I have this below gulpfile.js. When i run the app using \'gulp start\' then it is showing start is not a function. Glup-cli version i\'m using V4.0.0
const g
As Derek said, in v4 of gulp you have to use gulp.series or gulp.parallel;
Another differences for v4 is that you need to explicit when the task is done, you have 6 ways to do it.. I'll link this answer for a really good explanation Gulp error: The following tasks did not complete: Did you forget to signal async completion?
Mohamed explanation might be useful to understand in depth the reasons, you can find both: Migrate gulp.start function to Gulp v4
I have tried the solution but it needed more code, so I will write my main changes here:
gulp.task("build",
gulp.series(gulp.parallel("css", "js", "html"), function (done) {
done();
})
);
gulp.task("start", gulp.series("build", "browser-sync"), function (done) {
devMode = true;
gulp.watch(["./src/css/**/*.css'"], ["css"]);
gulp.watch(["./src/js/**/*.js"], ["js"]);
gulp.watch(["./src/templates/**/*.html"], ["html"]);
done();
});
in all of the other task, I simply added the done() function ( with the exception of the task html because it has a return that already explicit when the task is done)
Hope this will help someone else, have a nice day :)