How to using Gulp plugin notify with del?

Question

This should be easy... I am trying to create a notification that the del is done.

Notify = https://www.npmjs.com/package/gulp-notify

I have:

gulp.task('clean', function() {
    return del(['distFolder']);
});

That clears everything in the distFolder before it gets rebuilt.

What I am trying to do is something like below:

gulp.task('clean', function() {
    return del(['distFolder']).pipe(notify('Clean task finished'));
});

The above returns an error - "TypeError: del(...).pipe is not a function"

What happens when you try what you have? – Colin Marshall Dec 21 '15 at 06:44 — Colin Marshall, Dec 21 '15 at 06:44
@ColinMarshall - TypeError: del(...).pipe is not a function – RooksStrife Dec 21 '15 at 14:41 — RooksStrife, Dec 21 '15 at 14:41

score 3 · Accepted Answer · edited May 23 '17 at 10:28

The key to getting this done right is that del returns a promise. So you have to handle the promise.

I've created a gulpfile that has 3 tasks:

clean illustrates how to do it.
fail illustrates the point of being able to handle failures.
incorrect replicates the method in the OP's self-answer It is incorrect because del returns a promise object whether or not it is successful. So the && test will always evaluate the 2nd part of the expression and thus will always notify Clean Done! even if there was an error and nothing was deleted.

Here's the code:

var gulp = require("gulp");
var notifier = require("node-notifier");
var del = require("del");

// This is how you should do it.
gulp.task('clean', function(){
  return del("build").then(function () {
      notifier.notify({message:'Clean Done!'});
  }).catch(function () {
      notifier.notify({message:'Clean Failed!'});
  });
});

//
// Illustrates a failure to delete. You should first do:
//
// 1. mkdir protected
// 2. touch protected/foo.js
// 3. chmod a-rwx protected
//
gulp.task('fail', function(){
  return del("protected/**").then (function () {
      notifier.notify({message:'Clean Done!'});
  }).catch(function () {
      notifier.notify({message:'Clean Failed!'});
  });
});

// Contrary to what the OP has in the self-answer, this is not the
// correct way to do it. See the previous task for how you must setup
// your FS to get an error. This will fail to delete anything but
// you'll still get the "Clean Done" message.
gulp.task('incorrect', function(){
  return del("protected/**") && notifier.notify({message:'Clean Done!'});
});

BrettJephson · Answer 2 · 2015-12-21T16:53:21.337

If you look at the Del module it isn't returning a stream, so there will be no pipe function (as the error explains).

I'd probably use gulp-clean because it better integrates with gulp's streaming.

e.g.

var clean = require('gulp-clean');
var notify = require('gulp-notify');

gulp.task('clean', function() {
    return gulp.src('distFolder', {read: false})
         .pipe(clean())
         .pipe(notify('Clean task finished'));
});

score 0 · Answer 3 · answered Jan 27 '16 at 21:49

Solved it -

Node's notifier is a dependency of notify. So it should be in node_modules already. Depending on your NPM version it might not be in the root.

Add without NPM install - var notifier = require('node-notifier');

gulp.task('clean', function(){
  return del(dist) && notifier.notify({message:'Clean Done!'})
});

How to using Gulp plugin notify with del?

3 Answers3