I am attempting to re-run my gulp build when gulpfile.js changes, but I am having issues with the method all of my research has lead me to.
I have one watcher for all my less and javascript files and a configuration object that has the list of files to watch, how they are output, etc. This is a stripped-down example of what it looks like:
var $ = require('gulp-load-plugins')();
var config = {
root: rootPath,
output: {
app: 'app',
vendor: 'vendor'
}, // ...
};
gulp.task('default', ['build', 'watch']);
gulp.task('build', ['clean', 'less:app', 'less:theme', 'css:vendor', 'js:app', 'js:vendor', 'rev', 'css:copyfonts']);
gulp.task('watch', function () {
var allFiles = config.styles.appSrc
.concat(config.styles.vendorSrc)
.concat(config.scripts.appSrc)
.concat(config.scripts.vendorSrc);
$.watch(allFiles, function () {
gulp.start('default');
});
});
gulp.task('watch:gulp', function () {
var p;
gulp.watch('gulpfile.js', spawnUpdatedGulp);
spawnUpdatedGulp();
function spawnUpdatedGulp() {
if (p) {
p.kill();
}
p = spawn('gulp', ['default', '--color'], { stdio: 'inherit' });
}
});
// .. other build tasks ..
The above code shows how I tried the accepted answer to this:
How can Gulp be restarted upon each Gulpfile change?
However, it has a major issue. When I run watch:gulp, it runs the build just fine, and everything is great. The config.output.app variable is how the app specific css and js files are named, so my test case has been:
run gulp:watch, check that the css output is named according to config.output.app
change config.output.app, and perform step #1 again
save any random javascript file that it is watching, and see if it builds correctly
Step 3 is riddled with permission errors because of multiple watchers on the files, and this only gets worse the more I repeat steps 1 and 2. Visual Studio will even freeze.
I have not found a way to clean up the old watchers. I tried to manually kill them like this:
var appFileWatcher;
gulp.task('watch', function () {
var allFiles = config.styles.appSrc
.concat(config.styles.vendorSrc)
.concat(config.scripts.appSrc)
.concat(config.scripts.vendorSrc);
appFileWatcher = $.watch(allFiles, function () {
gulp.start('default');
});
});
gulp.task('watch:gulp', function () {
var p;
var gulpWatcher = $.watch('gulpfile.js', spawnUpdatedGulp);
spawnUpdatedGulp();
function spawnUpdatedGulp() {
if (p) {
p.kill();
}
if (appFileWatcher) {
appFileWatcher.unwatch();
}
gulpWatcher.unwatch();
p = spawn('gulp', ['default', '--color'], { stdio: 'inherit' });
}
});
This also does not work. I still get multiple watchers trying to perform the build when I perform my same test case.
How do I kill those watchers that stay around after the new gulp process is spawned?
I am a grunt and node noob but I managed to write a node script that does what I want it to and works from the command line. I don't want to publish the script as a node module but I would like to run it from my grunt file.
What changes (if any) do I need to make to the script for this to work?
The more I read about configuring grunt files and custom tasks the more confused I get. I currently have something that looks like this:
module.exports = function(grunt) {
grunt.initConfig({
'mytaskname': 'what goes here?'
});
grunt.loadNpmTasks('./node_modules/script_name');
grunt.registerTask('run-from-command-line', 'description', function() {
grunt.log.writeln('Not running...');
});
}
Any help would be greatly appreciated.
You could use the grunt-execute plugin for doing this which executes files in a node.js child process.
Example:
If your node script is in "node-scripts/script.js", Gruntfile.js would look something like this:
module.exports = function(grunt) {
grunt.initConfig({
execute: {
target: {
src: ["node-scripts/script.js"]
}
}
});
// Load the plugins
grunt.loadNpmTasks("grunt-execute");
grunt.registerTask("default", ["execute"]);
};
I'm new to Grunt and I'm trying to use the grunt-bower-concat node module to concat all my bower components into a single js file and css file respectively. It's working great, except that I want to force grunt-bower-concat to use the minified versions of my bower components instead of the uncompressed versions.
Luckily, it comes with a callback feature where I can customize this:
callback: function(mainFiles, component) {
return _.map(mainFiles, function(filepath) {
// Use minified files if available
var min = filepath.replace(/\.js$/, '.min.js');
return grunt.file.exists(min) ? min : filepath;
});
}
And I added it to my Gruntfile:
module.exports = function (grunt) {
grunt.initConfig({
pkg: grunt.file.readJSON('package.json'),
bower_concat: {
all: {
dest: "src/js/<%= pkg.name %>-bower.js",
cssDest: "src/css/<%= pkg.name %>-bower.css",
callback: function(mainFiles) {
return _.map(mainFiles, function(filepath) {
var min = filepath.replace(/\.js$/, '.min.js');
return grunt.file.exists(min) ? min : filepath;
});
}
}
},
...
And it fails with the following error:
$ /usr/local/bin/grunt --gruntfile /Applications/MAMP/htdocs/proj/Gruntfile.js bower_concat
Running "bower_concat:all" (bower_concat) task
Fatal error: _ is not defined
Process finished with exit code 3
This example is trying to use underscore's map function and it's clear Grunt does not have access to this library.
How can I load underscore or use it's functions inside of my Gruntfile?
Instead of requiring an extra library, simply replace
return _.map(mainFiles, function(filepath) {
With this:
return mainFiles.map(function(filepath) {
Doesn't look like you required underscore anywhere, unless you're not showing the whole file.
Any file in which you want to use underscore you need to do:
var _ = require('underscore');
before making use of _.
Oh, and of course you need to npm install underscore --save in the folder the gruntfile is in as well, to have the library there.
I am trying to use gulp-requirejs to build a demo project. I expect result to be a single file with all js dependencies and template included. Here is my gulpfile.js
var gulp = require('gulp');
var rjs = require('gulp-requirejs');
var paths = {
scripts: ['app/**/*.js'],
images: 'app/img/**/*'
};
gulp.task('requirejsBuild', function() {
rjs({
name: 'main',
baseUrl: './app',
out: 'result.js'
})
.pipe(gulp.dest('app/dist'));
});
// The default task (called when you run `gulp` from cli)
gulp.task('default', ['requirejsBuild']);
The above build file works with no error, but the result.js only contains the content of main.js and config.js. All the view files, jquery, underscore, backbone is not included.
How can I configure gulp-requirejs to put every js template into one js file?
If it is not the right way to go, can you please suggest other method?
Edit
config.js
require.config({
paths: {
"almond": "/bower_components/almond/almond",
"underscore": "/bower_components/lodash/dist/lodash.underscore",
"jquery": "/bower_components/jquery/dist/jquery",
"backbone": "/bower_components/backbone/backbone",
"text":"/bower_components/requirejs-text/text",
"book": "./model-book"
}
});
main.js
// Break out the application running from the configuration definition to
// assist with testing.
require(["config"], function() {
// Kick off the application.
require(["app", "router"], function(app, Router) {
// Define your master router on the application namespace and trigger all
// navigation from this instance.
app.router = new Router();
// Trigger the initial route and enable HTML5 History API support, set the
// root folder to '/' by default. Change in app.js.
Backbone.history.start({ pushState: false, root: '/' });
});
});
The output is just a combination this two files, which is not what I expected.
gulp-requirejs has been blacklisted by the gulp folks. They see the RequireJS optimizer as its own build system, incompatible with gulp. I don't know much about that, but I did find an alternative in amd-optimize that worked for me.
npm install amd-optimize --save-dev
Then in your gulpfile:
var amdOptimize = require('amd-optimize');
var concat = require('gulp-concat');
gulp.task('bundle', function ()
{
return gulp.src('**/*.js')
.pipe(amdOptimize('main'))
.pipe(concat('main-bundle.js'))
.pipe(gulp.dest('dist'));
});
The output of amdOptimize is a stream which contains the dependencies of the primary module (main in the above example) in an order that resolves correctly when loaded. These files are then concatenated together via concat into a single file main-bundle.js before being written into the dist folder.
You could also minify this file and perform other transformations as needed.
As an aside, in my case I was compiling TypeScript into AMD modules for bundling. Thinking this through further I realized that when bundling everything I don't need the asynchronous loading provided by AMD/RequireJS. I am going to experiment with having TypeScript compile CommonJS modules instead, then bundling them using webpack or browserify, both of which seem to have good support within gulp.
UPDATE
My previous answer always reported taskReady even if requirejs reported an error. I reconsidered this approach and added error logging. Also I try to fail the build completely as described here gulp-jshint: How to fail the build? because a silent fail really eats your time.
See updated code below.
Drew's comment about blacklist was very helpfull and gulp folks suggest using requirejs directly. So I post my direct requirejs solution:
var DIST = './dist';
var requirejs = require('requirejs');
var requirejsConfig = require('./requireConfig.js').RJSConfig;
gulp.task('requirejs', function (taskReady) {
requirejsConfig.name = 'index';
requirejsConfig.out = DIST + 'app.js';
requirejsConfig.optimize = 'uglify';
requirejs.optimize(requirejsConfig, function () {
taskReady();
}, function (error) {
console.error('requirejs task failed', JSON.stringify(error))
process.exit(1);
});
});
The file at ./dist/app.js is built and uglified. And this way gulp will know when require has finished building. So the task can be used as a dependency.
My solution works like this:
./client/js/main.js:
require.config({
paths: {
jquery: "../vendor/jquery/dist/jquery",
...
},
shim: {
...
}
});
define(["jquery"], function($) {
console.log($);
});
./gulpfile.js:
var gulp = require('gulp'),
....
amdOptimize = require("amd-optimize"),
concat = require('gulp-concat'),
...
gulp.task('scripts', function(cb) {
var js = gulp.src(path.scripts + '.js')
.pipe(cached('scripts'))
.pipe(jshint())
.pipe(jshint.reporter('default'))
.pipe(remember('scripts'))
.pipe(amdOptimize("main",
{
name: "main",
configFile: "./client/js/main.js",
baseUrl: './client/js'
}
))
.pipe(concat('main.js'));
.pipe(gulp.dest(path.destScripts));
}
...
This part was important:
configFile: "./client/js/main.js",
baseUrl: './client/js'
This allowed me to keep my configuration in one place. Otherwise I was having to duplicate my paths and shims into gulpfile.js.
This works for me. I seems that one ought to add in uglification etc via gulp if desired. .pipe(uglify()) ...
Currently I have to duplicate the config in main.js to run asynchronously.
....
var amdOptimize = require("amd-optimize");
...
var js = gulp.src(path.scripts + '.js')
.pipe(cached('scripts'))
.pipe(jshint())
.pipe(jshint.reporter('default'))
.pipe(remember('scripts'))
.pipe(amdOptimize("main",
{
name: "main",
paths: {
jquery: "client/vendor/jquery/dist/jquery",
jqueryColor: "client/vendor/jquery-color/jquery.color",
bootstrap: "client/vendor/bootstrap/dist/js/bootstrap",
underscore: "client/vendor/underscore-amd/underscore"
},
shim: {
jqueryColor : {
deps: ["jquery"]
},
bootstrap: {
deps: ["jquery"]
},
app: {
deps: ["bootstrap", "jqueryColor", "jquery"]
}
}
}
))
.pipe(concat('main.js'));
Try this code in your gulpfile:
// Node modules
var
fs = require('fs'),
vm = require('vm'),
merge = require('deeply');
// Gulp and plugins
var
gulp = require('gulp'),
gulprjs= require('gulp-requirejs-bundler');
// Config
var
requireJsRuntimeConfig = vm.runInNewContext(fs.readFileSync('app/config.js') + '; require;'),
requireJsOptimizerConfig = merge(requireJsRuntimeConfig, {
name: 'main',
baseUrl: './app',
out: 'result.js',
paths: {
requireLib: 'bower_modules/requirejs/require'
},
insertRequire: ['main'],
// aliases from config.js - libs will be included to result.js
include: [
'requireLib',
"almond",
"underscore",
"jquery",
"backbone",
"text",
"book"
]
});
gulp.task('requirejsBuild', ['component-scripts', 'external-scripts'], function (cb) {
return gulprjs(requireJsOptimizerConfig)
.pipe(gulp.dest('app/dist'));
});
Sorry for my english. This solution works for me. (I used gulp-requirejs at my job)
I think you've forgotten to set mainConfigFile in your gulpfile.js. So, this code will be work
gulp.task('requirejsBuild', function() {
rjs({
name: 'main',
mainConfigFile: 'path_to_config/config.js',
baseUrl: './app',
out: 'result.js'
})
.pipe(gulp.dest('app/dist'));
});
In addition, I think when you run that task in gulp, require can not find its config file and
This is not gulp-requirejs fault.
The reason why only main.js and config.js is in the output is because you're not requiring/defining any other files. Without doing so, the require optimizer wont understand which files to add, the paths in your config-file isn't a way to require them!
For example you could load a main.js file from your config file and in main define all your files (not optimal but just a an example).
In the bottom of your config-file:
// Load the main app module to start the app
requirejs(["main"]);
The main.js-file: (just adding jquery to show the technique.
define(["jquery"], function($) {});
I might also recommend gulp-requirejs-optimize instead, mainly because it adds the minification/obfuscation functions gulp-requirejs lacks: https://github.com/jlouns/gulp-requirejs-optimize
How to implement it:
var requirejsOptimize = require('gulp-requirejs-optimize');
gulp.task('requirejsoptimize', function () {
return gulp.src('src/js/require.config.js')
.pipe(requirejsOptimize(function(file) {
return {
baseUrl: "src/js",
mainConfigFile: 'src/js/require.config.js',
paths: {
requireLib: "vendor/require/require"
},
include: "requireLib",
name: "require.config",
out: "dist/js/bundle2.js"
};
})).pipe(gulp.dest(''));
});
I have RequireJS implemented fine, and a Grunt based build process which is optimizing the all the JS files app into one file via r.js which is also working fine. All my app files are concatenated into one big JS file for efficient production deployment.
Now I'm having the following requirements:
I need to write a plugin for requirejs, that will not load(not include the file) into the optimized file in the build process, but will required on demand:
Meaning in my code I'll have:
var myObj = require("myplugIn!jsFile");
So in the end when this line runs, it will runs in 2 options:
on build process, the file is not included in the optimized file
The application is running, it will be request the file on demand.
I wrote the following plugin, but is not working:
define(function () {
"use strict";
return {
load : function (name, req, onload, config) {
// we go inside here we are running the application not in build process
if (!config.isBuild) {
req([name], function () {
onload(arguments[0]);
});
}
}
};
});
What I'm missing here.
In your build configuration you can exclude files that you don't want to bundle. They will still be loaded on demand when needed. You may also do something like this:
define(function (){
// module code...
if (condition){
require(['mymodule'], function () {
// execute when mymodule has loaded.
});
}
}):
This way mymodule will be loaded only if condition is met. And only once, if you use same module dependency elsewhere it will return loaded module.
It was more simpler that I though, if helps someone, I'm posting the solution, I create a plugin , that in build process return nothing and in run time, returns the required file, hope helps someone.
define(function () {
"use strict";
return {
load : function (name, req, onload, config) {
if (config.isBuild) {
onload(null);
} else {
req([name], function () {
onload(arguments[0]);
});
}
}
};
});