I have the following gulpfile.js, which I'm executing via the command line gulp message:
var gulp = require('gulp');
gulp.task('message', function() {
console.log("HTTP Server Started");
});
I'm getting the following error message:
[14:14:41] Using gulpfile ~\Documents\node\first\gulpfile.js
[14:14:41] Starting 'message'...
HTTP Server Started
[14:14:41] The following tasks did not complete: message
[14:14:41] Did you forget to signal async completion?
I'm using gulp 4 on a Windows 10 system. Here is the output from gulp --version:
[14:15:15] CLI version 0.4.0
[14:15:15] Local version 4.0.0-alpha.2
Since your task might contain asynchronous code you have to signal gulp when your task has finished executing (= "async completion").
In Gulp 3.x you could get away without doing this. If you didn't explicitly signal async completion gulp would just assume that your task is synchronous and that it is finished as soon as your task function returns. Gulp 4.x is stricter in this regard. You have to explicitly signal task completion.
You can do that in six ways:
1. Return a Stream
This is not really an option if you're only trying to print something, but it's probably the most frequently used async completion mechanism since you're usually working with gulp streams. Here's a (rather contrived) example demonstrating it for your use case:
var print = require('gulp-print');
gulp.task('message', function() {
return gulp.src('package.json')
.pipe(print(function() { return 'HTTP Server Started'; }));
});
The important part here is the return statement. If you don't return the stream, gulp can't determine when the stream has finished.
2. Return a Promise
This is a much more fitting mechanism for your use case. Note that most of the time you won't have to create the Promise object yourself, it will usually be provided by a package (e.g. the frequently used del package returns a Promise).
gulp.task('message', function() {
return new Promise(function(resolve, reject) {
console.log("HTTP Server Started");
resolve();
});
});
Using async/await syntax this can be simplified even further. All functions marked async implicitly return a Promise so the following works too (if your node.js version supports it):
gulp.task('message', async function() {
console.log("HTTP Server Started");
});
3. Call the callback function
This is probably the easiest way for your use case: gulp automatically passes a callback function to your task as its first argument. Just call that function when you're done:
gulp.task('message', function(done) {
console.log("HTTP Server Started");
done();
});
4. Return a child process
This is mostly useful if you have to invoke a command line tool directly because there's no node.js wrapper available. It works for your use case but obviously I wouldn't recommend it (especially since it's not very portable):
var spawn = require('child_process').spawn;
gulp.task('message', function() {
return spawn('echo', ['HTTP', 'Server', 'Started'], { stdio: 'inherit' });
});
5. Return a RxJS Observable.
I've never used this mechanism, but if you're using RxJS it might be useful. It's kind of overkill if you just want to print something:
var of = require('rxjs').of;
gulp.task('message', function() {
var o = of('HTTP Server Started');
o.subscribe(function(msg) { console.log(msg); });
return o;
});
6. Return an EventEmitter
Like the previous one I'm including this for completeness sake, but it's not really something you're going to use unless you're already using an EventEmitter for some reason.
gulp.task('message3', function() {
var e = new EventEmitter();
e.on('msg', function(msg) { console.log(msg); });
setTimeout(() => { e.emit('msg', 'HTTP Server Started'); e.emit('finish'); });
return e;
});
An issue with Gulp 4.
For solving this problem try to change your current code:
gulp.task('simpleTaskName', function() {
// code...
});
for example into this:
gulp.task('simpleTaskName', async function() {
// code...
});
or into this:
gulp.task('simpleTaskName', done => {
// code...
done();
});
You need to do one thing:
Add async before function.
const gulp = require('gulp');
gulp.task('message', async function() {
console.log("Gulp is running...");
});
THIS WORKED!
The latest update on Feb 18, 2021, I found the problem after using the elder solution below, then I have fixed it by using the following instead for the next gulp version.
File: Package.json
...,
"devDependencies": {
"del": "^6.0.0",
"gulp": "^4.0.2",
},
...
File: gulpfile.js Example
const {task} = require('gulp');
const del = require('del');
async function clean() {
console.log('processing ... clean');
return del([__dirname + '/dist']);
}
task(clean)
...
Elder Version
gulp.task('script', done => {
// ... code gulp.src( ... )
done();
});
gulp.task('css', done => {
// ... code gulp.src( ... )
done();
});
gulp.task('default', gulp.parallel(
'script',
'css'
)
);
I was getting this same error trying to run a very simple SASS/CSS build.
My solution (which may solve this same or similar errors) was simply to add done as a parameter in the default task function, and to call it at the end of the default task:
// Sass configuration
var gulp = require('gulp');
var sass = require('gulp-sass');
gulp.task('sass', function () {
gulp.src('*.scss')
.pipe(sass())
.pipe(gulp.dest(function (f) {
return f.base;
}))
});
gulp.task('clean', function() {
})
gulp.task('watch', function() {
gulp.watch('*.scss', ['sass']);
})
gulp.task('default', function(done) { // <--- Insert `done` as a parameter here...
gulp.series('clean','sass', 'watch')
done(); // <--- ...and call it here.
})
Hope this helps!
This is an issue when migrating from gulp version 3 to 4, Simply you can add a parameter done to the call back function , see example,
const gulp = require("gulp")
gulp.task("message", function(done) {
console.log("Gulp is running...")
done()
});
I cannot claim to be very knowledgeable on this but I had the same problem and have resolved it.
There is a 7th way to resolve this, by using an async function.
Write your function but add the prefix async.
By doing this Gulp wraps the function in a promise, and the task will run without errors.
Example:
async function() {
// do something
};
Resources:
Last section on the Gulp page Async Completion: Using async/await.
Mozilla async functions docs.
You need to do two things:
Add async before function.
Start your function with return.
var gulp = require('gulp');
gulp.task('message', async function() {
return console.log("HTTP Server Started");
});
Workaround: We need to call the callback functions (Task and Anonymous):
function electronTask(callbackA)
{
return gulp.series(myFirstTask, mySeccondTask, (callbackB) =>
{
callbackA();
callbackB();
})();
}
Basically v3.X was simpler but v4.x is strict in these means of synchronous & asynchronous tasks.
The async/await is pretty simple & helpful way to understand the workflow & issue.
Use this simple approach
const gulp = require('gulp')
gulp.task('message',async function(){
return console.log('Gulp is running...')
})
Here you go: No synchronous tasks.
No synchronous tasks
Synchronous tasks are no longer supported. They often led to subtle mistakes that were hard to debug, like forgetting to return your streams from a task.
When you see the Did you forget to signal async completion? warning, none of the techniques mentioned above were used. You'll need to use the error-first callback or return a stream, promise, event emitter, child process, or observable to resolve the issue.
Using async/await
When not using any of the previous options, you can define your task as an async function, which wraps your task in a promise. This allows you to work with promises synchronously using await and use other synchronous code.
const fs = require('fs');
async function asyncAwaitTask() {
const { version } = fs.readFileSync('package.json');
console.log(version);
await Promise.resolve('some result');
}
exports.default = asyncAwaitTask;
My solution: put everything with async and await gulp.
async function min_css() {
return await gulp
.src(cssFiles, { base: "." })
.pipe(concat(cssOutput))
.pipe(cssmin())
.pipe(gulp.dest("."));
}
async function min_js() {
return await gulp
.src(jsFiles, { base: "." })
.pipe(concat(jsOutput))
.pipe(uglify())
.pipe(gulp.dest("."));
}
const min = async () => await gulp.series(min_css, min_js);
exports.min = min;
Solution is simple, but I outline the changes I made, the error I was getting, my gulpfile before and after, and the package versions--therefore making it appear very long.
I solved this by following the directions of multiple previous answers, in addition to following the error outputted when I would save my .scss file.
In short:
I changed how gulp-sass was imported—see (A)
I changed all functions to ASYNC functions—see (B)
(A) Changes made to gulp-sass import:
Before: var sass = require('gulp-sass)
After: var sass = require('gulp-sass')(require('sass'));
(B) Simply convert functions to ASYNC—
What my gulpfile looked like before:
'use strict';
// dependencies
var gulp = require('gulp');
var sass = require('gulp-sass');
var minifyCSS = require('gulp-clean-css');
var uglify = require('gulp-uglify');
var rename = require('gulp-rename');
var changed = require('gulp-changed');
var SCSS_SRC = './src/Assets/scss/**/*.scss';
var SCSS_DEST = './src/Assets/css';
function compile_scss() {
return gulp.src(SCSS_SRC)
.pipe(sass().on('error', sass.logError))
.pipe(minifyCSS())
.pipe(rename({ suffix: '.min' }))
.pipe(changed(SCSS_DEST))
.pipe(gulp.dest(SCSS_DEST));
}
function watch_scss() {
gulp.watch(SCSS_SRC, compile_scss);
}
gulp.task('default', watch_scss); //Run tasks
exports.compile_scss = compile_scss;
exports.watch_scss = watch_scss;
What my gulpfile looked like after:
'use strict';
// dependencies
var gulp = require('gulp');
//var sass = require('gulp-sass');
var sass = require('gulp-sass')(require('sass'));
var minifyCSS = require('gulp-clean-css');
var uglify = require('gulp-uglify');
var rename = require('gulp-rename');
var changed = require('gulp-changed');
var SCSS_SRC = './src/Assets/scss/**/*.scss';
var SCSS_DEST = './src/Assets/css';
async function compile_scss() {
return gulp.src(SCSS_SRC)
.pipe(sass().on('error', sass.logError))
.pipe(minifyCSS())
.pipe(rename({ suffix: '.min' }))
.pipe(changed(SCSS_DEST))
.pipe(gulp.dest(SCSS_DEST));
}
async function watch_scss() {
gulp.watch(SCSS_SRC, compile_scss);
}
gulp.task('default', watch_scss); // Run tasks
exports.compile_scss = compile_scss;
exports.watch_scss = watch_scss;
Package Versions:
"gulp": "^4.0.2",
"gulp-changed": "^4.0.3",
"gulp-rename": "^2.0.0",
"gulp-uglify": "^3.0.2",
"gulp-clean-css": "^4.3.0",
"gulp-sass": "^5.0.0",
"sass": "^1.38.0"
Error I was getting:
Error in plugin "gulp-sass"
Message:
gulp-sass 5 does not have a default Sass compiler; please set one yourself.
Both the `sass` and `node-sass` packages are permitted.
For example, in your gulpfile:
var sass = require('gulp-sass')(require('sass'));
[14:00:37] The following tasks did not complete: default, compile_scss
[14:00:37] Did you forget to signal async completion?
I got that solved, It's Pretty simple just add the below code snippet.
var gulp = require('gulp');
gulp.task('message', async function() {
console.log("HTTP Server Started");
});
I was struggling with this recently, and found the right way to create a default task that runs sass then sass:watch was:
gulp.task('default', gulp.series('sass', 'sass:watch'));
In gulp version 4 and over, it is required that all gulp tasks tell Gulp where the task will end. We do this by calling a function that is passed as the first argument in our task function
var gulp = require('gulp');
gulp.task('first_task', function(callback) {
console.log('My First Task');
callback();
})
Add done as a parameter in default function. That will do.
For those who are trying to use gulp for swagger local deployment, following code will help
var gulp = require("gulp");
var yaml = require("js-yaml");
var path = require("path");
var fs = require("fs");
//Converts yaml to json
gulp.task("swagger", done => {
var doc = yaml.safeLoad(fs.readFileSync(path.join(__dirname,"api/swagger/swagger.yaml")));
fs.writeFileSync(
path.join(__dirname,"../yourjsonfile.json"),
JSON.stringify(doc, null, " ")
);
done();
});
//Watches for changes
gulp.task('watch', function() {
gulp.watch('api/swagger/swagger.yaml', gulp.series('swagger'));
});
For me the issue was different: Angular-cli was not installed (I installed a new Node version using NVM and simply forgot to reinstall angular cli)
You can check running "ng version".
If you don't have it just run "npm install -g #angular/cli"
I know the problem was presented 6 years ago but probabily you missed return in your function.
I fixed this issue this morning with eslint, that gave me the same message after running "gulp lint" in my working directory.
Example:
function runLinter(callback)
{
return src(['**/*.js', '!node_modules/**'])
.pipe(eslint())
.on('end', ()=>
{
callback();
});
}
exports.lint = runLinter;
So I got the same error with Gulp 4, but the solution was different. I had the error:
"Did you forget to signal async completion?"
but before the error it also says:
"gulp-sass no longer has a default Sass compiler; please set one yourself"
I completely missed that part at first.
So I had this in the gulfile.js:
const sass = require('gulp-sass')
This should be changed to:
const sass = require('gulp-sass')(require('sass'));
Now it works.
I am aware of the ability to declare dependencies that will run before the task, e.g.
gulp.task('a', () => {});
gulp.task('b', () => {});
gulp.task('c', ['a', 'c'], () => {});
Tasks 'a' and 'b' will be run every time after task 'c' is called and before task 'c' is executed.
However, how do I programmatically call an arbitrary task from within a gulp.task?
I did it using gulp.start(); like this:
gulp.task('test1', function(){
gulp.start('test2');
})
gulp.task('test2', function(){
// do something
})
I was using gulp 3.9.1 if it matters. Looks like gulp.start() may be removed or deprecated; but it hasn't happened yet.
Update
If I were to break things out into separate functions, as Novellizator suggested, it would be something like this:
function doSomething(){
// do something
}
gulp.task('test1', function(){
doSomething();
})
gulp.task('test2', function(){
doSomething();
})
It is simple and avoids the use of gulp.start().
From Gulp 4+ I found this to work well as a replacement for gulp.start:
const task1 = () => { /*do stuff*/ };
const task2 = () => { /*do stuff*/ };
// Register the tasks with gulp. They will be named the same as their function
gulp.task(task1);
gulp.task(task2);
...
// Elsewhere in your gulfile you can run these tasks immediately by calling them like this
(gulp.series("task1", "task2")());
// OR
(gulp.parallel("task1", "task2")());
On the general theme of "don't" from the other answers, it depends what you're doing.
If you're thinking:
gulp.task("a", () => {
doSomethingInstant()
return doTaskB()
})
You want:
gulp.task("meta-a", () => {
doSomethingInstant()
})
gulp.task("a", ["meta-a", "b"])
This works because all the tasks are started in order, so if you're not doing anything async in "meta-a", it will finish before "b" starts. If it's async then you need to be much more explicit:
gulp.task("meta-a", () => {
return doSomethingAsync()
})
function taskB() {
// Original task B code...
}
gulp.task("b", taskB)
gulp.task("a", ["meta-a"], taskB)
On a related note, you can initiate the tasks using either gulp.series or gulp.parallel in gulp v4. Although it's not a direct replacement for gulp.start, it achieves the same objective.
Example:
Before v4
gulp.task('protractor', ['protractor:src']);
After v4
gulp.task('protractor', gulp.series('protractor:src'));
As JeffryHouser stated above using the following method works but is deprecated and will be removed in future versions. I just attempted the following on 3.9.1 and it works fine.
Code:
// Variables
var gulp = require('gulp');
var less = require('gulp-less');
var watch = require('gulp-watch');
// Task to compile less -> css: Method 1- The bad way
gulp.task('default', function(){
gulp.src('src/*.less')
gulp.start('lessToCss');
});
gulp.task('lessToCss', function() {
gulp.src('src/*.less')
.pipe(less())
.pipe(gulp.dest('src/main.css'));
});
The second way mentioned in the comment by Novellizator is:
"Break tasks out into functions, then reuse them in other tasks if you need to" piece of advice from the aforementioned issue
I do this by use of lazypipe() https://www.npmjs.com/package/lazypipe
Code:
// Variables
var gulp = require('gulp');
var less = require('gulp-less');
var watch = require('gulp-watch');
var lazypipe = require('lazypipe');
var fixMyCSS = lazypipe()
.pipe(less); // Be CAREFUL here do not use less() it will give you an errror
// Task to compile less -> css: Method 2 - The nice way
gulp.task('default', function(){
//gulp.start('lessToCss');
gulp.src('src/*.less')
.pipe(fixMyCSS())
.pipe(gulp.dest('src/main.css'));
});
A brief comparison
Method 1 yielding the following:
Starting 'default'...
[22:59:12] Starting 'lessToCss'...
[22:59:12] Finished 'lessToCss' after 13 ms
[22:59:12] Finished 'default' after 48 ms
For a total of 61 ms to complete
Method 2?
- [23:03:24] Starting 'default'...
- [23:03:24] Finished 'default' after 38 ms
Practically half that at 38 ms
For a total difference of 23 ms
Now why this is the case? I honestly don't know enough about gulp to say, but let's say method 2 is clearly the more readable,maintainable and even the faster choice.
They are both easy to write so long as you understand not to call a stream directly within the lazypipe().
The best way form me is:
task('task1', () => {
// Buy potatoes
});
task('task2', ['task1'], () => {
// Cut potatoes
});
The result of the task2 it will be: Buy potatoes and Cut potatoes
There are a couple of ways to do this:
The first one: just invoke a new task call for the other:
gulp.task('foo', function() {
gulp.task('caller', ['bar']); // <- this calls the other existing task!
}
gulp.task('bar', function() {
//do something
}
The second, less cleaner one: Gulp is Javascript, so you just set a couple of instructions as a random function and call it!
gulp.task('task', function() {
recurrent_task();
}
gulp.task('task2', function() {
recurrent_task();
}
function recurrent_task() {
//do something
}
I'm sure there's a way to do this, but I couldn't find it. What I want is to pass multiple files into browserify and output multiple files - this is useful for a case where a site has multiple SPA's, with each having its own requires.
Say I have app1.js, app2.js, etc, with each loading in different pages and having independent require('..') statements. I'm looking for a task that does something like this:
gulp.task('browserify', function() {
return
gulp.src('src/**/*.js')
.pipe(browserify) //
.pipe(gulp.dest('dist'));
});
Any idea what's a simple way to accomplish this? thanks.
I stumbled upon this problem actually earlier this week. The problem of "creating multiple bundles". This should work:
var gulp = require('gulp'),
source = require('vinyl-source-stream'),
browserify = require('browserify'),
es = require('event-stream');
gulp.task('default', function() {
// Your main files
var files = [
'./app/main-a.js',
'./app/main-b.js'
];
// Create a stream array
var tasks = files.map(function(entry) {
return browserify({ entries: [entry] })
.bundle()
.pipe(source(entry))
.pipe(gulp.dest('./dist'));
});
return es.merge.apply(null, tasks);
});
Please do not use the gulp-browserify plugin, as it's blacklisted by now. Use browserify itself instead.
Same with Globs:
gulp.task('withglob', function() {
return glob('./app/main-**.js', function(err, files) {
var tasks = files.map(function(entry) {
return browserify({ entries: [entry] })
.bundle()
.pipe(source(entry))
.pipe(rename({
extname: '.bundle.js'
}))
.pipe(gulp.dest('./dist'));
});
return es.merge.apply(null, tasks);
})
});
Btw.: That's the reason
I'm using Browserify in my gulpfile.js like so:
browserify("src/main.js").bundle()
.on("end", function() {
console.log("compiled JavaScript via Browserify");
})
.on("error", function(err) {
console.warn("ERROR: Browserify failed", err);
})
.pipe(source("bundle.js"))
.pipe(gulp.dest("./dist"));
However, it's easy to miss compilation errors - so I'd prefer to blank dist/bundle.js (or insert an alert("ERROR");) instead of just reporting errors in the terminal.
My attempts to implement this failed due to what I assume is a race condition (pipe conclusion overwriting file operations in the "error" event handler), so I'd be grateful for some advice on how to do this properly.
If you think your issue is due to the asynchronicity of the code, you can use async or q to dictate the order in which your code executes (try async.waterfall, for example).
If you want to blank out a file on error, you can write a gulp stream handler that handles control structures that get generated in the stream and blanks the file out, otherwise just passes it through. Here is the code of the handler:
var through2 = require('through2');
var gutil = require('gulp-util');
var blank = function () {
var blank = false;
return through2.obj(
function (file, encoding, cb) {
if (!file.control && !blank) {
this.push(file);
} else if (file.control) {
blank = true;
} else if (file.path && blank) {
var newFile = new gutil.File({
base: file.base,
cwd: file.cwd,
path: file.path,
contents: new Buffer('')
});
this.push(newFile);
}
cb();
}, function (cb) {
cb();
});
};
Then you need to catch the error and generate a control structure. This can be done like this, in the on("error" handler, you place the following line of code:
this.push({"control": "failed"});
If you make the blank handler the last handler in the stream before the output, like this:
.pipe(blank())
.pipe(gulp.dest('./dist'));
Then you will have an empty file instead of the processed one. Of course you could modify this to write the error information to the file instead or do any number of things.
That having been said, why would you simply not want the processing to stop when you encounter this error? Any unhandled stream error should stop processing.