CODEWITHSUNDEEP
gulp
Geeky Guy
Geeky Guy

Reputation: 9379

Gulp task does not run - becomes a callback for other tasks instead

I have a task that goes like this:

var gulp = require('gulp');
gulp.task('doStuff', ['a', 'b', 'c'], function () {
    console.log("success");
});

I have read the docs, and from what I understood, this is what should happen:

This is what actually happens:

I was surprised to find out that if I accept a callback in the other functions, though, that callback is my function. I.e. if I do:

var gulp = require('gulp');
gulp.task('a', function (callback) {
    callback();
});

This will run my code.

I'd like my function to be executed only once, and only after all other tasks have run. What am I missing, or misinterpreting? What can I do?

By the way, this may be relevant: most of my other tasks do things like generating files, and they return such files. Like:

var gulp = require('gulp');
var rename = require('gulp-rename');
var someLogic = require('someLogicIImplementedSomewhereElse');
var path = "../../somepath";

gulp.task('foo', function() {
    return gulp.src(path)
        .pipe(someLogic())
        .pipe(rename("renamed.js"))
        .pipe(gulp.dest(path));
});

Upvotes: 1

Views: 4908

Answers (1)

Raspo
Raspo

Reputation: 1087

Make sure that a, b and c either return something or execute the callback.

Try the following example:

gulp.task('a', function (done) {
    done();
});

gulp.task('b', function (done) {
    return gulp.src('.');
});

gulp.task('c', function (done) {
    done();
});

gulp.task('stuff', ['a', 'b', 'c'], function () {
    console.log("success");
});

Here is the output of gulp stuff

- Starting 'a'...
- Finished 'a' after 48 μs
- Starting 'b'...
- Starting 'c'...
- Finished 'c' after 6.03 μs
- Finished 'b' after 9.14 ms
- Starting 'stuff'...
- success
- Finished 'stuff' after 38 μs

Upvotes: 3

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