Is it possible, given a loaded module, to get its file path?
const MyModule = require('./MyModule');
const MyOtherModule = require('../otherfolder/MyOtherModule');
function print(){
console.log(thisIsThePathTo(MyModule)); <--- Should print the absolute path of the loaded module
console.log(thisIsThePathTo(MyOtherModule)); <--- Should print the absolute path of the loaded module
}
I saw require.resolve but I need the opposite lookup...
Any ideas?
Thanks!
The documentation for require.main describes the module object.
The module has an id and a path, however those are not exported. You can add those properties to the module.exports object to export them. Then, in a separate module, you can access them via MyOtherModule.id or MyOtherModule.path
For example,
In MyOtherModule/index.js:
myOtherModuleFunction = function() {
console.log('This is module 2')
}
module.exports = {
// spread all properties in module.exports
...module,
// then add the exports
exports: myOtherModuleFunction
}
and in MyModule/MyModule.js,
module.exports = {
...module,
exports: { someFunction: () => console.log('MyModule') }
}
and in MyModule/index.js:
const MyModule = require('./MyModule');
const MyOtherModule = require('../../MyOtherModule/');
function thisIsThePathTo(module) {
return module.path
}
function print(){
console.log(thisIsThePathTo(MyModule))
console.log(thisIsThePathTo(MyOtherModule))
}
print()
Running node src/MyModule/index.js outputs:
/.../stackoverflow/62043302/src/MyModule/
/.../stackoverflow/62043302/src/MyOtherModule
And if you print module.id instead of module.path, you'll get:
/.../stackoverflow/62043302/src/MyModule/index.js
/.../stackoverflow/62043302/src/MyOtherModule/index.js
However, spreading all properties includes module.children and module.parent, and you'll also have to use module.exports when accessing the so you probably only want to include id or path, like so:
myOtherModuleFunction = function() {
console.log('This is module 2')
}
const { id, path } = module
module.exports = {
id,
path,
myOtherModuleFunction,
}```
and require like so:
```js
const {id: otherModuleId, myOtherModuleFunction } = require('MyOtherModule')
This can get messy. If you're importing modules you did not author, you will not have the option to lookup the id or path (unless the authors added it to module.exports).
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>
}
Let's say I have two files, dir/a.js and lib/b.js
a.js:
b = require('../lib/b');
b.someFn();
b.js:
var fallback = "./config.json";
module.exports = {
someFn = function(jsonFile) {
console.log(require(jsonFile || fallback);
}
}
The entire purpose of b.js in this example is to read a json file, so I might call it as b.someFn("path/to/file.json").
But I want there to be a default, like a config file. But the default should be relative to a.js and not b.js. In other words, I should be able to call b.someFn() from a.js, and it should say, "since you didn't pass me the path, I will assume a default path of config.json." But the default should be relative to a.js, i.e. should be dir/config.json and not lib/config.json, which I would get if I did require(jsonFile).
I could get the cwd, but that will only work if I launch the script from within dir/.
Is there any way for b.js to say, inside someFn(), "give me the __dirname of the function that called me?"
Use callsite, then:
b.js:
var path = require('path'),
callsite = require('callsite');
module.exports = {
someFn: function () {
var stack = callsite(),
requester = stack[1].getFileName();
console.log(path.dirname(requester));
}
};
Alternatively, using parent-module:
const path = require('path');
const parentModule = require('parent-module');
// get caller of current script
console.log(path.dirname(parentModule()));
// get caller of module, change './index.js' to your "main" script
console.log(path.dirname(parentModule(require.resolve('./index.js'))));
If you want to get the directory of the script of the caller function, then use the stacktrace as the above answer shows, otherwise, what's the problem of hardcoding the directory of a.js?
var fallback = "dir_a/config.json";
module.exports = {
someFn = function(jsonFile) {
console.log(require(jsonFile || fallback);
}
}
maybe this question is a little silly, but is it possible to load multiple .js files with one require statement? like this:
var mylib = require('./lib/mylibfiles');
and use:
mylib.foo(); //return "hello from one"
mylib.bar(): //return "hello from two"
And in the folder mylibfiles will have two files:
One.js
exports.foo= function(){return "hello from one";}
Two.js
exports.bar= function(){return "hello from two";}
I was thinking to put a package.json in the folder that say to load all the files, but I don't know how. Other aproach that I was thinking is to have a index.js that exports everything again but I will be duplicating work.
Thanks!!
P.D: I'm working with nodejs v0.611 on a windows 7 machine
First of all using require does not duplicate anything. It loads the module and it caches it, so calling require again will get it from memory (thus you can modify module at fly without interacting with its source code - this is sometimes desirable, for example when you want to store db connection inside module).
Also package.json does not load anything and does not interact with your app at all. It is only used for npm.
Now you cannot require multiple modules at once. For example what will happen if both One.js and Two.js have defined function with the same name?? There are more problems.
But what you can do, is to write additional file, say modules.js with the following content
module.exports = {
one : require('./one.js'),
two : require('./two.js'),
/* some other modules you want */
}
and then you can simply use
var modules = require('./modules.js');
modules.one.foo();
modules.two.bar();
I have a snippet of code that requires more than one module, but it doesn't clump them together as your post suggests. However, that can be overcome with a trick that I found.
function requireMany () {
return Array.prototype.slice.call(arguments).map(function (value) {
try {
return require(value)
}
catch (event) {
return console.log(event)
}
})
}
And you use it as such
requireMany("fs", "socket.io", "path")
Which will return
[ fs {}, socketio {}, path {} ]
If a module is not found, an error will be sent to the console. It won't break the programme. The error will be shown in the array as undefined. The array will not be shorter because one of the modules failed to load.
Then you can bind those each of those array elements to a variable name, like so:
var [fs, socketio, path] = requireMany("fs", "socket.io", "path")
It essentially works like an object, but assigns the keys and their values to the global namespace. So, in your case, you could do:
var [foo, bar] = requireMany("./foo.js", "./bar.js")
foo() //return "hello from one"
bar() //return "hello from two"
And if you do want it to break the programme on error, just use this modified version, which is smaller
function requireMany () {
return Array.prototype.slice.call(arguments).map(require)
}
Yes, you may require a folder as a module, according to the node docs. Let's say you want to require() a folder called ./mypack/.
Inside ./mypack/, create a package.json file with the name of the folder and a main javascript file with the same name, inside a ./lib/ directory.
{
"name" : "mypack",
"main" : "./lib/mypack.js"
}
Now you can use require('./mypack') and node will load ./mypack/lib/mypack.js.
However if you do not include this package.json file, it may still work. Without the file, node will attempt to load ./mypack/index.js, or if that's not there, ./mypack/index.node.
My understanding is that this could be beneficial if you have split your program into many javascript files but do not want to concatenate them for deployment.
You can use destructuring assignment to map an array of exported modules from require statements in one line:
const requires = (...modules) => modules.map(module => require(module));
const [fs, path] = requires('fs', 'path');
I was doing something similar to what #freakish suggests in his answer with a project where I've a list of test scripts that are pulled into a Puppeteer + Jest testing setup. My test files follow the naming convention testname1.js - testnameN.js and I was able use a generator function to require N number of files from the particular directory with the approach below:
const fs = require('fs');
const path = require('path');
module.exports = class FilesInDirectory {
constructor(directory) {
this.fid = fs.readdirSync(path.resolve(directory));
this.requiredFiles = (this.fid.map((fileId) => {
let resolvedPath = path.resolve(directory, fileId);
return require(resolvedPath);
})).filter(file => !!file);
}
printRetrievedFiles() {
console.log(this.requiredFiles);
}
nextFileGenerator() {
const parent = this;
const fidLength = parent.requiredFiles.length;
function* iterate(index) {
while (index < fidLength) {
yield parent.requiredFiles[index++];
}
}
return iterate(0);
}
}
Then use like so:
//Use in test
const FilesInDirectory = require('./utilities/getfilesindirectory');
const StepsCollection = new FilesInDirectory('./test-steps');
const StepsGenerator = StepsCollection.nextFileGenerator();
//Assuming we're in an async function
await StepsGenerator.next().value.FUNCTION_REQUIRED_FROM_FILE(someArg);