app
And, then the file you're trying to get to is in:
/home/user1/Desktop/app/dataLocation
Then, from routes.js, you can get the directory of server.js with this:
const serverjsDir = path.join(__dirname, ".."); // go up one level
Or, you could get the whole path your edit seems to be looking for (in a dataLocation subdirectory below where server.js is) with this:
const pathAndFileName = path.join(__dirname, '../dataLocation', fileName).
More detail
If you're using CommonJS modules (that use require()), then you can get the current script's directory with __dirname and then it's up to you to know where other resources are relative to your own directory so you can navigate up or down the directory hierarchy from __dirname. For example, if you want to get access to a file main.css in another directory at the same directory level as your script file, you could construct at path like this:
let filePath = path.join(__dirname, "../css/main.css");
But what is the current script?
The current script is the script that contains the code that is accessing __dirname.
A web app makes use of multiple files. Since the place where i want to embed the location (in its code) is located inside routes.js, i guess this is where is should target, am i correct?
It's not clear to me exactly what you mean here. If you can provide a specific example showing what code located in what directory wants to access what file located in what other directory, then we could offer you the precise code to get to that file.
If the code that wants to access this other file is in routes.js, then the location of routes.js would be in __dirname and you would build a relative path from that location as shown above.
To help you understand, __dirname is not a global variable. It's a module-specific variable that contains a different value for every single module location in your project. So, whatever code is running in whatever module will have access to it's own copy of __dirname that specifies the location of that specific code.
To get access routes.js, if it is a module, Im assuming you are using Import or Require, and you could use
require(./app/routes);
If you were to go back a directory, you use ../ as many times as needed. If this is not the answer you were looking for, either reply or someone else may answer
I am creating a VSCode extension, and following the getting started guide (https://code.visualstudio.com/api/get-started/your-first-extension) have used yeoman scaffold to get started. I created a new file, newModule.js in the same directory and want to import it as a module for use in the main extension.js script. I then do:
const newModule = require('./newModule.js');
This throws an error:
cannot find module 'newModule' require stack: -
This problem disappears if I copy my file to the node_modules folder created by default. I would like to know what is going on here, and what the best way of handling imports is when working with javascript/Node.js/vs-extensions.
I also notice that node_modules folder is not pushed to github by default, why?
The node_modules folder is for storing all the code from the libraries and packages you are using. It is excluded from git because it is a waste of space and a distraction to store them all in your versioning-control, as you can just re-download them anytime.
Just put your module in the same /src directory, and use the import syntax to import it, instead of require.
import newModule from './newModule';
For example, see how it is done in this sample code.
Please instead
const newModule = require('./newModule.js');
Try this
import newModule from './newModule');
// ^^ Do not use file extension
Also make sure that the file you are calling is in the same directory
I'm having a problem with node.js require.resolve the method that makes no sense having with the test I'm running first.
So my Code
let checkPluginCanLoad = (pluginName, pluginFile) => {
return new Promise((res, rej) => {
// build the plugin's class files path
console.log(`looking for file at ${path}/${pluginName}/${pluginFile}`)
let req = `${path}/${pluginName}/${pluginFile}`
// check it exists blocking
let fileState = fs.existsSync(`${req}.js`);
if(fileState){
console.log(`File ${req}.js ${fileState?"exists.":"does not exist."}`);
// it exists try to load it
let plugin = require.resolve(`${req}`);
res(plugin);
}else{
// could not find or load the plugin
rej(`Plugin is invalid can't find or load class for ${pluginFile}`);
}
});
}
The vars are set to , path = "Plugins", pluginName = "TestPlugin", pluginFile = "TestPlugin";
My output is
Plugins from 'Plugins' have been loaded. index.js:36
looking for file at Plugins/TestPlugin/TestPlugin Plugins.js:133
File Plugins/TestPlugin/TestPlugin.js exists. Plugins.js:138 failed to
load plugin 'TestPlugin' with error 'Error: Cannot find module 'Plugins/TestPlugin/TestPlugin''
The final line
load plugin 'TestPlugin' ... comes from the system above this one catching the rejection.
So the file Exists according to FileSystem but the resolver can't find it.
I have tried prepending it with ../ before the path in case it's resolving from the file that is running it and not the application directory,
I have also tried prepending it with ./ in case it's running it from the application directory.
you need to add ./ to tell node that you are targeting local file.
let plugin = require.resolve(`./${req}`);
if you forgot node will search in steps described here
So this is a bug caused by symlinks on windows with junctioned directories,
So for my dev environment it junctions the directories from my Git Repository directory. but Node is running it through the junction directory but seeing the original directory.
E.G
Git repo is in D:\Projects\NodeAppRepo
then the dev environment is running inside D:\Project\NodeApp\
I have index.js, Core/, node_modules/ all running via a junction inside the NodeApp to the paths in the git repo.
so i'm running node index.js inside D:\Project\NodeApp\ wich is in turn loading D:\Project\NodeApp\Core\Core.js this then imports ./Plugins.js this is where the break happens because the path node has for plugins is not D:\Project\NodeApp\Core\Plugins.js it is D:\Projects\NodeAppRepo\Core\Plugin.js
I wanna create a node package modules, but I have difficulty to require a file from root project directory to use inside my node package module I created.
If I have directory structure like this
- node_modules
- library_name
- lib
- index.js
- bin
- run.sh
- config.js
If the run.sh called, it will run index.js. Inside index.js, how do I resolve to root directory which later I can require config.js inside index.js?
Package binary can accept configuration path explicitly as an argument.
If package binary doesn't run as NPM script, it shouldn't rely on parent project structure.
If package binary runs via NPM script:
"scripts": {
"foo": "library_name"
}
This will set current working directory to project root, so it could be required as:
const config = require(path.join(process.cwd(), 'config'));
Both approaches can be combined; this is often used to provide configuration files with default locations to third-party CLI (Mocha, etc).
If you're in index.js and config.js is in the directory above node_modules in your diagram, then you can build a path to config.js like this:
const path = require('path');
let configFilename = path.join(__dirname, "../../../", "config.js");
__dirname is the directory that index.js is in.
The first ../ takes you up to the library_name directory.
The second ../ takes you up to the node_modules directory.
The third ../ takes you up to the parent of node_modules (what you call project root) where config.js appears to be.
If you really want your module to be independent of how it is installed or how NPM might change in the future, then you need to somehow pass in the location of the config file in any number of ways:
By making sure the current working directory is set to the project root so you can use process.cwd() to get access to the config file.
By setting an environment variable to the root directory when starting your project.
By passing the root directory in to a module constructor function.
By loading and passing the config object itself in to a module constructor function.
I create module same your module.
And I call const config = require('../config'), it work.