Gulp 错误：以下任务未完成：您是否忘记发出异步完成信号？

Question

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

Answer 1

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 (eg 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;
});

Answer 2

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();
});

Answer 3

THIS WORKED!

Last updated on Feb 18, 2021, I found the problem after using the above solution then I have fixed it by using the following instead for the following 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'
  )
);

Answer 4

You need to do one thing:

• Add async before function.

 const gulp = require('gulp'); gulp.task('message', async function() { console.log("Gulp is running..."); });

Answer 5

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!

Answer 6

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()
    });

Answer 7

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:

1. Last section on the Gulp page Async Completion : Using async/await .

2. Mozilla async functions docs .

Answer 8

You need to do two things:

1. Add async before function.

2. Start your function with return .

    var gulp = require('gulp'); gulp.task('message', async function() { return console.log("HTTP Server Started"); });

Answer 9

Workaround: We need to call the callback functions (Task and Anonymous):

function electronTask(callbackA)
{
    return gulp.series(myFirstTask, mySeccondTask, (callbackB) =>
    {
        callbackA();
        callbackB();
    })();    
}

Answer 10

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...')
})

Answer 11

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;

Answer 12

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;

Answer 13

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");
});

Answer 14

我最近在为此苦苦挣扎，并找到了创建运行sass然后sass:watch的default任务的正确方法：

gulp.task('default', gulp.series('sass', 'sass:watch'));

Answer 15

Add done as a parameter in default function. That will do.

Answer 16

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'));  
});

Answer 17

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"

Answer 18

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?

Answer 19

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();
})

Answer 20

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;