Instead of mention every js seperatly,
is this the way to minify and concatinate a whole js folder?
module.exports = function(grunt) {
grunt.initConfig({
min: {
dist: {
src: ['scripts/*.js'],
dest: 'dist/built.min.js'
}
}
});
};
Yes, that's correct if you only want to concatenate and minify all .js files in the scripts directory one level deep.
For example, if scripts/ contains a.js and b.js and the foo/ directory, you'd get the concatenation and minified result of a.js + b.js but nothing in the foo/ directory.
What if you want to get everything in the foo/ directory (and all other nested directories) as well? Change the expression from ['scripts/*.js'] to ['scripts/**/*.js'] -- or any minimatch expression:
https://github.com/gruntjs/grunt/blob/master/docs/api_file.md#gruntfileexpand
You're able to use any minimatch expression since the grunt min task uses the expandFiles function:
https://github.com/gruntjs/grunt/blob/master/tasks/min.js#L21
The downside to using a minimatch expression with this task is it's hard to understand what order the files will be concatenated in, which is often very important. Be careful if this matters.
Also, please note that a new version of grunt (0.4) is coming out very soon. This will make this answer obsolete, as the min task has been changed in 0.4 (but will still support minimatch expression).
If your folder consist only js files you are right but if your folder have nested folders such as, foo is our main js folder in which we have another nested folder loo and inside it we also have some js files such as :
foo:
mu.js
su.js
loo:
ku.js
wu.js
In this case you have to modify your code in such manner :
module.exports = function(grunt) {
grunt.initConfig({
min: {
dist: {
src: 'foo/**/*.js',
dest: 'dist/foo.min.js'
}
}
});
};
by doing in such a way you can minimize your all js files of foo folder even nested folder files too. I suggest cocat js files before minimizing.
Related
I've written a gulp task to rename files so that they can be versioned. The problem is that the filenames of the files that the index.html scripts reference are not changed.
For example, in my index.html:
<script src=pub/main_v1.js"></script>
But if you actually navigate through the build folder to the subdirectory pub, you will find main.js.
Here is the custom gulp task:
const gulpConcat = require('gulp-concat');
const gulpReplace = require('gulp-replace');
const version = require('./package.json').version;
gulp.task('version', function () {
var vsn = '_' + version + '.js';
gulp.src('scripts/**/*.js')
.pipe(gulpConcat(vsn))
.pipe(gulp.dest('./prodBuild'));
return gulp.src('./prodBuild/index.html', { base: './prodBuild' })
.pipe(gulpReplace(/* some regex */, /* append vsn */))
.pipe(gulp.dest('./prodBuild'));
});
What do I need to fix/add so that the original filename changes to match that in the script tag?
Note: According to the gulp-concat docs, I should be able to find the concated files at prodBuild/[vsn], where [vsn] is _v1.js. However, it is no where to be found.
Update: The files rename properly in index.html, but I can't seem to get the renaming of the original files to work. Here's a snapshot of my build directory:
prodBuild/
pub/
main.js
someDir/
subDirA/
// unimportant stuff
subDirB/
file2.js
file3.js
// ...other files and folders...
EDIT:
The issue is that you return only one of the two tasks. The first task is simply ignored by gulp, since it is not returned. A simple solutions: Split it into two tasks, and reference the one from the other, like in this SO answer.
Old Answer
This looks like a perfect case for the gulp-rename. You could simply pipe your scripts through gulp-rename, like this:
.pipe(rename(function (path) {
path.basename += vsn;
path.extname = ".js"
}))
Gulp concat is, AFAIK, made for the concatination of files, not particularly for the renaming of them.
This seems like a very simple question, but spent the last 3 hours researching it, discovering it can be slow on every save on a new file if not using watchify.
This is my directory tree:
gulpfile.js
package.json
www/
default.htm
<script src="toBundleJsHere/file123.js"></script>
toBundletheseJs/
componentX/
file1.js
componentY/
file2.js
componentZ/
file3.js
toPutBundledJsHere/
file123.js
Requirements.
On every creation or save of a file within the folder toBundleTheseJs/ I want this file to be rebundled into toBundleJsHere/
What do I need to include in my package.json file?
And whats the minimum I need to write into my gulp file?
This should be as fast as possible so think I should be using browserify and watchify. I want to understand the minimum steps so using package manager like jspm is overkill a this point.
thanks
First you should listen to changes in the desired dir:
watch(['toBundletheseJs/**/*.js'], function () {
gulp.run('bundle-js');
});
Then the bundle-js task should bundle your files. A recommended way is gulp-concat:
var concat = require('gulp-concat');
var gulp = require('gulp');
gulp.task('bundle-js', function() {
return gulp.src('toBundletheseJs/**/*.js')
.pipe(concat('file123.js'))
.pipe(gulp.dest('./toPutBundledJsHere/'));
});
The right answer is: there is no legit need for concatenating JS files using gulp. Therefore you should never do that.
Instead, look into proper JS bundlers that will properly concatenate your files organizing them according to some established format, like commonsjs, amd, umd, etc.
Here's a list of more appropriate tools:
Webpack
Rollup
Parcel
Note that my answer is around end of 2020, so if you're reading this in a somewhat distant future keep in mind the javascript community travels fast so that new and better tools may be around.
var gulp = require('gulp');
var concat = require('gulp-concat');
gulp.task('js', function (done) {
// array of all the js paths you want to bundle.
var scriptSources = ['./node_modules/idb/lib/idb.js', 'js/**/*.js'];
gulp.src(scriptSources)
// name of the new file all your js files are to be bundled to.
.pipe(concat('all.js'))
// the destination where the new bundled file is going to be saved to.
.pipe(gulp.dest('dist/js'));
done();
});
Use this code to bundle several files into one.
gulp.task('scripts', function() {
return gulp.src(['./lib/file3.js', './lib/file1.js', './lib/file2.js']) //files separated by comma
.pipe(concat('script.js')) //resultant file name
.pipe(gulp.dest('./dist/')); //Destination where file to be exported
});
I’m trying to get grunt to concat some .css files from 2 different directories into a single css file
I got it working ok and am trying to write a similar piece of code to concat html files in the order I specify in a simple external .json file rather than editing the Grunfile.js directly each time but it keeps throwing errors
The file locations and permissions are all ok, I can do it with php or concat the css files ok, I’m not great on js syntax os using functions. I only learned to use Grunt in the last few days so am pretty sure have not called the file with the correct code??
Any advice kindly appreciated!
This is my Gruntfile.js exactly
module.exports = function(grunt) {
// 1. All configuration goes here
grunt.initConfig({
pkg: grunt.file.readJSON('package.json'),
concat: {
// 2. Configuration for concatinating files goes here
file.readJSON(‘html-files.json’);
},
watch: {
files: ['../../pages/**/*.html’],
tasks: ['concat'],
options : { nospawn : true }
}
});
// 3. Where we tell Grunt we plan to use this plug-in
grunt.loadNpmTasks('grunt-contrib-watch');
grunt.loadNpmTasks('grunt-contrib-concat');
// 4. Where we tell Grunt what to do when we type "grunt" into the terminal.
grunt.registerTask('default', ['concat']);
};
This is the html-files.json file contents exactly
html: {
src: [
'../../pages/**/this-block.html',
'../../pages/**/that-block.html',
],
dest: '../../somepage/content.html',
}
I am using Browserify to compile a large Node.js application into a single file (using options --bare and --ignore-missing [to avoid troubles with lib-cov in Express]). I have some code to dynamically load modules based on what is available in a directory:
var fs = require('fs'),
path = require('path');
fs.readdirSync(__dirname).forEach(function (file) {
if (file !== 'index.js' && fs.statSync(path.join(__dirname, file)).isFile()) {
module.exports[file.substring(0, file.length-3)] = require(path.join(__dirname, file));
}
});
I'm getting strange errors in my application where aribtrary text files are being loaded from the directory my compiled file is loaded in. I think it's because paths are no longer set correctly, and because Browserify won't be able to require() the correct files that are dynamically loaded like this.
Short of making a static index.js file, is there a preferred method of dynamically requiring a directory of modules that is out-of-the-box compatible with Browserify?
This plugin allows to require Glob patterns: require-globify
Then, with a little hack you can add all the files on compilation and not executing them:
// Hack to compile Glob files. Don´t call this function!
function ಠ_ಠ() {
require('views/**/*.js', { glob: true })
}
And, for example, you could require and execute a specific file when you need it :D
var homePage = require('views/'+currentView)
Browserify does not support dynamic requires - see GH issue 377.
The only method for dynamically requiring a directory I am aware of: a build step to list the directory files and write the "static" index.js file.
There's also the bulkify transform, as documented here:
https://github.com/chrisdavies/tech-thoughts/blob/master/browserify-include-directory.md
Basically, you can do this in your app.js or whatever:
var bulk = require('bulk-require');
// Require all of the scripts in the controllers directory
bulk(__dirname, ['controllers/**/*.js']);
And my gulpfile has something like this in it:
gulp.task('js', function () {
return gulp.src('./src/js/init.js')
.pipe(browserify({
transform: ['bulkify']
}))
.pipe(rename('app.js'))
.pipe(uglify())
.pipe(gulp.dest('./dest/js'));
});
I am currently using Grunt, and as I was trying Gulp, the same problem I encountered first with Grunt occurred to me.
I am trying to process some js files (concat, uglify and minify them), but I don't want all of them to compile into one big file, I want multiple output files, each from the processing of some input files :
scripts =
firstOutput:
outputFilename: 'first.min.js',
inputFiles: ['one.js', 'two.js']
secondOutput:
outputFilename: 'second.min.js',
inputFiles: ['three.js']
thirdOutput:
outputFilename: 'third.min.js',
inputFiles: ['four.js', 'five.js']
The only way I found (for now) to achieve that with Grunt is with multiple watches and multiple uglify tasks (or one uglify task and a listener on watch change to dynamically modify the uglify task src and dest) :
module.exports = (grunt) ->
grunt.loadNpmTasks 'grunt-contrib-watch'
grunt.loadNpmTasks 'grunt-contrib-uglify'
grunt.initConfig
watch:
firstOutput:
files: scripts.firstOutput.inputFiles
tasks: ['uglify:firstOutput']
options :
spawn : false
secondOutput:
files: scripts.secondOutput.inputFiles
tasks: ['uglify:secondOutput']
options :
spawn : false
thirdOutput:
files: scripts.thirdOutput.inputFiles
tasks: ['uglify:thirdOutput']
options :
spawn : false
uglify:
firstOutput:
files: scripts.firstOutput.inputFiles
dest: scripts.firstOutput.outputFilename
secondOutput:
files: scripts.secondOutput.inputFiles
dest: scripts.secondOutput.outputFilename
thirdOutput:
files: scripts.thirdOutput.inputFiles
dest: scripts.thirdOutput.outputFilename
grunt.registerTask 'default', 'watch'
And, as you can imagine, this is just an example, in my case of a big web application, there's a lot more than just three output js files, and I also process a few less files into some css files
My Gruntfile is really huge, and I find it has a lot of duplicate code, is there any way to have this code refactored to have one watch and one uglify task, with an automatically guessed src and dest with some kind of dependency (to know that if the four.js file is modified, it has to process the third output) ?
If you have some way to do it with Gulp I'll take it with great pleasure, as I would like to test it in my usual workflow.
Here's how you can do this with gulp + vanilla javascript:
var _ = require("underscore")
, gulp = require("gulp")
, uglify = require("gulp-uglify")
var scripts = [
{
output: 'first.min.js',
input: ['one.js', 'two.js']
}
, {
output: 'second.min.js',
input: ['three.js']
}
, {
output: 'third.min.js',
input: ['four.js', 'five.js']
}
];
function build(files, dest) {
return gulp.src(files)
.pipe(uglify())
.pipe(gulp.dest(dest));
}
gulp.task("watch", function () {
_.each(scripts, function (script, i) {
gulp.watch(script.input, function () {
build(script.input, script.output);
});
});
});
Even better if you can use globs to match sets of files so you don't have to write out the path for every single input set. Something like input: ["one/**/*.js, "other/**/*.js"]
"I am trying to process some js files (concat, uglify and minify
them), but I don't want all of them to compile into one big file"
Can I ask why? The benefit of one larger file is that you save on HTTP requests, every resource you load will cause some slowdown of your website. May I suggest using proper dependency management with RequireJS? That way the optimiser can walk your dependency graph and output optimised files for you.
http://requirejs.org/
There's a grunt task for this too:
https://github.com/gruntjs/grunt-contrib-requirejs