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I have a gulp task setup like so:

gulp.task('dev', ['lint', 'dev:js', 'dev:css', 'fonts', 'connect', 'watch', 'browser']);

I would like for the lint task to be run before any of the other tasks, but they all run asynchronously.

I understand that the fn argument for gulp.task would only be run after the dependencies run, so that seems like the ideal place to run all of the other tasks after lint, I'm just not sure how to do that with gulp.

mariocatch
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    Possible duplicate of [How to run Gulp tasks synchronously/one after the other](http://stackoverflow.com/questions/22824546/how-to-run-gulp-tasks-synchronously-one-after-the-other) – Sven Schoenung Jun 09 '16 at 14:24

1 Answers1

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You can use gulp-sequence or run-sequence module.

Here's a example of gulp-sequence

Source: https://www.npmjs.com/package/gulp-sequence

var gulpSequence = require('gulp-sequence');

gulp.task( 'dev', gulpSequence( 'lint', [ 'dev:js', 'dev:css', 'fonts', 'connect', 'watch', 'browser' ] ) );

It will run lint first, then all other async.

Priyesh Kumar
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