Let each of the task finish before running the next about gulp HOT 6 CLOSED

You can add dep when define a task.

g.task('copy', ['clean'], function() {

});

from gulp.

Tasks run with maximum concurrency.

What you actually want is:

var g = require('gulp');
var util = require('util');

g.task('clean', function(cb) {
  setTimeout(function() {
    util.log('clean task executed.');
    cb(null);
  }, 3000);
});

g.task('copy', ['clean'], function(cb) {
  util.log('copy task executed.');
  cb();
});

g.task('default', ['clean', 'copy'], function() {
  util.log('default task executed.');
});

Notice the only change is making copy depend on clean

from gulp.

I agree, the docs need work here. By default, the tasks run with maximum concurrency -- e.g. it launches all the tasks at once and waits for nothing. If you want to create a series where tasks run in a particular order, you need to do two things: give it a hint to tell it when the task is done, and give it a hint that a task depends on completion of another.

For these examples, let's presume you have two tasks, "one" and "two" that you specifically want to run in this order:

1. In task "one" you add a hint to tell it when the task is done. Either take in a callback and call it when you're done or return a promise or stream that the engine should wait to resolve or end respectively.
2. In task "two" you add a hint telling the engine that it depends on completion of the first task.

So this example would look like this:

var gulp = require('gulp');

// takes in a callback so the engine knows when it'll be done
gulp.task('one', function (cb) {
    // do stuff -- async or otherwise
    cb(err); // if you pass non-null, non-undefined, gulp will stop, and note that the build failed
});

// identifies a dependent task must be complete before this one begins
gulp.task('two', ['one'], function () {
    // task 'one' is done now
});

from gulp.

thank you robrich, that was what I was looking for

from gulp.

Just a quick note:

When doing things in a stream, this seems to work great for me:

gulp.task('do-stuff', function (done) {
    gulp.src(someFiles)
        .pipe(doStuffToThem())
        .on('end', function () { done(); });
});

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I guess there is no way to use on the task, that has a dependency the returned stream of that dependency?

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