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);
}
}
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');