Gulp error The following tasks did not complete Did you forget to signal async completion

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

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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      gt  await gulp series min css  min js    exports min   min

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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   gt        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

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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        lt --- Insert  done  as a parameter here        gulp series  clean   sass    watch       done        lt ---    and call it here       Hope this helps

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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

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Add done as a parameter in default function  That will do

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Basically v3 X was simpler but v4 x is strict in these means of synchronous  amp  asynchronous tasks   The async await is pretty simple  amp  helpful way to understand the workflow  amp  issue   Use this simple approach  const gulp   require  gulp    gulp task  message  async function    return console log  Gulp is running

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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

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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

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Workaround  We need to call the callback functions  Task and Anonymous    function electronTask callbackA        return gulp series myFirstTask  mySeccondTask   callbackB    gt                callbackA            callbackB

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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

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This worked   gulp task  script   done   gt               code gulp src            done         gulp task  css   done   gt               code gulp src            done         gulp task  default   gulp parallel           script            css

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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

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You need to do one thing   Add async before function    x000D   x000D  const gulp   require  gulp     gulp task  message   async function         console log  Gulp is running           x000D   x000D   x000D

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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      gt    e emit  msg    HTTP Server Started    e emit  finish          return e

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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   gt         code      done

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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

[gulp] Gulp error: The following tasks did not complete: Did you forget to signal async completion?

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