I prefer to import dependencies without lots of relative filesystem navigation like ../../../foo/bar.
In the front-end I have traditionally used RequireJS to configure a default base path which enables "absolute" pathing e.g. myapp/foo/bar.
How can I achieve this in Node.js?
What you can do is use require.main.require which would always be the module path of the starting file. See https://nodejs.org/api/modules.html#modules_accessing_the_main_module
This means, that when you run
const x = require.main.require('some/file/path/y')
It would require it based on the base path of your starting file (that was invoked , node app.js meaning the app.js file).
Other options include using process.cwd() to get the starting path of your app, but that would be depending on where you invoke your node process, not the location of the file. Meaning, node app.js would be different than of you would start it one level up, node src/app.js.
Just to add to the other answer, you can put this few lines in your main file:
(function (module) {
let path = require('path')
let __require = module.constructor.prototype.require
let __cpath = __dirname
module.constructor.prototype.require = function (str) {
if (str.startsWith('package:')) {
let package = str.substr(8)
let p = path.resolve(__dirname, package)
return __require(p)
} else {
return __require(str)
}
}
})(module)
It extends your require in such way:
if path begins with package: (i.e. package:src/foo/bar) it translates it to require('/absolute/path/to/your/app/src/foo/bar'),
if it doesn't start with package: then it behaves like a normal require would.
Related
I am setting a gulp development environment, but when I used an absolute path in gulp watcher, it didn't work.Does gulp watcher support absolute path?
gulpfile.js
gulp.task('watch', function() {
gulp.watch(
path.join(__dirname, 'public/css/**/*.css'),
gulp.parallel('dev:css', 'dev:cssImgs')
);
});
I found the same error, and sorry for necroing this question in 2019., but this problem isn't solved yet.
It seems that watch process only accept relative paths written in unix style. It won't accept absolute paths or even relative paths with back slashes (windows style).
Using node path to clear out the path won't help on windows, so only solution would be to use node path.normalize('..\\your\\relative\\path\\') and then replace \\ with /.
I wrote one simple function that I will normalize path in unix way no matter on which OS it's used.
/*
This plugin is used to fix path separators for dynamically created links used
for `gulp watch` command.
It converts links like
`../../some//path\with/wrong//\separators/`
to
`../../some/path/with/wrong/separators/`
It also replaces `~` with user home directory path
*/
module.exports = (oldPath) => {
const pathString = oldPath;
const fix = /\\{1,}|\/{1,}/;
const user_home = require('os').homedir();
return pathString.replace(new RegExp(fix, 'gm'), '/').replace(new RegExp(fix, 'gm'), '/').replace('~', user_home)
}
Save that code as fixPath.js and then use it in this manner:
const fixPath = require('./fixPath')
let badPath = '../../some//path\with/wrong//\separators/'
let goodPath = fixPath(badPath) // "../../some/path/with/wrong/separators/"
I have a problem, i am trying to create an object that i can use over and over in a separate javascript file. I have a file called foo.js and another file called boo.js. I want to create the object in boo.js In my server.js file i required foo.js, and it works fine i can access foo.js. I want to be able to access boo.js from foo.js. I keep getting errors when i require boo.js in foo.js and i cant access it. Is There a way to do this?
here is my code
//boo.js
var obj = function () {
return {
data: 'data'
}
}
module.exports = {
obj: obj
}
foo.js
//foo.js
var request = require('request');
var x = require('./modules/boo')
var random= function() {
return x.obj();
}
module.exports = {
random: random
}
If they are in the same directory you will want to require like so var x = require('./boo'). The ./ is relative to the current directory.
they are both in the same directory
In that case, you'll want to remove the modules/ from the path:
var x = require('./boo');
require() is aware of the current script's location and bases relative paths from the script's own parent directory.
The ./ at the start of the path will refer to the same directory as __dirname, which seems to be modules/.
console.log(__dirname);
// "/project-path/modules"
Including modules/ in the path will result in doubling it:
var path = require('path');
console.log(path.resolve(__dirname, './modules/boo'));
// "/project-path/modules/modules/boo"
(Side note: The fs module does not behave the same way. Relative paths for it are based from the current working directory or process.cwd().)
I'm trying to build a local library of JS modules to use in Node projects.
If a new project lives in /Users/me/projects/path/to/new/project/ and my library files are located in /Users/me/projects/library/*.js is there a way to access those files without using a relative path?
In /Users/me/projects/path/to/new/project/app.js you can require foo.js like so:
var foo = require('../../../../../library/foo') and that will work but that's clunky and if files move you'd have to update your relative paths.
I've tried requireFrom and app-module-path with no luck as they are relative to a project root.
Any ideas for how to require files from outside of your project dir?
Thanks in advance!
var librarypath = '/Users/me/projects/library/';
// or if you prefer...
// var librarypath = '../../../../../library/';
var foo = require(librarypath + 'foo.js');
... or dressed up a bit more ...
function requirelib(lib){ return require('/Users/me/projects/library/'+lib+'.js'); }
var foo = requirelib('foo');
var bar = requirelib('bar');
I had the same problem many times. This can be solved by using the basetag npm package. It doesn't have to be required itself, only installed as it creates a symlink inside node_modules to your base path.
const localFile = require('$/local/file')
// instead of
const localFile = require('../../local/file')
Using the $/... prefix will always reference files relative to your apps root directory.
Disclaimer: I created basetag to solve this problem
Given the structure directory structure:
.
├── alpha.js
└── foo
└── beta.js
And file contents
alpha.js
module.exports = "alpha"
foo/beta.js
var cwd = process.cwd()
process.chdir('../')
var alpha = require('./alpha.js')
console.log(alpha)
process.chdir(cwd)
From within foo/beta.js. I'd like to be able to trick require into thinking that the current working directory is the project root. The example above does not work when the following is run.
node ./foo/beta.js
However if I switch over the code to within foo/beta.js to the following. Reading the file from the system and passing it to the npm module _eval.
updated foo/beta.js
var path = require('path')
var cwd = process.cwd()
process.chdir(path.join(__dirname, '..'))
var fs = require('fs')
var _eval = require('eval')
var alpha = _eval(fs.readFileSync('./alpha.js'))
console.log(alpha)
process.chdir(cwd)
This does work, which proves it should be possible with require as well. No matter where you run if from it will always require the file. node ./foo/beta.js or cd foo && node ./beta.js
Is there any way that I can prepend or set the directory that require uses from within the file?
From the node.js doc for require():
If the module identifier passed to require() is not a native module,
and does not begin with '/', '../', or './', then node starts at the
parent directory of the current module, and adds /node_modules, and
attempts to load the module from that location.
If it is not found there, then it moves to the parent directory, and
so on, until the root of the file system is reached.
From this, you can see that the current directory is not used in loading modules. The main takeaway here should be that modules paths are relative to the location of the current module. This allows modules to load sub-modules in a self-contained fashion without having to know anything about where the parent is placed in the directory structure.
Here's a work-around function that loads a module file descriptor from the current directory if the filename passed to it does not start with a path separator or a ..
var path = require('path');
function requireCWD(fname) {
var fullname = fname;
if (fname && fname.length &&
!path.isAbsolute(fname) &&
fname.charAt(0) !== '.') {
fullname = path.join(process.cwd(), fname);
}
return require(fullname);
}
Then, any filename you give to requireCWD that is not relative and does not start with a "." will be loaded relative to the current working directory. If you want to allow even "." and ".." to be relative to the current working directory, then you can remove that test for '.' from the function.
My project has got many folders and I often load my own modules in node.js in the following way:
var request = require("request"),
config = require("../../../modules/config"),
urls = require("../../../modules/urls");
I sometimes move the folders around and the path changes, so I need to adjust the ../ part manually.
I don't want to move my modules into the node_modules folder, but I'd like to load the modules in the following way:
var request = require("request"),
config = require("config"),
urls = require("urls");
or
var request = require("request"),
config = require("modules/config"),
urls = require("modules/urls");
What are my options?
New Answer:
Relative paths seem to be the simplest choice after all, it allows you to run your script from any location.
var config = require("../../config");
Old answer:
I found out that, while not ideal, there's also the possibility to use process.cwd().
var request = require("request"),
config = require(process.cwd() + "/modules/config");
or, if the process.cwd() is set to a global variable in the main js file
global.cwd = process.cwd();
then
var request = require("request"),
config = require(global.cwd + "/modules/config"),
urls = require(global.cwd + "/modules/urls");
You can try to do the following, based on some conditions
if the scripts are exclusively written for your application, meaning it won't work with any other application, and the scripts don't have any dependencies place them under modules directory and try to create expose a variable such as global.basepath and using path.join to construct the filepath and require it.You could also inject module variable after you require them at the main script of your app.
main.js
var config = require('modules/config.js')(_,mongoose,app);
modules/config.js
module.exports=function(_,mongoose,app){
return function(){
// _,mongoose,app
this.loadConfigVariable..
}
}
if the scripts are not one-files that have separate tests, require other vendor modules in order to be executed then create individual node modules and publish them to a registry for convenience when installing the main application.
Just as express does, there is the main application express and there are modules that express uses such as express-validation which in turn has its own dependencies.
You could add a symlink in node_modules, pointing to wherever your modules are.
For example, add a symlink named "my_modules", referencing '../foo/bar/my_modules'. Then, your requires will look like
var config = require('my_modules/config');
var urls = require('my_modules/urls');