Gulp：您是否忘记发出异步完成错误信号

Question

I have had to recently upgrade to gulp 4 and I found this error:

Did you forget to signal async completion?

My gulpfile looks as follows:

(function () {
  "use strict";

  const demoPath = "demo/node_modules/style";
  var cleanCSS = require("gulp-clean-css");
  const del = require("del");
  const fs = require("fs");
  var gulp = require("gulp");
  var sass = require("gulp-sass");
  var path = require("path");
  var rename = require("gulp-rename");

  gulp.task("clean", () => {
    return del.sync([
      "**.js", "!gulpfile.js"
    ]);
  }, async  function(done){
    console.log("This is clean task!");
    done();
  });

  function clean() {
    return gulp.src("src/**/*.scss")
      .pipe(sass())
      .pipe(cleanCSS({}))
      .pipe(rename({
        extname: ".min.css"
      }))
      .pipe(gulp.dest("./dist"));
  }

  gulp.task("build", gulp.series("clean"), function(done) {
    console.log("Started");
    done();
  });

}());

I read that with gulp 4 we would need to return a promise, a stream or callback etc. I did try to add the callback in the function and execute them but I still see the error.

Could anyone point me to a direction here?

Thanks for reading.

Answer 1

This code is gulp v3 syntax:

  gulp.task("build", gulp.series("clean"), function(done) {
    console.log("Started");
    done();
  });

gulp.task() no longer takes 3 arguments, only two. So change to:

  gulp.task("build", gulp.series("clean", function(done) {
    console.log("Started");
    done();
  }));

where gulp.series has the two (or more) functions inside it.