Good day!
I want to automate the conversion of images to webp format. To avoid doing it manually or via online converters, I decided to use gulp 4 and gulp-webp.
This is the structure of nesting folders in my project:
I want Gulp, when it finds a picture, so that it creates a folder called "webp" at the same nesting level and places the converted picture in this folder.
I want the following result:
My Gulpfile.js:
let gulp = require('gulp'),
webp = require('gulp-webp');
gulp.task('webp', () => {
// './dev/img/**/*.{png,gif,jpg}' - all files in img and all files in subfolders in img
return gulp.src('./dev/img/**/*.{png,gif,jpg}')
.pipe(webp())
.pipe(gulp.dest(gulp.src)) //something like this, but it doesn't work
}
);
This can be done with the help of gulp-rename.
To install gulp-rename run:
npm i -D gulp-rename
Then in your gulpfile.js add the import:
const rename = require('gulp-rename');
Then change your stream like this:
return gulp.src('./dev/img/**/*.{png,gif,jpg}')
.pipe(webp())
.pipe(rename({ prefix: 'webp/' }))
.pipe(gulp.dest('./dev/img'));
The prefix option inserts "webp/" before the bare filename after the dest target "./dev/img".
I'm new comer in Gulp, and i got stock when i try to output my files to their parent's path.
My directories like this:
app/page_a/less/a.less
app/page_a/css
My goal is:
app/page_a/less/a.less
app/page_a/css/a.css
Here's what i'm doing:
I use ** because i want to match all of my page_* module.
gulp.task('test', function () {
var files = ['./app/**/*.less'];
return gulp.src(files, {
base: '.' // Now the base url becomes where the *.less is.
})
.pipe(less())
.pipe(gulp.dest('.'));
});
My result:( which is obviously wrong :( )
app/page_a/less/a.less
app/page_a/less/a.css
app/page_a/css
So how can i output *.css into the corresponding app/page_*/css path?
The path you specify in the base option to gulp.src() isn't relative to each respective .less file, but rather the current working directory. This is usually the directory your Gulpfile.js resides in.
What you actually want to do is alter the path for each resulting .css file after your less() plugin has run, so when the .css files are written to the directory specified by gulp.dest() the /less/ part of the path has been replaced with /css/.
This is exactly what the gulp-rename plugin is for:
var gulp = require('gulp');
var less = require('gulp-less');
var rename = require('gulp-rename');
gulp.task('default', function () {
return gulp.src( './app/**/*.less')
.pipe(less())
.pipe(rename(function(file) {
file.dirname = file.dirname.replace(/less$/, "css");
}))
.pipe(gulp.dest('./app/'));
});
With gulp.dest('.') you say 'here'
Try with:
.pipe(gulp.dest('./app/page_a/css/'));
This should be work
I have an app. My app source code is structured like this:
./
gulpfile.js
src
img
bg.png
logo.png
data
list.json
favicon.ico
web.config
index.html
deploy
I am trying to use Gulp to copy two files: ./img/bg.png and ./data/list.json. I want to copy these two files to the root of the deploy directory. In other words, the result of the task should have:
./
deploy
imgs
bg.png
data
list.json
How do I write a Gulp task to do this type of copying? The thing that is confusing me is the fact that I want my task to copy two seperate files instead of files that fit a pattern. I know if I had a pattern, I could do this:
var copy = require('gulp-copy');
gulp.task('copy-resources', function() {
return gulp.src('./src/img/*.png')
.pipe(gulp.dest('./deploy'))
;
});
Yet, I'm still not sure how to do this with two seperate files.
Thanks
You can create separate tasks for each target directory, and then combine them using a general "copy-resources" task.
gulp.task('copy-img', function() {
return gulp.src('./src/img/*.png')
.pipe(gulp.dest('./deploy/imgs'));
});
gulp.task('copy-data', function() {
return gulp.src('./src/data/*.json')
.pipe(gulp.dest('./deploy/data'));
});
gulp.task('copy-resources', ['copy-img', 'copy-data']);
You could also use merge-stream
Install dependency:
npm i -D merge-stream
Load the depedency in your gulp file and use it:
const merge = require("merge-stream");
gulp.task('copy-resources', function() {
return merge([
gulp.src('./src/img/*.png').pipe(gulp.dest('./deploy/imgs')),
gulp.src('./src/data/*.json').pipe(gulp.dest('./deploy/data'))
]);
});
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);
});
});
});
Say, for example, you are building a project on Backbone or whatever and you need to load scripts in a certain order, e.g. underscore.js needs to be loaded before backbone.js.
How do I get it to concat the scripts so that they’re in order?
// JS concat, strip debugging and minify
gulp.task('scripts', function() {
gulp.src(['./source/js/*.js', './source/js/**/*.js'])
.pipe(concat('script.js'))
.pipe(stripDebug())
.pipe(uglify())
.pipe(gulp.dest('./build/js/'));
});
I have the right order of scripts in my source/index.html, but since files are organized by alphabetic order, gulp will concat underscore.js after backbone.js, and the order of the scripts in my source/index.html does not matter, it looks at the files in the directory.
So does anyone have an idea on this?
Best idea I have is to rename the vendor scripts with 1, 2, 3 to give them the proper order, but I am not sure if I like this.
As I learned more I found Browserify is a great solution, it can be a pain at first but it’s great.
I had a similar problem recently with Grunt when building my AngularJS app. Here's a question I posted.
What I ended up doing is to explicitly list the files in order in the grunt config. The config file will then look like this:
[
'/path/to/app.js',
'/path/to/mymodule/mymodule.js',
'/path/to/mymodule/mymodule/*.js'
]
Grunt is able to figure out which files are duplicates and not include them. The same technique will work with Gulp as well.
Another thing that helps if you need some files to come after a blob of files, is to exclude specific files from your glob, like so:
[
'/src/**/!(foobar)*.js', // all files that end in .js EXCEPT foobar*.js
'/src/js/foobar.js',
]
You can combine this with specifying files that need to come first as explained in Chad Johnson's answer.
I have used the gulp-order plugin but it is not always successful as you can see by my stack overflow post gulp-order node module with merged streams. When browsing through the Gulp docs I came across the streamque module which has worked quite well for specifying order of in my case concatenation. https://github.com/gulpjs/gulp/blob/master/docs/recipes/using-multiple-sources-in-one-task.md
Example of how I used it is below
var gulp = require('gulp');
var concat = require('gulp-concat');
var handleErrors = require('../util/handleErrors');
var streamqueue = require('streamqueue');
gulp.task('scripts', function() {
return streamqueue({ objectMode: true },
gulp.src('./public/angular/config/*.js'),
gulp.src('./public/angular/services/**/*.js'),
gulp.src('./public/angular/modules/**/*.js'),
gulp.src('./public/angular/primitives/**/*.js'),
gulp.src('./public/js/**/*.js')
)
.pipe(concat('app.js'))
.pipe(gulp.dest('./public/build/js'))
.on('error', handleErrors);
});
With gulp-useref you can concatenate every script declared in your index file, in the order in which you declare it.
https://www.npmjs.com/package/gulp-useref
var $ = require('gulp-load-plugins')();
gulp.task('jsbuild', function () {
var assets = $.useref.assets({searchPath: '{.tmp,app}'});
return gulp.src('app/**/*.html')
.pipe(assets)
.pipe($.if('*.js', $.uglify({preserveComments: 'some'})))
.pipe(gulp.dest('dist'))
.pipe($.size({title: 'html'}));
});
And in the HTML you have to declare the name of the build file you want to generate, like this:
<!-- build:js js/main.min.js -->
<script src="js/vendor/vendor.js"></script>
<script src="js/modules/test.js"></script>
<script src="js/main.js"></script>
In your build directory you will have the reference to main.min.js which will contain vendor.js, test.js, and main.js
The sort-stream may also be used to ensure specific order of files with gulp.src. Sample code that puts the backbone.js always as the last file to process:
var gulp = require('gulp');
var sort = require('sort-stream');
gulp.task('scripts', function() {
gulp.src(['./source/js/*.js', './source/js/**/*.js'])
.pipe(sort(function(a, b){
aScore = a.path.match(/backbone.js$/) ? 1 : 0;
bScore = b.path.match(/backbone.js$/) ? 1 : 0;
return aScore - bScore;
}))
.pipe(concat('script.js'))
.pipe(stripDebug())
.pipe(uglify())
.pipe(gulp.dest('./build/js/'));
});
I just add numbers to the beginning of file name:
0_normalize.scss
1_tikitaka.scss
main.scss
It works in gulp without any problems.
I have my scripts organized in different folders for each package I pull in from bower, plus my own script for my app. Since you are going to list the order of these scripts somewhere, why not just list them in your gulp file? For new developers on your project, it's nice that all your script end-points are listed here. You can do this with gulp-add-src:
gulpfile.js
var gulp = require('gulp'),
less = require('gulp-less'),
minifyCSS = require('gulp-minify-css'),
uglify = require('gulp-uglify'),
concat = require('gulp-concat'),
addsrc = require('gulp-add-src'),
sourcemaps = require('gulp-sourcemaps');
// CSS & Less
gulp.task('css', function(){
gulp.src('less/all.less')
.pipe(sourcemaps.init())
.pipe(less())
.pipe(minifyCSS())
.pipe(sourcemaps.write('source-maps'))
.pipe(gulp.dest('public/css'));
});
// JS
gulp.task('js', function() {
gulp.src('resources/assets/bower/jquery/dist/jquery.js')
.pipe(addsrc.append('resources/assets/bower/bootstrap/dist/js/bootstrap.js'))
.pipe(addsrc.append('resources/assets/bower/blahblah/dist/js/blah.js'))
.pipe(addsrc.append('resources/assets/js/my-script.js'))
.pipe(sourcemaps.init())
.pipe(concat('all.js'))
.pipe(uglify())
.pipe(sourcemaps.write('source-maps'))
.pipe(gulp.dest('public/js'));
});
gulp.task('default',['css','js']);
Note: jQuery and Bootstrap added for demonstration purposes of order. Probably better to use CDNs for those since they are so widely used and browsers could have them cached from other sites already.
Try stream-series. It works like merge-stream/event-stream.merge() except that instead of interleaving, it appends to the end. It doesn't require you to specify the object mode like streamqueue, so your code comes out cleaner.
var series = require('stream-series');
gulp.task('minifyInOrder', function() {
return series(gulp.src('vendor/*'),gulp.src('extra'),gulp.src('house/*'))
.pipe(concat('a.js'))
.pipe(uglify())
.pipe(gulp.dest('dest'))
});
merge2 looks like the only working and maintained ordered stream merging tool at the moment.
Update 2020
The APIs are always changing, some libraries become unusable or contain vulnerabilities, or their dependencies contain vulnerabilities, that are not fixed for years. For text files manipulations you'd better use custom NodeJS scripts and popular libraries like globby and fs-extra along with other libraries without Gulp, Grunt, etc wrappers.
import globby from 'globby';
import fs from 'fs-extra';
async function bundleScripts() {
const rootPaths = await globby('./source/js/*.js');
const otherPaths = (await globby('./source/**/*.js'))
.filter(f => !rootFiles.includes(f));
const paths = rootPaths.concat(otherPaths);
const files = Promise.all(
paths.map(
// Returns a Promise
path => fs.readFile(path, {encoding: 'utf8'})
)
);
let bundle = files.join('\n');
bundle = uglify(bundle);
bundle = whatever(bundle);
bundle = bundle.replace(/\/\*.*?\*\//g, '');
await fs.outputFile('./build/js/script.js', bundle, {encoding: 'utf8'});
}
bundleScripts.then(() => console.log('done');
An alternative method is to use a Gulp plugin created specifically for this problem. https://www.npmjs.com/package/gulp-ng-module-sort
It allows you to sort your scripts by adding in a .pipe(ngModuleSort()) as such:
var ngModuleSort = require('gulp-ng-module-sort');
var concat = require('gulp-concat');
gulp.task('angular-scripts', function() {
return gulp.src('./src/app/**/*.js')
.pipe(ngModuleSort())
.pipe(concat('angularAppScripts.js))
.pipe(gulp.dest('./dist/));
});
Assuming a directory convention of:
|——— src/
| |——— app/
| |——— module1/
| |——— sub-module1/
| |——— sub-module1.js
| |——— module1.js
| |——— module2/
| |——— sub-module2/
| |——— sub-module2.js
| |——— sub-module3/
| |——— sub-module3.js
| |——— module2.js
| |——— app.js
Hope this helps!
For me I had natualSort() and angularFileSort() in pipe which was reordering the files. I removed it and now it works fine for me
$.inject( // app/**/*.js files
gulp.src(paths.jsFiles)
.pipe($.plumber()), // use plumber so watch can start despite js errors
//.pipe($.naturalSort())
//.pipe($.angularFilesort()),
{relative: true}))
I just use gulp-angular-filesort
function concatOrder() {
return gulp.src('./build/src/app/**/*.js')
.pipe(sort())
.pipe(plug.concat('concat.js'))
.pipe(gulp.dest('./output/'));
}
I'm in a module environnement where all are core-dependents using gulp.
So, the core module needs to be appended before the others.
What I did:
Move all the scripts to an src folder
Just gulp-rename your core directory to _core
gulp is keeping the order of your gulp.src, my concat src looks like this:
concat: ['./client/src/js/*.js', './client/src/js/**/*.js', './client/src/js/**/**/*.js']
It'll obviously take the _ as the first directory from the list (natural sort?).
Note (angularjs):
I then use gulp-angular-extender to dynamically add the modules to the core module.
Compiled it looks like this:
angular.module('Core', ["ui.router","mm.foundation",(...),"Admin","Products"])
Where Admin and Products are two modules.
if you would like to order third party libraries dependencies, try wiredep. This package basically checks each package dependency in bower.json then wire them up for you.
I tried several solutions from this page, but none worked. I had a series of numbered files which I simply wanted be ordered by alphabetical foldername so when piped to concat() they'd be in the same order. That is, preserve the order of the globbing input. Easy, right?
Here's my specific proof-of-concept code (print is just to see the order printed to the cli):
var order = require('gulp-order');
var gulp = require('gulp');
var print = require('gulp-print').default;
var options = {};
options.rootPath = {
inputDir: process.env.INIT_CWD + '/Draft',
inputGlob: '/**/*.md',
};
gulp.task('default', function(){
gulp.src(options.rootPath.inputDir + options.rootPath.inputGlob, {base: '.'})
.pipe(order([options.rootPath.inputDir + options.rootPath.inputGlob]))
.pipe(print());
});
The reason for the madness of gulp.src? I determined that gulp.src was running async when I was able to use a sleep() function (using a .map with sleeptime incremented by index) to order the stream output properly.
The upshot of the async of src mean dirs with more files in it came after dirs with fewer files, because they took longer to process.
In my gulp setup, I'm specifying the vendor files first and then specifying the (more general) everything, second. And it successfully puts the vendor js before the other custom stuff.
gulp.src([
// vendor folder first
path.join(folder, '/vendor/**/*.js'),
// custom js after vendor
path.join(folder, '/**/*.js')
])
Apparently you can pass in the "nosort" option to gulp.src gulp.src.