Related
I want to use module-alias with esm. I have already found an answer here. The problem is that I am using it manually, like this:
import * as path from 'path';
import * as moduleAlias from 'module-alias';
moduleAlias.addAlias('#', path.join(process.cwd(), 'dist', 'server'));
How can I fix it if it is not directly called by module-alias/register but from this code?
I solved the problem. To do it I just stop using module-alias which is at this point an obsolete npm package (+3 year of inactivity as of 19/07/2022) and useful only for the commonjs module resolution.
Citing the right answer from this github error, the solution is creating a file custom-loader.mjs and add it as a loader when calling node
import path from 'node:path';
export default function loadAliases(aliasesToAdd) {
const getAliases = () => {
const base = process.cwd();
const absoluteAliases = Object.keys(aliasesToAdd).reduce((acc, key) =>
aliasesToAdd[key][0] === '/'
? acc
: { ...acc, [key]: path.join(base, aliasesToAdd[key]) },
aliasesToAdd)
return absoluteAliases;
}
const isAliasInSpecifier = (path, alias) => {
return path.indexOf(alias) === 0
&& (path.length === alias.length || path[alias.length] === '/')
}
const aliases = getAliases();
return (specifier, parentModuleURL, defaultResolve) => {
const alias = Object.keys(aliases).find((key) => isAliasInSpecifier(specifier, key));
const newSpecifier = alias === undefined
? specifier
: path.join(aliases[alias], specifier.substr(alias.length));
return defaultResolve(newSpecifier, parentModuleURL);
}
}
export const resolve = loadAliases({
"#": "./dist/source",
"#src": "./dist/source",
"#test": "./dist/test"
});
Then, when calling the script, add --loader=./custom-loader.mjs
node --no-warnings --loader=./custom-loader.mjs myscript.js
UPDATE: I created this npm module to automatically take care of this.
Let's say I have a file called app.js. Pretty simple:
var express = require('express');
var app = express.createServer();
app.set('views', __dirname + '/views');
app.set('view engine', 'ejs');
app.get('/', function(req, res){
res.render('index', {locals: {
title: 'NowJS + Express Example'
}});
});
app.listen(8080);
What if I have a functions inside "tools.js". How would I import them to use in apps.js?
Or...am I supposed to turn "tools" into a module, and then require it? << seems hard, I rather do the basic import of the tools.js file.
You can require any js file, you just need to declare what you want to expose.
// tools.js
// ========
module.exports = {
foo: function () {
// whatever
},
bar: function () {
// whatever
}
};
var zemba = function () {
}
And in your app file:
// app.js
// ======
var tools = require('./tools');
console.log(typeof tools.foo); // => 'function'
console.log(typeof tools.bar); // => 'function'
console.log(typeof tools.zemba); // => undefined
If, despite all the other answers, you still want to traditionally include a file in a node.js source file, you can use this:
var fs = require('fs');
// file is included here:
eval(fs.readFileSync('tools.js')+'');
The empty string concatenation +'' is necessary to get the file content as a string and not an object (you can also use .toString() if you prefer).
The eval() can't be used inside a function and must be called inside the global scope otherwise no functions or variables will be accessible (i.e. you can't create a include() utility function or something like that).
Please note that in most cases this is bad practice and you should instead write a module. However, there are rare situations, where pollution of your local context/namespace is what you really want.
Update 2015-08-06
Please also note this won't work with "use strict"; (when you are in "strict mode") because functions and variables defined in the "imported" file can't be accessed by the code that does the import. Strict mode enforces some rules defined by newer versions of the language standard. This may be another reason to avoid the solution described here.
You need no new functions nor new modules.
You simply need to execute the module you're calling if you don't want to use namespace.
in tools.js
module.exports = function() {
this.sum = function(a,b) { return a+b };
this.multiply = function(a,b) { return a*b };
//etc
}
in app.js
or in any other .js like myController.js :
instead of
var tools = require('tools.js') which force us to use a namespace and call tools like tools.sum(1,2);
we can simply call
require('tools.js')();
and then
sum(1,2);
in my case I have a file with controllers ctrls.js
module.exports = function() {
this.Categories = require('categories.js');
}
and I can use Categories in every context as public class after require('ctrls.js')()
Create two js files
// File cal.js
module.exports = {
sum: function(a,b) {
return a+b
},
multiply: function(a,b) {
return a*b
}
};
Main js file
// File app.js
var tools = require("./cal.js");
var value = tools.sum(10,20);
console.log("Value: "+value);
Console Output
Value: 30
create two files e.g app.js and tools.js
app.js
const tools= require("./tools.js")
var x = tools.add(4,2) ;
var y = tools.subtract(4,2);
console.log(x);
console.log(y);
tools.js
const add = function(x, y){
return x+y;
}
const subtract = function(x, y){
return x-y;
}
module.exports ={
add,subtract
}
output
6
2
Here is a plain and simple explanation:
Server.js content:
// Include the public functions from 'helpers.js'
var helpers = require('./helpers');
// Let's assume this is the data which comes from the database or somewhere else
var databaseName = 'Walter';
var databaseSurname = 'Heisenberg';
// Use the function from 'helpers.js' in the main file, which is server.js
var fullname = helpers.concatenateNames(databaseName, databaseSurname);
Helpers.js content:
// 'module.exports' is a node.JS specific feature, it does not work with regular JavaScript
module.exports =
{
// This is the function which will be called in the main file, which is server.js
// The parameters 'name' and 'surname' will be provided inside the function
// when the function is called in the main file.
// Example: concatenameNames('John,'Doe');
concatenateNames: function (name, surname)
{
var wholeName = name + " " + surname;
return wholeName;
},
sampleFunctionTwo: function ()
{
}
};
// Private variables and functions which will not be accessible outside this file
var privateFunction = function ()
{
};
I was also looking for a NodeJS 'include' function and I checked the solution proposed by Udo G - see message https://stackoverflow.com/a/8744519/2979590. His code doesn't work with my included JS files.
Finally I solved the problem like that:
var fs = require("fs");
function read(f) {
return fs.readFileSync(f).toString();
}
function include(f) {
eval.apply(global, [read(f)]);
}
include('somefile_with_some_declarations.js');
Sure, that helps.
Create two JavaScript files. E.g. import_functions.js and main.js
1.) import_functions.js
// Declaration --------------------------------------
module.exports =
{
add,
subtract
// ...
}
// Implementation ----------------------------------
function add(x, y)
{
return x + y;
}
function subtract(x, y)
{
return x - y;
}
// ...
2.) main.js
// include ---------------------------------------
const sf= require("./import_functions.js")
// use -------------------------------------------
var x = sf.add(4,2);
console.log(x);
var y = sf.subtract(4,2);
console.log(y);
output
6
2
The vm module in Node.js provides the ability to execute JavaScript code within the current context (including global object). See http://nodejs.org/docs/latest/api/vm.html#vm_vm_runinthiscontext_code_filename
Note that, as of today, there's a bug in the vm module that prevenst runInThisContext from doing the right when invoked from a new context. This only matters if your main program executes code within a new context and then that code calls runInThisContext. See https://github.com/joyent/node/issues/898
Sadly, the with(global) approach that Fernando suggested doesn't work for named functions like "function foo() {}"
In short, here's an include() function that works for me:
function include(path) {
var code = fs.readFileSync(path, 'utf-8');
vm.runInThisContext(code, path);
}
say we wants to call function ping() and add(30,20) which is in lib.js file
from main.js
main.js
lib = require("./lib.js")
output = lib.ping();
console.log(output);
//Passing Parameters
console.log("Sum of A and B = " + lib.add(20,30))
lib.js
this.ping=function ()
{
return "Ping Success"
}
//Functions with parameters
this.add=function(a,b)
{
return a+b
}
Udo G. said:
The eval() can't be used inside a function and must be called inside
the global scope otherwise no functions or variables will be
accessible (i.e. you can't create a include() utility function or
something like that).
He's right, but there's a way to affect the global scope from a function. Improving his example:
function include(file_) {
with (global) {
eval(fs.readFileSync(file_) + '');
};
};
include('somefile_with_some_declarations.js');
// the declarations are now accessible here.
Hope, that helps.
app.js
let { func_name } = require('path_to_tools.js');
func_name(); //function calling
tools.js
let func_name = function() {
...
//function body
...
};
module.exports = { func_name };
It worked with me like the following....
Lib1.js
//Any other private code here
// Code you want to export
exports.function1 = function(params) {.......};
exports.function2 = function(params) {.......};
// Again any private code
now in the Main.js file you need to include Lib1.js
var mylib = requires('lib1.js');
mylib.function1(params);
mylib.function2(params);
Please remember to put the Lib1.js in node_modules folder.
Another way to do this in my opinion, is to execute everything in the lib file when you call require() function using (function(/* things here */){})(); doing this will make all these functions global scope, exactly like the eval() solution
src/lib.js
(function () {
funcOne = function() {
console.log('mlt funcOne here');
}
funcThree = function(firstName) {
console.log(firstName, 'calls funcThree here');
}
name = "Mulatinho";
myobject = {
title: 'Node.JS is cool',
funcFour: function() {
return console.log('internal funcFour() called here');
}
}
})();
And then in your main code you can call your functions by name like:
main.js
require('./src/lib')
funcOne();
funcThree('Alex');
console.log(name);
console.log(myobject);
console.log(myobject.funcFour());
Will make this output
bash-3.2$ node -v
v7.2.1
bash-3.2$ node main.js
mlt funcOne here
Alex calls funcThree here
Mulatinho
{ title: 'Node.JS is cool', funcFour: [Function: funcFour] }
internal funcFour() called here
undefined
Pay atention to the undefined when you call my object.funcFour(), it will be the same if you load with eval(). Hope it helps :)
You can put your functions in global variables, but it's better practice to just turn your tools script into a module. It's really not too hard – just attach your public API to the exports object. Take a look at Understanding Node.js' exports module for some more detail.
I just want to add, in case you need just certain functions imported from your tools.js, then you can use a destructuring assignment which is supported in node.js since version 6.4 - see node.green.
Example:
(both files are in the same folder)
tools.js
module.exports = {
sum: function(a,b) {
return a + b;
},
isEven: function(a) {
return a % 2 == 0;
}
};
main.js
const { isEven } = require('./tools.js');
console.log(isEven(10));
output: true
This also avoids that you assign those functions as properties of another object as its the case in the following (common) assignment:
const tools = require('./tools.js');
where you need to call tools.isEven(10).
NOTE:
Don't forget to prefix your file name with the correct path - even if both files are in the same folder, you need to prefix with ./
From Node.js docs:
Without a leading '/', './', or '../' to indicate a file, the module
must either be a core module or is loaded from a node_modules folder.
Include file and run it in given (non-global) context
fileToInclude.js
define({
"data": "XYZ"
});
main.js
var fs = require("fs");
var vm = require("vm");
function include(path, context) {
var code = fs.readFileSync(path, 'utf-8');
vm.runInContext(code, vm.createContext(context));
}
// Include file
var customContext = {
"define": function (data) {
console.log(data);
}
};
include('./fileToInclude.js', customContext);
Using the ESM module system:
a.js:
export default function foo() {};
export function bar() {};
b.js:
import foo, {bar} from './a.js';
This is the best way i have created so far.
var fs = require('fs'),
includedFiles_ = {};
global.include = function (fileName) {
var sys = require('sys');
sys.puts('Loading file: ' + fileName);
var ev = require(fileName);
for (var prop in ev) {
global[prop] = ev[prop];
}
includedFiles_[fileName] = true;
};
global.includeOnce = function (fileName) {
if (!includedFiles_[fileName]) {
include(fileName);
}
};
global.includeFolderOnce = function (folder) {
var file, fileName,
sys = require('sys'),
files = fs.readdirSync(folder);
var getFileName = function(str) {
var splited = str.split('.');
splited.pop();
return splited.join('.');
},
getExtension = function(str) {
var splited = str.split('.');
return splited[splited.length - 1];
};
for (var i = 0; i < files.length; i++) {
file = files[i];
if (getExtension(file) === 'js') {
fileName = getFileName(file);
try {
includeOnce(folder + '/' + file);
} catch (err) {
// if (ext.vars) {
// console.log(ext.vars.dump(err));
// } else {
sys.puts(err);
// }
}
}
}
};
includeFolderOnce('./extensions');
includeOnce('./bin/Lara.js');
var lara = new Lara();
You still need to inform what you want to export
includeOnce('./bin/WebServer.js');
function Lara() {
this.webServer = new WebServer();
this.webServer.start();
}
Lara.prototype.webServer = null;
module.exports.Lara = Lara;
You can simple just require('./filename').
Eg.
// file: index.js
var express = require('express');
var app = express();
var child = require('./child');
app.use('/child', child);
app.get('/', function (req, res) {
res.send('parent');
});
app.listen(process.env.PORT, function () {
console.log('Example app listening on port '+process.env.PORT+'!');
});
// file: child.js
var express = require('express'),
child = express.Router();
console.log('child');
child.get('/child', function(req, res){
res.send('Child2');
});
child.get('/', function(req, res){
res.send('Child');
});
module.exports = child;
Please note that:
you can't listen PORT on the child file, only parent express module has PORT listener
Child is using 'Router', not parent Express moudle.
Node works based on commonjs modules and more recently, esm modules. Basically, you should create modules in separated .js files and make use of imports/exports (module.exports and require).
Javascript on the browser works differently, based on scope. There is the global scope, and through clojures (functions inside other functions) you have private scopes.
So,in node, export functions and objects that you will consume in other modules.
The cleanest way IMO is the following, In tools.js:
function A(){
.
.
.
}
function B(){
.
.
.
}
module.exports = {
A,
B
}
Then, in app.js, just require the tools.js as following: const tools = require("tools");
I was as well searching for an option to include code without writing modules, resp. use the same tested standalone sources from a different project for a Node.js service - and jmparattes answer did it for me.
The benefit is, you don't pollute the namespace, I don't have trouble with "use strict"; and it works well.
Here a full sample:
Script to load - /lib/foo.js
"use strict";
(function(){
var Foo = function(e){
this.foo = e;
}
Foo.prototype.x = 1;
return Foo;
}())
SampleModule - index.js
"use strict";
const fs = require('fs');
const path = require('path');
var SampleModule = module.exports = {
instAFoo: function(){
var Foo = eval.apply(
this, [fs.readFileSync(path.join(__dirname, '/lib/foo.js')).toString()]
);
var instance = new Foo('bar');
console.log(instance.foo); // 'bar'
console.log(instance.x); // '1'
}
}
Hope this was helpfull somehow.
Like you are having a file abc.txt and many more?
Create 2 files: fileread.js and fetchingfile.js, then in fileread.js write this code:
function fileread(filename) {
var contents= fs.readFileSync(filename);
return contents;
}
var fs = require("fs"); // file system
//var data = fileread("abc.txt");
module.exports.fileread = fileread;
//data.say();
//console.log(data.toString());
}
In fetchingfile.js write this code:
function myerror(){
console.log("Hey need some help");
console.log("type file=abc.txt");
}
var ags = require("minimist")(process.argv.slice(2), { string: "file" });
if(ags.help || !ags.file) {
myerror();
process.exit(1);
}
var hello = require("./fileread.js");
var data = hello.fileread(ags.file); // importing module here
console.log(data.toString());
Now, in a terminal:
$ node fetchingfile.js --file=abc.txt
You are passing the file name as an argument, moreover include all files in readfile.js instead of passing it.
Thanks
Another method when using node.js and express.js framework
var f1 = function(){
console.log("f1");
}
var f2 = function(){
console.log("f2");
}
module.exports = {
f1 : f1,
f2 : f2
}
store this in a js file named s and in the folder statics
Now to use the function
var s = require('../statics/s');
s.f1();
s.f2();
To turn "tools" into a module, I don't see hard at all. Despite all the other answers I would still recommend use of module.exports:
//util.js
module.exports = {
myFunction: function () {
// your logic in here
let message = "I am message from myFunction";
return message;
}
}
Now we need to assign this exports to global scope (in your app|index|server.js )
var util = require('./util');
Now you can refer and call function as:
//util.myFunction();
console.log(util.myFunction()); // prints in console :I am message from myFunction
To interactively test the module ./test.js in a Unix environment, something like this could be used:
>> node -e "eval(''+require('fs').readFileSync('./test.js'))" -i
...
Use:
var mymodule = require("./tools.js")
app.js:
module.exports.<your function> = function () {
<what should the function do>
}
I am trying copy my vendor files to my dev folder using gulp. When I was in development mode, I want copy only the unminified files, if unminified is not present copy minified files. And in production mode I want copy minifed files if files are not present minify the normal files.
my folder structure
js
app.js
jquery
jquery.min.js
jquery.js
fontawesome
fontawesome.min.js
fontawesome.min.css
fonts.ttf...
Here my basic I had written.
var scriptsPath = '../vendor/';
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
gulp.task('vendor', function() {
var folders = getFolders(scriptsPath);
var cssFilter = $.filter('**/*.css')
var tasks = folders.map(function(folder) {
var jsFilter;
if (isProduction) {
jsFilter = $.filter('**/*.min.js');
} else {
jsFilter = $.filter(['**/*.js', '!**/*.min.js']);
}
return gulp.src(path.join(scriptsPath, '**/'))
.pipe(jsFilter)
.pipe($.if(useSourceMaps, $.sourcemaps.init()))
.pipe($.if(isProduction, $.uglify({preserveComments: 'some'})))
.on('error', handleError)
.pipe(jsFilter.restore())
.pipe(cssFilter)
.pipe($.if( isProduction, $.minifyCss() ))
.on('error', handleError)
.pipe(cssFilter.restore())
.on('error', handleError)
.pipe(gulp.dest(build.vendor.js));
});
return es.concat.apply(null, tasks);
});
I am trying the last two days using gulp-if& some methods. But not yet get the solution.Thanks in advance.
You are trying to cram way to much into your vendor task. The stuff you do with your JS files is completely unrelated to the stuff you do with your CSS files. That's hard to read.
Instead of using gulp-filter try splitting vendor up into smaller tasks like vendor-js, vendor-css, etc... and then declare them as dependencies for your vendor task:
gulp.task('vendor', ['vendor-js', 'vendor-css' /* etc ... */]);
Your vendor-js task could then look like this:
var glob = require('glob');
gulp.task('vendor-js', function () {
var js = glob.sync('../vendor/**/*.js');
if (isProduction) {
// use <file>.min.js, unless there is only <file>.js
js = js.filter(function(file) {
return file.match(/\.min\.js$/) ||
js.indexOf(file.replace(/\.js$/, '.min.js')) < 0;
});
} else {
// use <file>.js, unless there is only <file>.min.js
js = js.filter(function(file) {
return !file.match(/\.min\.js$/) ||
js.indexOf(file.replace(/\.min\.js$/, '.js')) < 0;
});
}
gulp.src(js, { base: '../vendor' })
.pipe($.if(isProduction, // only minify for prod and when
$.if("!**/*.min.js", uglify()))) // the file isn't minified already
.pipe(gulp.dest('build'));
});
Adapting this to you specific needs should be fairly trivial from here on.
I'm trying to create a full path if it doesn't exist.
The code looks like this:
var fs = require('fs');
if (!fs.existsSync(newDest)) fs.mkdirSync(newDest);
This code works great as long as there is only one subdirectory (a newDest like 'dir1') however when there is a directory path like ('dir1/dir2') it fails with
Error: ENOENT, no such file or directory
I'd like to be able to create the full path with as few lines of code as necessary.
I read there is a recursive option on fs and tried it like this
var fs = require('fs');
if (!fs.existsSync(newDest)) fs.mkdirSync(newDest,'0777', true);
I feel like it should be that simple to recursively create a directory that doesn't exist. Am I missing something or do I need to parse the path and check each directory and create it if it doesn't already exist?
I'm pretty new to Node. Maybe I'm using an old version of FS?
Update
NodeJS version 10.12.0 has added a native support for both mkdir and mkdirSync to create a directory recursively with recursive: true option as the following:
fs.mkdirSync(targetDir, { recursive: true });
And if you prefer fs Promises API, you can write
fs.promises.mkdir(targetDir, { recursive: true });
Original Answer
Create directories recursively if they do not exist! (Zero dependencies)
const fs = require('fs');
const path = require('path');
function mkDirByPathSync(targetDir, { isRelativeToScript = false } = {}) {
const sep = path.sep;
const initDir = path.isAbsolute(targetDir) ? sep : '';
const baseDir = isRelativeToScript ? __dirname : '.';
return targetDir.split(sep).reduce((parentDir, childDir) => {
const curDir = path.resolve(baseDir, parentDir, childDir);
try {
fs.mkdirSync(curDir);
} catch (err) {
if (err.code === 'EEXIST') { // curDir already exists!
return curDir;
}
// To avoid `EISDIR` error on Mac and `EACCES`-->`ENOENT` and `EPERM` on Windows.
if (err.code === 'ENOENT') { // Throw the original parentDir error on curDir `ENOENT` failure.
throw new Error(`EACCES: permission denied, mkdir '${parentDir}'`);
}
const caughtErr = ['EACCES', 'EPERM', 'EISDIR'].indexOf(err.code) > -1;
if (!caughtErr || caughtErr && curDir === path.resolve(targetDir)) {
throw err; // Throw if it's just the last created dir.
}
}
return curDir;
}, initDir);
}
Usage
// Default, make directories relative to current working directory.
mkDirByPathSync('path/to/dir');
// Make directories relative to the current script.
mkDirByPathSync('path/to/dir', {isRelativeToScript: true});
// Make directories with an absolute path.
mkDirByPathSync('/path/to/dir');
Demo
Try It!
Explanations
[UPDATE] This solution handles platform-specific errors like EISDIR for Mac and EPERM and EACCES for Windows. Thanks to all the reporting comments by #PediT., #JohnQ, #deed02392, #robyoder and #Almenon.
This solution handles both relative and absolute paths. Thanks to #john comment.
In the case of relative paths, target directories will be created (resolved) in the current working directory. To Resolve them relative to the current script dir, pass {isRelativeToScript: true}.
Using path.sep and path.resolve(), not just / concatenation, to avoid cross-platform issues.
Using fs.mkdirSync and handling the error with try/catch if thrown to handle race conditions: another process may add the file between the calls to fs.existsSync() and fs.mkdirSync() and causes an exception.
The other way to achieve that could be checking if a file exists then creating it, I.e, if (!fs.existsSync(curDir) fs.mkdirSync(curDir);. But this is an anti-pattern that leaves the code vulnerable to race conditions. Thanks to #GershomMaes comment about the directory existence check.
Requires Node v6 and newer to support destructuring. (If you have problems implementing this solution with old Node versions, just leave me a comment)
A more robust answer is to use use mkdirp.
var mkdirp = require('mkdirp');
mkdirp('/path/to/dir', function (err) {
if (err) console.error(err)
else console.log('dir created')
});
Then proceed to write the file into the full path with:
fs.writeFile ('/path/to/dir/file.dat'....
One option is to use shelljs module
npm install shelljs
var shell = require('shelljs');
shell.mkdir('-p', fullPath);
From that page:
Available options:
p: full path (will create intermediate dirs if necessary)
As others have noted, there's other more focused modules. But, outside of mkdirp, it has tons of other useful shell operations (like which, grep etc...) and it works on windows and *nix
Edit: comments suggest this doesn't work on systems that don't have mkdir cli instances. That is not the case. That's the point shelljs - create a portable cross platform set of shell like functions. It works on even windows.
fs-extra adds file system methods that aren't included in the native fs module. It is a drop in replacement for fs.
Install fs-extra
$ npm install --save fs-extra
const fs = require("fs-extra");
// Make sure the output directory is there.
fs.ensureDirSync(newDest);
There are sync and async options.
https://github.com/jprichardson/node-fs-extra/blob/master/docs/ensureDir.md
Using reduce we can verify if each path exists and create it if necessary, also this way I think it is easier to follow. Edited, thanks #Arvin, we should use path.sep to get the proper platform-specific path segment separator.
const path = require('path');
// Path separators could change depending on the platform
const pathToCreate = 'path/to/dir';
pathToCreate
.split(path.sep)
.reduce((prevPath, folder) => {
const currentPath = path.join(prevPath, folder, path.sep);
if (!fs.existsSync(currentPath)){
fs.mkdirSync(currentPath);
}
return currentPath;
}, '');
This feature has been added to node.js in version 10.12.0, so it's as easy as passing an option {recursive: true} as second argument to the fs.mkdir() call.
See the example in the official docs.
No need for external modules or your own implementation.
i know this is an old question, but nodejs v10.12.0 now supports this natively with the recursive option set to true. fs.mkdir
// Creates /tmp/a/apple, regardless of whether `/tmp` and /tmp/a exist.
fs.mkdir('/tmp/a/apple', { recursive: true }, (err) => {
if (err) throw err;
});
Now with NodeJS >= 10.12.0, you can use fs.mkdirSync(path, { recursive: true }) fs.mkdirSync
Example for Windows (no extra dependencies and error handling)
const path = require('path');
const fs = require('fs');
let dir = "C:\\temp\\dir1\\dir2\\dir3";
function createDirRecursively(dir) {
if (!fs.existsSync(dir)) {
createDirRecursively(path.join(dir, ".."));
fs.mkdirSync(dir);
}
}
createDirRecursively(dir); //creates dir1\dir2\dir3 in C:\temp
You can simply check folder exist or not in path recursively and make the folder as you check if they are not present. (NO EXTERNAL LIBRARY)
function checkAndCreateDestinationPath (fileDestination) {
const dirPath = fileDestination.split('/');
dirPath.forEach((element, index) => {
if(!fs.existsSync(dirPath.slice(0, index + 1).join('/'))){
fs.mkdirSync(dirPath.slice(0, index + 1).join('/'));
}
});
}
You can use the next function
const recursiveUpload = (path: string) => {
const paths = path.split("/")
const fullPath = paths.reduce((accumulator, current) => {
fs.mkdirSync(accumulator)
return `${accumulator}/${current}`
})
fs.mkdirSync(fullPath)
return fullPath
}
So what it does:
Create paths variable, where it stores every path by itself as an element of the array.
Adds "/" at the end of each element in the array.
Makes for the cycle:
Creates a directory from the concatenation of array elements which indexes are from 0 to current iteration. Basically, it is recursive.
Hope that helps!
By the way, in Node v10.12.0 you can use recursive path creation by giving it as the additional argument.
fs.mkdir('/tmp/a/apple', { recursive: true }, (err) => {
if (err) throw err;
});
https://nodejs.org/api/fs.html#fs_fs_mkdirsync_path_options
Too many answers, but here's a solution without recursion that works by splitting the path and then left-to-right building it back up again
function mkdirRecursiveSync(path) {
let paths = path.split(path.delimiter);
let fullPath = '';
paths.forEach((path) => {
if (fullPath === '') {
fullPath = path;
} else {
fullPath = fullPath + '/' + path;
}
if (!fs.existsSync(fullPath)) {
fs.mkdirSync(fullPath);
}
});
};
For those concerned about windows vs Linux compatibility, simply replace the forward slash with double backslash '\' in both occurrence above but TBH we are talking about node fs not windows command line and the former is pretty forgiving and the above code will simply work on Windows and is more a complete solution cross platform.
const fs = require('fs');
try {
fs.mkdirSync(path, { recursive: true });
} catch (error) {
// this make script keep running, even when folder already exist
console.log(error);
}
An asynchronous way to create directories recursively:
import fs from 'fs'
const mkdirRecursive = function(path, callback) {
let controlledPaths = []
let paths = path.split(
'/' // Put each path in an array
).filter(
p => p != '.' // Skip root path indicator (.)
).reduce((memo, item) => {
// Previous item prepended to each item so we preserve realpaths
const prevItem = memo.length > 0 ? memo.join('/').replace(/\.\//g, '')+'/' : ''
controlledPaths.push('./'+prevItem+item)
return [...memo, './'+prevItem+item]
}, []).map(dir => {
fs.mkdir(dir, err => {
if (err && err.code != 'EEXIST') throw err
// Delete created directory (or skipped) from controlledPath
controlledPaths.splice(controlledPaths.indexOf(dir), 1)
if (controlledPaths.length === 0) {
return callback()
}
})
})
}
// Usage
mkdirRecursive('./photos/recent', () => {
console.log('Directories created succesfully!')
})
Here's my imperative version of mkdirp for nodejs.
function mkdirSyncP(location) {
let normalizedPath = path.normalize(location);
let parsedPathObj = path.parse(normalizedPath);
let curDir = parsedPathObj.root;
let folders = parsedPathObj.dir.split(path.sep);
folders.push(parsedPathObj.base);
for(let part of folders) {
curDir = path.join(curDir, part);
if (!fs.existsSync(curDir)) {
fs.mkdirSync(curDir);
}
}
}
How about this approach :
if (!fs.existsSync(pathToFile)) {
var dirName = "";
var filePathSplit = pathToFile.split('/');
for (var index = 0; index < filePathSplit.length; index++) {
dirName += filePathSplit[index]+'/';
if (!fs.existsSync(dirName))
fs.mkdirSync(dirName);
}
}
This works for relative path.
Based on mouneer's zero-dependencies answer, here's a slightly more beginner friendly Typescript variant, as a module:
import * as fs from 'fs';
import * as path from 'path';
/**
* Recursively creates directories until `targetDir` is valid.
* #param targetDir target directory path to be created recursively.
* #param isRelative is the provided `targetDir` a relative path?
*/
export function mkdirRecursiveSync(targetDir: string, isRelative = false) {
const sep = path.sep;
const initDir = path.isAbsolute(targetDir) ? sep : '';
const baseDir = isRelative ? __dirname : '.';
targetDir.split(sep).reduce((prevDirPath, dirToCreate) => {
const curDirPathToCreate = path.resolve(baseDir, prevDirPath, dirToCreate);
try {
fs.mkdirSync(curDirPathToCreate);
} catch (err) {
if (err.code !== 'EEXIST') {
throw err;
}
// caught EEXIST error if curDirPathToCreate already existed (not a problem for us).
}
return curDirPathToCreate; // becomes prevDirPath on next call to reduce
}, initDir);
}
As clean as this :)
function makedir(fullpath) {
let destination_split = fullpath.replace('/', '\\').split('\\')
let path_builder = destination_split[0]
$.each(destination_split, function (i, path_segment) {
if (i < 1) return true
path_builder += '\\' + path_segment
if (!fs.existsSync(path_builder)) {
fs.mkdirSync(path_builder)
}
})
}
I had issues with the recursive option of fs.mkdir so I made a function that does the following:
Creates a list of all directories, starting with the final target dir and working up to the root parent.
Creates a new list of needed directories for the mkdir function to work
Makes each directory needed, including the final
function createDirectoryIfNotExistsRecursive(dirname) {
return new Promise((resolve, reject) => {
const fs = require('fs');
var slash = '/';
// backward slashes for windows
if(require('os').platform() === 'win32') {
slash = '\\';
}
// initialize directories with final directory
var directories_backwards = [dirname];
var minimize_dir = dirname;
while (minimize_dir = minimize_dir.substring(0, minimize_dir.lastIndexOf(slash))) {
directories_backwards.push(minimize_dir);
}
var directories_needed = [];
//stop on first directory found
for(const d in directories_backwards) {
if(!(fs.existsSync(directories_backwards[d]))) {
directories_needed.push(directories_backwards[d]);
} else {
break;
}
}
//no directories missing
if(!directories_needed.length) {
return resolve();
}
// make all directories in ascending order
var directories_forwards = directories_needed.reverse();
for(const d in directories_forwards) {
fs.mkdirSync(directories_forwards[d]);
}
return resolve();
});
}
I solved the problem this way - similar to other recursive answers but to me this is much easier to understand and read.
const path = require('path');
const fs = require('fs');
function mkdirRecurse(inputPath) {
if (fs.existsSync(inputPath)) {
return;
}
const basePath = path.dirname(inputPath);
if (fs.existsSync(basePath)) {
fs.mkdirSync(inputPath);
}
mkdirRecurse(basePath);
}
Could not find an example to create dir's with the required permissions.
Create Directories async recursively with the permissions you want.
Heres a plain nodejs solution
node v18.12.1
Ubuntu 18
//-----------------------------
const fs = require('fs');
const fsPromises = fs.promises;
const checkDirAccess = async (userDir) => {
try {
await fsPromises.access(userDir, fs.constants.R_OK | fs.constants.W_OK);
console.log(` ${userDir} Dir existss`);
return userDir;
} catch (err) {
if(err.errno = -2)
return await crDir(userDir);
else
throw err;
}
}
const crDir = async (userDir) => {
try {
let newDir = await fsPromises.mkdir(userDir, { recursive: true, mode: 0o700});
// When userDir is created; newDir = undefined;
console.log(` Created new Dir ${newDir}`);
return newDir;
} catch (err) {
throw err;
}
}
const directoryPath = ['uploads/xvc/xvc/xvc/specs', 'uploads/testDir11', 'uploads/xsXa/', 'uploads//xsb//', 'uploads//xsV/'];
const findDir = async() => {
try {
for (const iterator of directoryPath) {
let dirOK = await checkDirAccess(iterator);
if(dirOK)
console.log(`found ${dirOK}`)
}
} catch (error) {
console.error('Error : ', error);
}
}
Exec can be messy on windows. There is a more "nodie" solution. Fundamentally, you have a recursive call to see if a directory exists and dive into the child (if it does exist) or create it. Here is a function that will create the children and call a function when finished:
fs = require('fs');
makedirs = function(path, func) {
var pth = path.replace(/['\\]+/g, '/');
var els = pth.split('/');
var all = "";
(function insertOne() {
var el = els.splice(0, 1)[0];
if (!fs.existsSync(all + el)) {
fs.mkdirSync(all + el);
}
all += el + "/";
if (els.length == 0) {
func();
} else {
insertOne();
}
})();
}
This version works better on Windows than the top answer because it understands both / and path.sep so that forward slashes work on Windows as they should. Supports absolute and relative paths (relative to the process.cwd).
/**
* Creates a folder and if necessary, parent folders also. Returns true
* if any folders were created. Understands both '/' and path.sep as
* path separators. Doesn't try to create folders that already exist,
* which could cause a permissions error. Gracefully handles the race
* condition if two processes are creating a folder. Throws on error.
* #param targetDir Name of folder to create
*/
export function mkdirSyncRecursive(targetDir) {
if (!fs.existsSync(targetDir)) {
for (var i = targetDir.length-2; i >= 0; i--) {
if (targetDir.charAt(i) == '/' || targetDir.charAt(i) == path.sep) {
mkdirSyncRecursive(targetDir.slice(0, i));
break;
}
}
try {
fs.mkdirSync(targetDir);
return true;
} catch (err) {
if (err.code !== 'EEXIST') throw err;
}
}
return false;
}
How do I require all files in a folder in node.js?
need something like:
files.forEach(function (v,k){
// require routes
require('./routes/'+v);
}};
When require is given the path of a folder, it'll look for an index.js file in that folder; if there is one, it uses that, and if there isn't, it fails.
It would probably make most sense (if you have control over the folder) to create an index.js file and then assign all the "modules" and then simply require that.
yourfile.js
var routes = require("./routes");
index.js
exports.something = require("./routes/something.js");
exports.others = require("./routes/others.js");
If you don't know the filenames you should write some kind of loader.
Working example of a loader:
var normalizedPath = require("path").join(__dirname, "routes");
require("fs").readdirSync(normalizedPath).forEach(function(file) {
require("./routes/" + file);
});
// Continue application logic here
I recommend using glob to accomplish that task.
var glob = require( 'glob' )
, path = require( 'path' );
glob.sync( './routes/**/*.js' ).forEach( function( file ) {
require( path.resolve( file ) );
});
Base on #tbranyen's solution, I create an index.js file that load arbitrary javascripts under current folder as part of the exports.
// Load `*.js` under current directory as properties
// i.e., `User.js` will become `exports['User']` or `exports.User`
require('fs').readdirSync(__dirname + '/').forEach(function(file) {
if (file.match(/\.js$/) !== null && file !== 'index.js') {
var name = file.replace('.js', '');
exports[name] = require('./' + file);
}
});
Then you can require this directory from any where else.
Another option is to use the package require-dir which let's you do the following. It supports recursion as well.
var requireDir = require('require-dir');
var dir = requireDir('./path/to/dir');
I have a folder /fields full of files with a single class each, ex:
fields/Text.js -> Test class
fields/Checkbox.js -> Checkbox class
Drop this in fields/index.js to export each class:
var collectExports, fs, path,
__hasProp = {}.hasOwnProperty;
fs = require('fs');
path = require('path');
collectExports = function(file) {
var func, include, _results;
if (path.extname(file) === '.js' && file !== 'index.js') {
include = require('./' + file);
_results = [];
for (func in include) {
if (!__hasProp.call(include, func)) continue;
_results.push(exports[func] = include[func]);
}
return _results;
}
};
fs.readdirSync('./fields/').forEach(collectExports);
This makes the modules act more like they would in Python:
var text = new Fields.Text()
var checkbox = new Fields.Checkbox()
One more option is require-dir-all combining features from most popular packages.
Most popular require-dir does not have options to filter the files/dirs and does not have map function (see below), but uses small trick to find module's current path.
Second by popularity require-all has regexp filtering and preprocessing, but lacks relative path, so you need to use __dirname (this has pros and contras) like:
var libs = require('require-all')(__dirname + '/lib');
Mentioned here require-index is quite minimalistic.
With map you may do some preprocessing, like create objects and pass config values (assuming modules below exports constructors):
// Store config for each module in config object properties
// with property names corresponding to module names
var config = {
module1: { value: 'config1' },
module2: { value: 'config2' }
};
// Require all files in modules subdirectory
var modules = require('require-dir-all')(
'modules', // Directory to require
{ // Options
// function to be post-processed over exported object for each require'd module
map: function(reqModule) {
// create new object with corresponding config passed to constructor
reqModule.exports = new reqModule.exports( config[reqModule.name] );
}
}
);
// Now `modules` object holds not exported constructors,
// but objects constructed using values provided in `config`.
I know this question is 5+ years old, and the given answers are good, but I wanted something a bit more powerful for express, so i created the express-map2 package for npm. I was going to name it simply express-map, however the people at yahoo already have a package with that name, so i had to rename my package.
1. basic usage:
app.js (or whatever you call it)
var app = require('express'); // 1. include express
app.set('controllers',__dirname+'/controllers/');// 2. set path to your controllers.
require('express-map2')(app); // 3. patch map() into express
app.map({
'GET /':'test',
'GET /foo':'middleware.foo,test',
'GET /bar':'middleware.bar,test'// seperate your handlers with a comma.
});
controller usage:
//single function
module.exports = function(req,res){
};
//export an object with multiple functions.
module.exports = {
foo: function(req,res){
},
bar: function(req,res){
}
};
2. advanced usage, with prefixes:
app.map('/api/v1/books',{
'GET /': 'books.list', // GET /api/v1/books
'GET /:id': 'books.loadOne', // GET /api/v1/books/5
'DELETE /:id': 'books.delete', // DELETE /api/v1/books/5
'PUT /:id': 'books.update', // PUT /api/v1/books/5
'POST /': 'books.create' // POST /api/v1/books
});
As you can see, this saves a ton of time and makes the routing of your application dead simple to write, maintain, and understand. it supports all of the http verbs that express supports, as well as the special .all() method.
npm package: https://www.npmjs.com/package/express-map2
github repo: https://github.com/r3wt/express-map
Expanding on this glob solution. Do this if you want to import all modules from a directory into index.js and then import that index.js in another part of the application. Note that template literals aren't supported by the highlighting engine used by stackoverflow so the code might look strange here.
const glob = require("glob");
let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
/* see note about this in example below */
allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;
Full Example
Directory structure
globExample/example.js
globExample/foobars/index.js
globExample/foobars/unexpected.js
globExample/foobars/barit.js
globExample/foobars/fooit.js
globExample/example.js
const { foo, bar, keepit } = require('./foobars/index');
const longStyle = require('./foobars/index');
console.log(foo()); // foo ran
console.log(bar()); // bar ran
console.log(keepit()); // keepit ran unexpected
console.log(longStyle.foo()); // foo ran
console.log(longStyle.bar()); // bar ran
console.log(longStyle.keepit()); // keepit ran unexpected
globExample/foobars/index.js
const glob = require("glob");
/*
Note the following style also works with multiple exports per file (barit.js example)
but will overwrite if you have 2 exports with the same
name (unexpected.js and barit.js have a keepit function) in the files being imported. As a result, this method is best used when
your exporting one module per file and use the filename to easily identify what is in it.
Also Note: This ignores itself (index.js) by default to prevent infinite loop.
*/
let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;
globExample/foobars/unexpected.js
exports.keepit = () => 'keepit ran unexpected';
globExample/foobars/barit.js
exports.bar = () => 'bar run';
exports.keepit = () => 'keepit ran';
globExample/foobars/fooit.js
exports.foo = () => 'foo ran';
From inside project with glob installed, run node example.js
$ node example.js
foo ran
bar run
keepit ran unexpected
foo ran
bar run
keepit ran unexpected
One module that I have been using for this exact use case is require-all.
It recursively requires all files in a given directory and its sub directories as long they don't match the excludeDirs property.
It also allows specifying a file filter and how to derive the keys of the returned hash from the filenames.
Require all files from routes folder and apply as middleware. No external modules needed.
// require
const { readdirSync } = require("fs");
// apply as middleware
readdirSync("./routes").map((r) => app.use("/api", require("./routes/" + r)));
I'm using node modules copy-to module to create a single file to require all the files in our NodeJS-based system.
The code for our utility file looks like this:
/**
* Module dependencies.
*/
var copy = require('copy-to');
copy(require('./module1'))
.and(require('./module2'))
.and(require('./module3'))
.to(module.exports);
In all of the files, most functions are written as exports, like so:
exports.function1 = function () { // function contents };
exports.function2 = function () { // function contents };
exports.function3 = function () { // function contents };
So, then to use any function from a file, you just call:
var utility = require('./utility');
var response = utility.function2(); // or whatever the name of the function is
Can use : https://www.npmjs.com/package/require-file-directory
Require selected files with name only or all files.
No need of absoulute path.
Easy to understand and use.
Using this function you can require a whole dir.
const GetAllModules = ( dirname ) => {
if ( dirname ) {
let dirItems = require( "fs" ).readdirSync( dirname );
return dirItems.reduce( ( acc, value, index ) => {
if ( PATH.extname( value ) == ".js" && value.toLowerCase() != "index.js" ) {
let moduleName = value.replace( /.js/g, '' );
acc[ moduleName ] = require( `${dirname}/${moduleName}` );
}
return acc;
}, {} );
}
}
// calling this function.
let dirModules = GetAllModules(__dirname);
Create an index.js file in your folder with this code :
const fs = require('fs')
const files = fs.readdirSync('./routes')
for (const file of files) {
require('./'+file)
}
And after that you can simply load all the folder with require("./routes")
If you include all files of *.js in directory example ("app/lib/*.js"):
In directory app/lib
example.js:
module.exports = function (example) { }
example-2.js:
module.exports = function (example2) { }
In directory app create index.js
index.js:
module.exports = require('./app/lib');