I have a very minimal gulpfile as follows, with a watch task registered:
var gulp = require("gulp");
var jshint = require("gulp-jshint");
gulp.task("lint", function() {
gulp.src("app/assets/**/*.js")
.pipe(jshint())
.pipe(jshint.reporter("default"));
});
gulp.task('watch', function() {
gulp.watch("app/assets/**/*.js", ["lint"]);
});
I cannot get the watch task to run continuously. As soon as I run gulp watch, it terminates immediately.
I've cleared my npm cache, reinstalled dependencies etc, but no dice.
$ gulp watch
[gulp] Using gulpfile gulpfile.js
[gulp] Starting 'watch'...
[gulp] Finished 'watch' after 23 ms
It's not exiting, per se, it's running the task synchronously.
You need to return the stream from the lint
task, otherwise gulp doesn't know when that task has completed.
gulp.task("lint", function() {
return gulp.src("./src/*.js")
^^^^^^
.pipe(jshint())
.pipe(jshint.reporter("default"));
});
Also, you might not want to use gulp.watch
and a task for this sort of watch. It probably makes more sense to use the gulp-watch
plugin so you can only process changed files, sort of like this:
var watch = require('gulp-watch');
gulp.task('watch', function() {
watch({glob: "app/assets/**/*.js"})
.pipe(jshint())
.pipe(jshint.reporter("default"));
});
This task will not only lint when a file changes, but also any new files that are added will be linted as well.
To add to OverZealous' answer which is correct.
gulp.watch
now allows you to pass a string array as the callback so you can have two separate tasks. For example, hint:watch
and 'hint'.
You can then do something like the following.
gulp.task('hint', function(event){
return gulp.src(sources.hint)
.pipe(plumber())
.pipe(hint())
.pipe(jshint.reporter("default"));
})
gulp.task('hint:watch', function(event) {
gulp.watch(sources.hint, ['hint']);
})
This is only an example though and ideally you'd define this to run on say a concatted dist file.