How to use broswerified node module into another JavaScript file - javascript

I have a node module file which I have broswerifed using browserify
// myMath.js file
// --------------
module.exports = {
add: function(a, b){
return a + b;
}
}
Now to use it in another JavaScript application, I am using following code
// index.js file
// -------------
var myMath = require("./myMath");
var result = myMath.add(1, 2);
console.log("the result is", result);
Now I am using broswerify:
browserify index.js -o app.js
But this apporach requires involving index.js file as well. Is there any way to directly include broswerified myMath.js file into my JavaScript application ??

Related

Error: requireJS config is not a function

My Project structure looks like this -
src
|- main.js
|- math.js
math.js is just an AMD module and I am requiring it in main.js
I have installed require.js using npm and requiring it in main.js
//main.js
var rjs = require('requirejs');
rjs.config({
//Pass the top-level main.js/index.js require
//function to requirejs so that node modules
//are loaded relative to the top-level JS file.
nodeRequire: require,
});
rjs(["./math.js"], function(math) {
console.log(math.add(2, 3));
});
//math.js
define("math", function() {
function add(a, b) {
return a + b;
}
function subtract(a, b) {
return a - b;
}
function multiply(a, b) {
return a * b;
}
function divide(a, b) {
return a / b;
}
return { add, subtract, multiply, divide };
});
While running r.js ./src/main.js, I am getting the below error.
rjs.config is not a function
at /Users/dheerajmahra/Desktop/learning/different-module-formats/src/main.js:3:5
at /Users/dheerajmahra/Desktop/learning/different-module-formats/src/main.js:14:2
at Script.runInThisContext (vm.js:96:20)
at Object.runInThisContext (vm.js:303:38)
....
Based of reading the requirejs-docs, it seems that you are conflating the two approaches here, one being through the node-package, the other one being through the r.js standalone file.
It seems like you either haven't done npm i requirejs or, you are not running through the node process.
If you are using the r.js file directly, replace require('requirejs') with require('./path/to/r.js').

How to use js-cookie in a standalone *.js script?

I am using gulp to concatenate and minify a number of standalone *.js scripts used on my web site. Basically this is just a catchall folder where I place little utility scripts that run on page load. For example, one of them starts a carousel slider, another adds a class to the header that shrinks it on scroll, etc. Each of these "features" has its own standalone *.js file.
Now, I would like to use the popular js-cookie library in one of those scripts. Unfortunately, since my project is not set up as an ES6 module, I am not able to able to import the js-cookie library the way it's specified in the docs, like this:
import Cookies from 'js-cookie'
When I do this, I get the error message Uncaught SyntaxError: Cannot use import statement outside a module.
I tried changing it to this:
window.Cookies = require('js-cookie')
but that gave me this error:
Uncaught ReferenceError: require is not defined
Here is my gulpfile, followed by the feature.js script in which I'm trying to use the js-cookie library:
gulpfile.js
// Initialize modules
const { src, dest, watch, series, parallel } = require('gulp');
const sourcemaps = require('gulp-sourcemaps');
const sass = require('gulp-sass');
const concat = require('gulp-concat');
const uglify = require('gulp-uglify');
const postcss = require('gulp-postcss');
const autoprefixer = require('autoprefixer');
const cssnano = require('cssnano');
var replace = require('gulp-replace');
var merge = require('merge-stream');
// File paths (note that src paths are arrays)
const files = {
scssSrcPath: [
'scss/*.scss',
'scss/_pageContentModules/*.scss'
],
jsSrcPath: [
'js/*.js',
'node_modules/slick-carousel/slick/slick.js'
],
scssDstPath: '../web/css',
jsDstPath: '../web/js'
}
// Sass task: compiles SCSS files into style.css
function scssTask(){
return merge(files.scssSrcPath.map(function (file) {
return src(file)
}))
.pipe(sourcemaps.init()) // initialize sourcemaps first
.pipe(sass()) // compile SCSS to CSS
.pipe(postcss([ autoprefixer(), cssnano() ])) // PostCSS plugins
.pipe(sourcemaps.write('.'))
.pipe(dest(files.scssDstPath));
}
// JS task: concatenates and uglifies JS files to script.js
function jsTask(){
return merge(files.jsSrcPath.map(function (file) {
return src(file)
}))
.pipe(concat('app.js'))
.pipe(uglify())
.pipe(dest(files.jsDstPath));
}
// Watch task: watch SCSS and JS files for changes
// If any change, run scss and js tasks simultaneously
function watchTask(){
watch(files.scssSrcPath, scssTask);
watch(files.jsSrcPath, jsTask);
}
// Export the default Gulp task so it can be run
// Runs the scss and js tasks simultaneously
// then watch task
exports.default = series(
parallel(scssTask, jsTask),
watchTask
);
js/feature.js
import Cookies from 'js-cookie';
const rs = cookies.get('referral_source');
if (typeof rs !== 'undefined') {
console.log('referral_source = ' + rs);
}
How can I get this working? Is there a way to do it using my simple Gulp setup, or do I need to go beyond and set up a full-on Webpack setup (with all the complexity that adds)?
Unfortunately, as far as I know, Gulp does not support the ability to use ES6 modules. If you want to use them, you will need to use Webpack.
But js-cookie does have a jsDelivr CDN: <script src="https://cdn.jsdelivr.net/npm/js-cookie#3.0.1/dist/js.cookie.min.js"></script>. By including this before your JS script, like this:
<script src="https://cdn.jsdelivr.net/npm/jscookie#3.0.1/dist/js.cookie.min.js"></script>
<script src="./js/feature.js"></script>

How can I achieve this using gulp?

I am enumerating the subdirectories in a directory. For each sub directory I would like to apply a number of gulp activities like less compilation, and then create an output file specific to that subdirectory.
I would like the gulp process to continue, as further transformation steps need to be performed later.
Can someone help me understand how I can create these files half way through the "gulp pipeline"?
This seems quite interesting to achieve and gulp has no limitations at all.
I will give you detailed example how I have managed to accomplish such a task a while ago.
Let assume that you have directoryA. Subdirectories childA, childB and childC are contained into directoryA. So basically your tree structure looks like:
directoryA
--childA
--childB
--childC
I am always looking for a flexible solutions so I would suggest to include a JSON file in each subdirectory naming the tasks you would like to running. Using fs you can access these files. You can also use run-sequence to execute gulp tasks synchronously.
For demo purposes place a file named manifest.json inside childA subdirectory.
Manifest.json contains the following declarations:
{
"filesToProccess" : ["./childA/*.js", "./childB/*.js"],
"tasksToRun" :["taskA", "taskB"]
}
Finally gulpfile would like this:
'use strict';
//dependencies declared into package.json
//install them using npm
var gulp = require('gulp'),
fs = require('fs'),
runSequence = require('run-sequence'),
path = require('path');
//these two array will keep the actions you have included into manifest file.
var filesHolder = [], tasksHolder = [];
gulp.task('taskA', function () {
return gulp.src(filesHolder)
.pipe(whatever)
.pipe(gulp.dest('whatever')); //chailed actions
});
gulp.task('taskB', function () {
return gulp.src(filesHolder)
.pipe(whatever)
.pipe(gulp.dest('whatever'));
});
//a simple utility function to read all subdirectories of directoryA
function getDirectories(srcpath) {
return fs.readdirSync(srcpath).filter(function(file) {
return fs.statSync(path.join(srcpath, file)).isDirectory();
});
}
//finally insert the default gulp task
gulp.task('default', function(){
var manifest;
//map directory's A subdirectories
var availableDirs = getDirectories("./directoryA");
//finally loop the available subdirectories, load each manifest file and
availableDirs.forEach(function(subdir) {
manifest = require("./directoryA/"+subdir+"manifest.json");
filesHolder = manifest.filesToProccess;
tasksHolder = manifest.tasksToRun;
runSequence( tasksHolder , function () {
console.log( " Task ended :" + tasksHolder + " for subdirectory : " + subdir);
});
});
});

Gulp js src : get deepest files first

I use Gulp to build a small web application based on the React framework.
To compile the client scripts, my task starts like that :
gulp.task( 'buildClientScripts', function () {
// Get all the js and jsx scripts
// Starts with the app.js file
return gulp.src( [
'app/app.js',
'app/clientScripts/**/*.jsx',
'app/clientScripts/**/*.js',
] )
The order of the files is my problem. For this file structure :
- clientScripts
---- components
-------- subComponents
------------ mysubComponent.js
-------- myComponent.js
---- main.js
The load order will be : main.js, myComponent.js, mysubComponent.js. But obviously, my top files need the deeper files to work.
How do I ask gulp to load the deepest files first ?
You can use gulp-sort to order the files after gulp.src reads them.
Count slashes in path as crude way to determine depth.
var sort = require('gulp-sort');
var path = require('path');
gulp.task( 'buildClientScripts', function () {
// Get all the js and jsx scripts
// Starts with the app.js file
return gulp.src( [
'app/app.js',
'app/clientScripts/**/*.jsx',
'app/clientScripts/**/*.js',
] )
.pipe(sort(function(file1, file2) {
return countSlashes(file2) - countSlashes(file1);
function countSlashes(file) {
return file.path.split(path.sep).length - 1;
}
})

Make browserify modules external with Gulp

I have a library lib.js that I want to create from lib/a.js and lib/b.js and to be able to use it from a script client.js using var a = require('lib/a.js'); and that it works when I just include the compiled lib.js library before client.js (therefore, lib.js has to declare a require function that knows about lib/a.js)
I guess I have to use external and alias but I am not sure what is the proper way to do it
Also, is it possible to have a Gulp file that creates all the alias automatically for the folders in my library? eg. creates an alias for all the files in the lib/ dir?
Here are a couple of gulp tasks that would help to build your common lib.js and the client.js bundles separately.
Note that you have to tell browserify to b.require() lib/*.js when bundling lib.js, and you have to tell it to b.external() the libraries that will be loaded separately when bundling client.js
var path = require('path');
var gulp = require('gulp');
var browserify = require('browserify');
var concat = require('gulp-concat');
var transform = require('vinyl-transform');
gulp.task('build-lib', function () {
// use `vinyl-transform` to wrap around the regular ReadableStream returned by b.bundle();
// so that we can use it down a vinyl pipeline as a vinyl file object.
// `vinyl-transform` takes care of creating both streaming and buffered vinyl file objects.
var browserified = transform(function(filename) {
// basename, for eg: 'a.js'
var basename = path.basename(filename);
// define the exposed name that your client.js would use to require();
// for eg: require('lib/a.js'); // -> exposed name should be 'lib/a.js'
var expose = 'lib/' + basename;
return browserify(filename)
.require(filename, { expose: expose})
.bundle();
});
return gulp.src(['./lib/*.js'])
.pipe(browserified)
.pipe(concat('lib.js'))
.pipe(gulp.dest('./dist'));
});
gulp.task('build-client', function () {
var browserified = transform(function(filename) {
// filename = './client.js'
// let browserify know that lib/a.js and and lib/b.js are external files
// and will be loaded externally (in your case, by loading the bundled lib.js
// for eg: <script src='dist/lib.js'>)
return browserify(filename)
.external('lib/a.js')
.external('lib/b.js')
.bundle();
});
return gulp.src(['./client.js'])
.pipe(browserified)
.pipe(gulp.dest('./dist'));
});
gulp.task('default', ['build-lib', 'build-client']);
Are you looking for external requires?
To use with gulp-browserify, check the README
.on('prebundle', function(bundle) {
bundle.external('domready');
bundle.external('react');
})
Should work with bundle.require as well.

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