Gulp run alternative - gulp

Everytime I run gulp, I see this message gulp.run() has been deprecated. Use task dependencies or gulp.watch task triggering instead.
Example code:
var watch = require('gulp-watch');
watch(['public/**/*.js','!public/**/*.min.js'],function(){
gulp.run('compressjs');
gulp.run('bs-reload');
});
How can I avoid using gulp.run() with gulp-watch?

You shouldn't use run. Here is an alternative (to address that part of your answer), but not what you need to do:
gulp
.start('default')
.once('task_stop', function(){
//do other stuff.
});
If you really must fire an ad hoc task, but can literally use run...You can use .start with the task name, and also attach to the task_stop handler to fire something when the task is complete. This is nice when writing tests for gulp tasks, but that's really it.
however in day to day gulp usage, this is an antipattern.
Normally, you build smaller tasks and composite them. This is the right way. See this:
var gulp = require('gulp'),
runSequence = require('run-sequence');
function a(){
//gulpstuff
}
function b(){
//gulpstuff
}
function d(callback){
runSequence('a', 'b', callback)
}
gulp
.task('a', a) // gulp a -runs a
.task('b', b) // gulp b runs b
.task('c', ['a', 'b']) //gulp c runs a and b at the same time
.task('d', d); //gulp d runs a, then b.
basically if c or d was a watch task, you'd achieve the same goal of firing the already registered smaller gulp tasks without .run

gulp.run() was deprecated because people were using it as a crutch. You are using it as a crutch!
I'm not sure why you're using gulp-watch, the built in gulp.watch would be far more appropriate for what you're using it for. Have a look at the documentation for .watch: https://github.com/gulpjs/gulp/blob/master/docs/API.md#gulpwatchglob--opts-tasks-or-gulpwatchglob--opts-cb
Here's what you should have written. Please understand why you're using it instead of just copying it:
gulp.watch(['public/**/*.js','!public/**/*.min.js'], ['compressjs', 'bs-reload'])

You can always just use plain old javascript functions. From what i've read this is considered to be a more "gulp-ish" way of doing things.
I ran into a similar situation once and basically solved it with something like this:
var watch = require('gulp-watch');
watch(['public/**/*.js','!public/**/*.min.js'], function(){
compress();
bsReload();
});
And then these functions are basically wrappers around the guts of what would have been your original gulp tasks:
var compress = function () {
return gulp.src("stuff/**")
.pipe(gulp-compress())
.pipe(gulp.dest("./the_end/");
};
Its easy to become caught up in the idea that one has to use gulp tasks for everything otherwise you are "doing it wrong" but if you need to use something like this go for it.
If you also want a gulp task with the same functionality then whip up something like this:
gulp.task("compress", function () {
return compress();
});
and you can still take advantage of gulp task dependecies when using the same code if you need it somewhere else.

Related

Gulp4. "AssertionError : Task never defined" when calling or importing tasks

Below you can see simplified view of an issue. Basically, I'm able to call task1.js using gulp.series in tasks task2,3.js, but once I add same code to call task1.js in task4.js - Task never defined: task1 error gets thrown.
There are more tasks in the tasks folder than in file structure example below.
I've got three tasks,
...
/tasks
build.js
clean.js
dev.js
gulpfile.babel.js
...
all of them required in gulpfile.babel.js using the require-dir package
import requireDir from 'require-dir';
requireDir('./tasks', {recurse: true});
This allows me to call a task from clean.js at dev.js, and it works fine.
import gulp from 'gulp';
gulp.task('dev', gulp.series('clean');
But after I add same code structure at build.js.
import gulp from 'gulp';
gulp.task('build', gulp.series('clean');
it somehow breaks gulp stream (I guess), so now on any task call I get:
$gulp dev
-AssertionError [ERR_ASSERTION]: Task never defined: clean.
$gulp -v
[11:50:11] CLI version 2.0.1
[11:50:11] Local version 4.0.0
For those migrating from gulp v3 to v4 or are using gulp.task() to define tasks in gulp v4 and get this error message: Task never defined, the problem usually lies here:
Forward references
A forward reference is when you compose tasks, using string
references, that haven't been registered yet. This was a common
practice in older versions, but this feature was removed to achieve
faster task runtime and promote the use of named functions. In newer
versions, you'll get an error, with the message "Task never defined",
if you try to use forward references. You may experience this when
trying to use exports for your task registration and composing tasks
by string. In this situation, use named functions instead of string
references.
During migration, you may need to use the forward reference registry.
This will add an extra closure to every task reference and
dramatically slow down your build. Don't rely on this fix for very
long.
From gulpjs documentation re: gulp.series and gulp.parallel documentation.
Here is what that means. There are two ways to create tasks:
1. gulp.task('someStringAsTask', function() {..})
2. function myNamedFunction () {…}
When you use version 1 (gulp.task…) you cannot refer to that task by its string name until it has been registered. So you cannot do this:
exports.sync = gulp.series('sass2css', serve, watch);
// or gulp.task('dev', gulp.series('sass2css', serve, watch); doesn't work either
gulp.task('sass2css', function() {
return gulp.src(paths.sass.stylesFile)
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest(paths.css.temp))
.pipe(reload({ stream: true }));
})
Results in
AssertionError [ERR_ASSERTION]: Task never defined: sass2css
That is a forward reference, composing a task (using gulp.series or gulp.parallel) and referring to a task by its string name (in the above case 'sass2css') before it has been registered. (calling "gulp.task(…..)" is the act of registering) Putting the gulp.task('sass2css',...) first fixes the problem.
If you use version two of defining a task:
function sass2css() {
return gulp.src(paths.sass.stylesFile)
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest(paths.css.temp))
.pipe(reload({ stream: true }));
}
you are now using a named function to register a task and do not need to use its name as a string. So this now works:
exports.sync = gulp.series(sass2css, serve, watch);
// gulp.task('dev', gulp.series(sass2css, serve, watch); this also works
followed by (or preceded by - either works):
function sass2css() {
return gulp.src(paths.sass.stylesFile)
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest(paths.css.temp))
.pipe(reload({ stream: true }));
}
The original OP used this and it worked:
import gulp from 'gulp';
gulp.task('dev', gulp.series('clean');
Noted that clean.js got imported before dev.js so that was okay.
This didn't work:
import gulp from 'gulp';
gulp.task('build', gulp.series('clean');
because the string-referenced task, 'clean' gets imported (and thus registered) after build.js where it is referenced - thus creating an illegal forward reference to a string-referenced task.
So there are two standard ways to fix this error:
Use named functions to define tasks not gulp.task('someTask',...). Then it doesn't matter the order of using those named functions when composing other tasks, i.e., when using gulp.series or gulp.parallel. And there are other advantages to using named functions, such as passing arguments, so this is the best option.
If you do use the older gulp v3 gulp.task method of creating tasks with string references, be careful to not refer to those tasks until after the task is actually created.
Also see my answer at task never defined error for fixing another problem which results in the same error message. Specifically using gulp.task('someTask', ['anotherTask'], function(){}) synatx in a gulp4 file.
The series and parallel functions of gulp 4 do not create a task definition as its README seems to suggest, but instead they both run the tasks in parameter. In order to work as intended, one need to surround the call with a closure.
So, to fix the excerpt
gulp.task('build', gulp.series('clean'));
it is necessary to add the closure:
// Older EcmaScripts:
gulp.task('build', function() { return gulp.series('clean') });
// EcmaScript 6:
gulp.task('build', () => gulp.series('clean'));
I had a similar setup where I had recursively require'd all tasks under a directory. And after updating to gulp 4 started getting error Task never defined.
I tried Pedro solution, but this caused another error:
The following tasks did not complete: default
Did you forget to signal async completion?
The solution was fairly simple for me, just import the missing tasks.
import gulp from 'gulp';
import './clean';
gulp.task('build', gulp.series('clean'));
The easiest solution might be using the official undertaker-forward-reference package:
const gulp = require("gulp");
const FwdRef = require("undertaker-forward-reference");
gulp.registry(FwdRef()); // Or gulp.registry(new FwdRef());
gulp.task("firstRegisteredTask", gulp.series("laterRegisteredTask")); // Works thanks to undertaker-forward-reference
gulp.task("laterRegisteredTask", () => {
return gulp.src("someGlob").pipe(gulp.dest("someFolder"));
});
This solution might negatively affect performance though (source):
This will add an extra closure to every task reference and dramatically slow down your build.

With gulp, are watch tasks guaranteed to run before other tasks?

I have a gulp watch task that uses gulp-sass to convert SASS to CSS. For development, separated CSS files are all I want.
For release, I have a min tasks that uses gulp-cssmin and concat to bundle and minify the CSS. However, the min task doesn't deal with the SASS files. It assumes they have already been converted to CSS.
Is this a valid assumption? What if an automated build gets the latest source (which does not have .scss files) and runs min? Is there a race condition here as to whether the watch task will run in time for min to pick up the SASS files?
If your min task is also watching the SASS files, your assumption is not necessarily valid. By default, all the tasks will run at the same time. You can however define dependencies in your task, so that a task does not start until another task is finished. This could solve your problem. The documentation about dependencies can be found here.
There are also two other solutions:
Watch the CSS files for the min task.
Run the logic in your min task at the end of the task that compiles the SASS.
Edward! Better use optional minification to prevent extra reading from the disk.
var env = require('minimist')(process.argv.slice(2));
var noop = require('readable-stream/passthrough').bind(null, { objectMode: true });
gulp.task('style', function () {
return gulp.src('...')
.pipe(process())
.pipe(env.min ? uglify() : noop())
.pipe(gulp.dest('...'))
});
Also I don't recommend these tools from gulp-util which will be deprecated
Even if gulp guaranteed that it would wait for watch tasks to complete before starting dependent tasks, a race could still occur during the interval from when a source file is changed until the OS notifies gulp. So I decided not to use the SASS → CSS transform output. For as infrequently as I need to build the minimized versions, there's no payoff for saving those milliseconds.
Instead I used stream-combiner2 to reuse the transform logic:
var combiner = require('stream-combiner2');
function scssTransform() {
return combiner(
sourcemaps.init(),
sass(),
sourcemaps.write());
}
I use the transform for my non-mimized "sandbox" for local development:
gulp.task('sandbox:css', function () {
return gulp.src(paths.scss)
.pipe(scssTransform())
.pipe(gulp.dest(dirs.styles));
});
My task for serving the site locally depends on and watches the task:
gulp.task('serve.sandbox', ['sandbox:css' /* ... more ... */], function () {
// ... Start the web server ...
gulp.watch(paths.scss, ['sandbox:css']);
// ... Watch more stuff ...
});
I also use the transform for my minimized deployment:
gulp.task('deploy:css', function () {
return gulp.src(paths.scss)
.pipe(scssTransform())
.pipe(concat(files.cssMin))
.pipe(cssmin())
.pipe(gulp.dest(dirs.deploy));
});

gulp, build multiple projects

I have a Gulp build process that runs through roughly 10 tasks, including browserify and watch. It currently builds a common-bundle.js, and common-libs.js. It uses browser-sync to give me sub-second rebuilds.
Now I want to also build a project that depends on the common project. I want to retain the live rebuilds of both common and this project so that I could work on both of them at the same time. I want to keep the build process itself as DRY as possible and reuse the tasks i created to build common.
For example, a sample task:
var config = require('../config');
gulp.task('styles', function () {
return gulp.src(config.styles.src) // if i could tell it to get config elsewhere...
...
I can't pass a parameter into each task to tell it, go run the task but use:
var config = require('../config').common;
vs.
var config = require('../config').projectA;
I don't think tasks can take parameters.
Is there a different way to structure this?
git/gist link would be highly appreciated.
For now I am trying this approach - each task js file has 2 tasks defined, but at least the logic of the task is reused. I still wish for something cleaner.
./gulp/task/style.js:
function styles(config){
return gulp.src(config.styles.src)
...
}
}
gulp.task('styles', function () {
styles(config.common);
});
gulp.task('stylesProject1', function () {
styles(config.project1);
});
devTask.js:
runSequence(['styles', 'stylesProject1], 'watch', callback);

Gulp task order

I'm having some trouble with a couple of gulp tasks. i've tried everything and followed the documentation but nothing seems to work. Hoping someone can help.
What i'm trying to do is have sass minification happen, then rsync the result to a local vagrant folder (NOT a remote server), then reload browser sync.
The problem i am having is that I'm fighting the fact that gulp wants to run all the tasks together. I need them to happen one after the other. All the tasks work on their own, i am just having trouble making them run sequentially. I've tried callbacks and playing with dependency but i'm obviously missing something.
My setup is complicated, all my tasks are in separate js files but i've tried to combine what i have into this single github gist so people can help. Thanks for any assistance.
https://gist.github.com/CodeStalker/9661725dcaf105d2ed6c
The only way I have got this to work is to pipe rsync onto the end of the sass magnification, and wrap it in gulp-if using a server: true variable so it only does the rsync if it knows its running on a VM. Not ideal.
This is a common issue people run into with Gulp, especially when coming from Grunt. Gulp is async to the max, it always wants to do as many things as it can all at the same time.
The first step to getting tasks to run sequentially is to let gulp know when your task ends, this can be done a couple different ways.
1) Return a stream, gulp will wait for the stream's "end" event as the trigger for that task being done. Example:
gulp.task( "stream-task", function(){
return gulp.src( srcGlob )
.pipe(sass())
.pipe(compressCSS())
.pipe(gulp.dest( destGlob ));
});
2) Return a Promise, gulp will wait for the promise to enter a resolved state before signaling the task as done. Example: (not perfect just to get the point across)
gulp.task( "promise-task", function() {
return new Promise(function(resolve, reject){
fs.readFile( "filename", function( err, data ){
if( err ){ return reject(err); }
return resolve( data );
});
});
});
3) Call the task callback, if the function doesn't return anything but the function signature takes an argument, that argument will be a callback function you can call to signal the task as done. Example:
gulp.task( "cb-task", function( done ){
fs.readFile( "filename", function( err, data ){
// do stuff, call done...
done();
});
});
Most often you are going to be returning a stream like example 1, which is what a typical gulp task looks like. Options 2 and 3 are more for when you are doing something that isn't really a traditional stream based gulp task.
The next thing is setting the "dependency" of one task for another. The gulp docs show this as the way you do that:
gulp.task( "some-task-with-deps", [ "array-of-dep", "task-names" ], function(){ /* task body */ });
Now I don't know the current status but there were some issues with these dependency tasks not running in the proper order. This was originally caused by a problem with one of gulp's dependencies (orchestrator I believe). One kind gentleman out there in NPM land made a nice little package to be used in the interim while the bugs were being worked out. I started using it to order my tasks and haven't looked back.
https://www.npmjs.com/package/run-sequence
The documentation is good so I won't go into a lot of detail here. Basically run-sequence lets explicitly order your gulp tasks, just remember it doesn't work if you don't implement one of the three options above for each of your tasks.
Looking at your gist adding a couple missing return statements in your tasks may just do the trick, but as an example this is what my "dev" task looks like for my project at work...
gulp.task( "dev", function( done ){
runSequence(
"build:dev",
"build:tests",
"server:dev",
[
"less:watch",
"jscs:watch",
"lint:watch",
"traceur:watch"
],
"jscs:dev",
"lint:dev",
"tdd",
done // <-- thats the callback method to gulp let know when this task ends
);
});
Also for reference my "build:dev" task is another use of run-sequence
gulp.task( "build:dev", function( done ){
runSequence(
"clean:dev",
[
"less:dev",
"symlink:dev",
"vendor:dev",
"traceur:dev"
],
done // <-- thats the callback method to let know when this task ends
);
});
If the tasks need to be run in order the task name gets added as its own argument to runSequence if they don't conflict send them in as an array to have the tasks run at the same time and speed up your build process.
One thing to note about watch tasks! Typically watch tasks run indefinitely so trying to return the "infinite" stream from them may make gulp think that the task never ends. In that case I'll use the callback method to make gulp think the task is done even if it is still running, something like...
gulp.task( "watch-stuff", function( done ){
gulp.watch( watchGlob )
.on( "change", function( event ){ /* process file */ });
done();
});
For more on watch tasks and how I do incremental builds check out an answer I wrote the other day, Incremental gulp less build
Here you go, I believe it should work now. If not let me know, I might have made a typo.
Your gist fixed
I think you are hitting (i don't have enough reputation points to post link, google issues #96) which is fixed in 4.x which isn't out yet :-). That said, check this hack out: My post about Gulp v3.x bug

Is gulp lying about how long a task takes?

Gulp says it took 13ms to run my "js" command, but then it hangs for another 2 seconds or so after it's "complete".
So either my task didn't complete in 13ms, or gulp is twiddling its thumbs for an extra two seconds after it's done. Which is it? Is there a way to fix this? I'd like to know how long my tasks are really taking; maybe I can trim some fat off.
First, you need to have async tasks set up correctly. This usually means simply returning the stream, but you can also accept a callback or return a promise:
gulp.task('foo', function() {
return gulp.src()...
});
gulp.task('foo', function(cb) {
// use the callback however
doAsyncThing().on('done', cb);
});
Second, tasks are always run simultaneously, unless they have a dependency on each other. If you need to run tasks in a series, rather than parallel, and cannot use the built-in dependency resolution, then use my library, run-sequence, designed specifically for this scenario..