Gulp returns 0 when tasks fail

Question

I\'m using Gulp in my small project in order to run tests and lint my code. When any of those tasks fail, Gulp always exits with return code 0. If I run jshint by hand, it e

Answer 1

gulp-jshint has been struggling with how to fail the build on jshint fail. On the one hand, we can crash the build inside jshint, but then you never get to the reporter. On the other hand, requiring the reporter to fail the build isn't part of the default reporters. I generally hook up my own reporter that keeps track of fails, and .on('end', function () { will process.exit(1). It's quite brute force, but it works like a charm. See https://github.com/wearefractal/gulp-jshint/issues/10

Answer 2

Similar to Andy Piper answer above, I have found this module stream-combiner2 to be useful when running a sequence of tasks to ensure and exit code is emitted if there is an error somewhere. Can be used something like this

var combiner = require('stream-combiner2');

var tasks = combiner.obj([
    gulp.src(files),
    task1(options),
    task2(options),
    gulp.dest('path/to/dest')
]);

tasks.on('error', function () {
    process.exit(1)
});

Answer 3

@robrich is right, you have to keep track of exit codes yourself, but there's no need for a forceful approach. The process global is an EventEmitter which you can bind your exit function to.

var exitCode = 0

gulp.task('test', function (done) {
  return require('child_process')
    .spawn('npm', ['test'], {stdio: 'pipe'})
    .on('close', function (code, signal) {
      if (code) exitCode = code
      done()
    })
})

gulp.on('err', function (err) {
  process.emit('exit') // or throw err
})

process.on('exit', function () {
  process.nextTick(function () {
    process.exit(exitCode)
  })
})

Answer 4

gulp.task('default', function(done) {
  return gulp.src( ... ).pipe(jasmine( ... )).on('error', function(err) { done(err); } )
});

Works for me

Answer 5

A fix has been introduced in this commit.

It works something like this:

gulp.src("src/**/*.js")
  .pipe(jshint())
  .pipe(jshint.reporter("default"))
  .pipe(jshint.reporter("fail"));

I have been using it on circleci and it works a treat!

Answer 6

You need to 'return gulp.src(...' so that the task waits for the returned stream.

EDIT

Gulp tasks are asynchronous by nature. The actual task is not executed yet at the time of 'gulp.src(...).pipe(...);'. In your example, the gulp tasks mark their results as success before the actual tasks are executed.

There are some ways to make gulp to wait for your actual task. Use a callback or return a stream or promise.

https://github.com/gulpjs/gulp/blob/master/docs/API.md#async-task-support

The easiest way is just returning the stream of 'gulp.src(...).pipe(...)'. If gulp task gets a stream, it will listen to 'end' event and 'error' event. They corresponds to return code 0 and 1. So, a full example for your 'lint' task would be:

gulp.task('lint', function () {
    return gulp.src('./*.js')
               .pipe(jshint('jshintrc.json'))
               .pipe(jshint.reporter('jshint-stylish'));
});

A bonus is that now you can measure the actual time spent on your tasks.