The following gulp watch task isn't getting triggered when I change any LESS file in the project. Can anyone spot what I'm doing wrong? Most the answers here say to NOT use the watch-less module, which I'm not. It's supposed to listen to changes in any LESS file in the project and when one changes, go to the app.less file to regenerate the CSS file (app.less has #includes to all the files).
var watch = require("gulp-watch");
var less = require("gulp-less");
gulp.watch(paths.source + "**/*.less", function(event){
gulp.src(paths.source + paths.assets + paths.less + "app.less")
.pipe(less().on("error", console.log))
.pipe(gulp.dest(paths.dev + paths.css));
});
Here are some issues:
require("gulp-watch"); is useless here. Actually gulp.watch is a core API of gulp.
The gulpfile.js consists of several gulp tasks.
Run gulp watch in your terminal.
For example:
var gulp = require('gulp');
var path = require('path');
var less = require('gulp-less');
var paths = {
// your paths
};
gulp.task('styles', function () {
return gulp.src(paths.source + paths.assets + paths.less + "app.less")
.pipe(less({
// paths to be used for #import directives
paths: [ path.join(__dirname, 'less', 'includes') ]
}))
.pipe(gulp.dest('./'));
});
gulp.task('watch', function() {
gulp.watch('less/**/*.less', ['styles']);
});
Related
So consider the following gulp file:
var gulp = require('gulp');
var source = require('vinyl-source-stream');
var browserify = require('browserify');
var babelify = require("babelify");
var watch = require('gulp-watch');
gulp.task('make:engine', function() {
return browserify({entries: [
'./src/index.js'
]})
.transform("babelify")
.bundle()
.pipe(source('engine.js'))
.pipe(gulp.dest('dist/'));
});
gulp.task('make:example', function() {
return browserify({entries: [
'./examples/index.js'
]})
.transform("babelify")
.bundle()
.pipe(source('compiled-example.js'))
.pipe(gulp.dest('dist/example/'));
});
gulp.task('watch', ['make:engine', 'make:example'], function(){
return watch('*.js', ['make:engine', 'make:example']);
});
gulp watch spits out:
[15:06:01] Using gulpfile ~/Documents/ice-cream-engine/gulpfile.js
[15:06:01] Starting 'make:engine'...
[15:06:01] Starting 'make:example'...
[15:06:03] Finished 'make:example' after 1.57 s
[15:06:03] Finished 'make:engine' after 1.6 s
[15:06:03] Starting 'watch'...
In ANY js file, if I make an edit and save the file, the terminal doesn't update to say that it is making the engine or the examples.
The project directory looks as follows:
What am I doing wrong such that it wont actually re-compile on ANY JS file change?
According to your screenshot, the only .js file that matches the glob you have passed to watch is gulpfile.js.
If you want to watch all .js files under the src and example directories, you could use the following:
gulp.task('watch', ['make:engine', 'make:example'], function(){
return watch(['src/**/*.js', 'examples/**/*.js'], ['make:engine', 'make:example']);
});
There is a project, its structure:
I do run gulp watch, when changing sass-file everything works correctly , but when I change js-file - assembly going infinite.
gulpfile:
var gulp = require('gulp');
var uglify = require('gulp-uglify');
var concat = require('gulp-concat');
var compass = require('gulp-compass');
gulp.task('compress-js', function() {
return gulp.src([
'./www/js/jquery/**/*.js',
'./www/js/vendor/**/*.js',
'./www/js/lib/**/*.js',
'./www/js/common/app.js',
'./www/js/pages/**/*.js',
'./www/js/common/main.js',
'!./www/js/combine.js'
])
.pipe(concat('combine.js'))
.pipe(uglify())
.pipe(gulp.dest('./www/js/'));
});
gulp.task('compress-css', function() {
return gulp.src('./scss/**/*.scss')
.pipe(compass({
config_file: './scss/config.rb',
css: './www/css',
sass: './scss'
}))
.pipe(gulp.dest('./www/css/'));
});
gulp.task('watch', function() {
gulp.watch(['./scss/**/*.scss'], ['compress-css']);
gulp.watch(['./www/js/**/*.js'], ['compress-js']);
});
Tell me please, what is wrong here?
The problem seems in your gulp.watch
gulp.watch(['./www/js/**/*.js'], ['compress-js']);
When you concat the js files for the build you exclude combine.js but you take it as good for the watch so he loop
I generate CSS from LESS files and want to give to all generated Bootstrap CSS files a prefix "bootsrtap", but not to the bootstrap.css. So, I set the prefix directly after the compilation, but all my attempts to do a futher rename are failing.
var gulp = require('gulp'),
less = require('gulp-less'),
watch = require('gulp-watch'),
prefix = require('gulp-autoprefixer'),
plumber = require('gulp-plumber'),
filter = require('gulp-filter'),
rename = require('gulp-rename'),
path = require('path')
;
// ...
gulp.task('build-vendors', function() {
gulp.src(['./public/components/bootstrap/less/theme.less', './public/components/bootstrap/less/bootstrap.less']) // path to less file
.pipe(plumber())
.pipe(less())
.pipe(rename({prefix: 'bootstrap-'}))
.pipe(gulp.dest('./public/css')) // path to css directory
;
});
gulp.task('clean-up', function() {
// const bootstrapFileRenameFilter = filter(['*', 'bootstrap-bootstrap.css']);
gulp.src('./public/css/bootstrap-bootstrap.css')
.pipe(plumber())
// .pipe(bootstrapFileRenameFilter)
.pipe(rename({basename: 'bootstrap.css'}))
.pipe(gulp.dest('./public/css'))
;
});
gulp.task('watch', function() {
gulp.watch('public/less/*.less', ['build-less', 'build-vendors'])
});
// gulp.task('default', ['watch', 'build-less', 'build-vendors']);
gulp.task('default', ['build-less', 'build-vendors', 'clean-up']);
What I expect:
./public/css/bootstrap-theme.css
./public/css/bootstrap.css
What I'm currently getting:
./public/css/bootstrap-theme.css
./public/css/bootstrap-bootstrap.css
How to rename a single file from a.foo to b.bar?
gulp-rename accepts a function as an argument to do the renaming. This allows you to target specific files for renaming using any criteria you like. In your case:
gulp.task('build-vendors', function() {
gulp.src(['./public/components/bootstrap/less/theme.less',
'./public/components/bootstrap/less/bootstrap.less'])
.pipe(plumber())
.pipe(less())
.pipe(rename(function(path) {
//rename all files except 'bootstrap.css'
if (path.basename + path.extname !== 'bootstrap.css') {
path.basename = 'bootstrap-' + path.basename;
}
}))
.pipe(gulp.dest('./public/css'));
});
Since you now only rename those files where you actually want to have the bootstrap- prefix, you don't have to clean-up your mess afterwards and can just drop the whole clean task altogether.
I have the following gulp task:
var es = require('event-stream'),
concat = require('gulp-concat'),
templateCache = require('gulp-angular-templatecache');
var scripts = gulp.src(paths.js + '/**/*.js'),
templates = gulp.src(paths.templates + '/**/*.html')
.pipe(templateCache('templates.js', {
module: 'spot.im.core',
standalone: false
}));
es.merge(templates, scripts)
.pipe(concat('all.js'))
.pipe(gulp.dest(paths.dist))
When running this, I don't get the templates inside all.js. But, when I am doing:
es.merge(templates)
.pipe(concat('all.js'))
.pipe(gulp.dest(paths.dist))
all.js contains the templates.
What am I doing wrong?
I think you forgot to define a task here; and, you can replace event-stream with merge-stream instead. I've had no trouble running merge-stream in many gulp tasks.
var merge = require('merge-stream'),
concat = require('gulp-concat'),
templateCache = require('gulp-angular-templatecache');
gulp.task('default', function () {
var scripts = gulp.src(paths.js + '/**/*.js');
var templates = gulp.src(paths.templates + '/**/*.html')
.pipe(templateCache('templates.js', {
module: 'spot.im.core',
standalone: false
}));
return merge(templates, scripts)
.pipe(concat('all.js'))
.pipe(gulp.dest(paths.dist));
});
Run on the command line with gulp.
I'm new to gulp, but I'm wondering if its possible to iterate through directories in a gulp task.
Here's what I mean, I know a lot of the tutorials / demos show processing a bunch of JavaScript files using something like "**/*.js" and then they compile it into a single JavaScript file. But I want to iterate over a set of directories, and compile each directory into it's own JS file.
For instance, I have a file structure like:
/js/feature1/something.js
/js/feature1/else.js
/js/feature1/foo/bar.js
/js/feature1/foo/bar2.js
/js/feature2/another-thing.js
/js/feature2/yet-again.js
...And I want two files: /js/feature1/feature1.min.js and /js/feature2/feature2.min.js where the first contains the first 4 files and the second contains the last 2 files.
Is this possible, or am I going to have to manually add those directories to a manifest? It would be really nice to pragmatically iterate over all the directories within /js/.
Thanks for any help you can give me.
-Nate
Edit: It should be noted that I don't only have 2 directories, but I have many (maybe 10-20) so I don't really want to write a task for each directory. I want to handle each directory the same way: get all of the JS inside of it (and any sub-directories) and compile it down to a feature-based minified JS file.
There's an official recipe for this: Generating a file per folder
var fs = require('fs');
var path = require('path');
var merge = require('merge-stream');
var gulp = require('gulp');
var concat = require('gulp-concat');
var rename = require('gulp-rename');
var uglify = require('gulp-uglify');
var scriptsPath = 'src/scripts';
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
gulp.task('scripts', function() {
var folders = getFolders(scriptsPath);
var tasks = folders.map(function(folder) {
return gulp.src(path.join(scriptsPath, folder, '/**/*.js'))
// concat into foldername.js
.pipe(concat(folder + '.js'))
// write to output
.pipe(gulp.dest(scriptsPath))
// minify
.pipe(uglify())
// rename to folder.min.js
.pipe(rename(folder + '.min.js'))
// write to output again
.pipe(gulp.dest(scriptsPath));
});
// process all remaining files in scriptsPath root into main.js and main.min.js files
var root = gulp.src(path.join(scriptsPath, '/*.js'))
.pipe(concat('main.js'))
.pipe(gulp.dest(scriptsPath))
.pipe(uglify())
.pipe(rename('main.min.js'))
.pipe(gulp.dest(scriptsPath));
return merge(tasks, root);
});
You could use glob to get a list of directories and iterate over them, using gulp.src to create a separate pipeline for each feature. You can then return a promise which is resolved when all of your streams have ended.
var fs = require('fs');
var Q = require('q');
var gulp = require('gulp');
var glob = require('glob');
gulp.task('minify-features', function() {
var promises = [];
glob.sync('/js/features/*').forEach(function(filePath) {
if (fs.statSync(filePath).isDirectory()) {
var defer = Q.defer();
var pipeline = gulp.src(filePath + '/**/*.js')
.pipe(uglify())
.pipe(concat(path.basename(filePath) + '.min.js'))
.pipe(gulp.dest(filePath));
pipeline.on('end', function() {
defer.resolve();
});
promises.push(defer.promise);
}
});
return Q.all(promises);
});
I am trying myself to get how streams work in node.
I made a simple example for you, on how to make a stream to filter folders and start a new given stream for them.
'use strict';
var gulp = require('gulp'),
es = require('event-stream'),
log = require('consologger');
// make a simple 'stream' that prints the path of whatever file it gets into
var printFileNames = function(){
return es.map(function(data, cb){
log.data(data.path);
cb(null, data);
});
};
// make a stream that identifies if the given 'file' is a directory, and if so
// it pipelines it with the stream given
var forEachFolder = function(stream){
return es.map(function(data, cb){
if(data.isDirectory()){
var pathToPass = data.path+'/*.*'; // change it to *.js if you want only js files for example
log.info('Piping files found in '+pathToPass);
if(stream !== undefined){
gulp.src([pathToPass])
.pipe(stream());
}
}
cb(null, data);
});
};
// let's make a dummy task to test our streams
gulp.task('dummy', function(){
// load some folder with some subfolders inside
gulp.src('js/*')
.pipe(forEachFolder(printFileNames));
// we should see all the file paths printed in the terminal
});
So in your case, you can make a stream with whatever you want to make with the files in a folder ( like minify them and concatenate them ) and then pass an instance of this stream to the forEachFolder stream I made. Like I do with the printFileNames custom stream.
Give it a try and let me know if it works for you.
First, install gulp-concat & gulp-uglify.
$ npm install gulp-concat
$ npm install gulp-uglify
Next, do something like:
//task for building feature1
gulp.task('minify-feature1', function() {
return gulp.src('/js/feature1/*')
.pipe(uglify()) //minify feature1 stuff
.pipe(concat('feature1.min.js')) //concat into single file
.pipe(gulp.dest('/js/feature1')); //output to dir
});
//task for building feature2
gulp.task('minify-feature2', function() { //do the same for feature2
return gulp.src('/js/feature2/*')
.pipe(uglify())
.pipe(concat('feature2.min.js'))
.pipe(gulp.dest('/js/feature2'));
});
//generic task for minifying features
gulp.task('minify-features', ['minify-feature1', 'minify-feature2']);
Now, all you have to do to minify everything from the CLI is:
$ gulp minify-features
I had trouble with the gulp recipe, perhaps because I'm using gulp 4 and/or because I did not want to merge all my folders' output anyway.
I adapted the recipe to generate (but not run) an anonymous function per folder and return the array of functions to enable them to be processed by gulp.parallel - in a way where the number of functions I would generate would be variable. The keys to this approach are:
Each generated function needs to be a function or composition (not a stream). In my case, each generated function was a series composition because I do lots of things when building each module folder.
The array of functions needs to passed into my build task using javascript apply() since every member of the array needs to be turned into an argument to gulp.parallel in my case.
Excerpts from my function that generates the array of functions:
function getModuleFunctions() {
//Get list of folders as per recipe above - in my case an array named modules
//For each module return a function or composition (gulp.series in this case).
return modules.map(function (m) {
var moduleDest = env.folder + 'modules/' + m;
return gulp.series(
//Illustrative functions... all must return a stream or call callback but you can have as many functions or compositions (gulp.series or gulp.parallel) as desired
function () {
return gulp.src('modules/' + m + '/img/*', { buffer: false })
.pipe(gulp.dest(moduleDest + '/img'));
},
function (done) {
console.log('In my function');
done();
}
);
});
}
//Illustrative build task, running two named tasks then processing all modules generated above in parallel as dynamic arguments to gulp.parallel, the gulp 4 way
gulp.task('build', gulp.series('clean', 'test', gulp.parallel.apply(gulp.parallel, getModuleFunctions())));
`