问题
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
回答1:
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.
回答2:
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.
来源:https://stackoverflow.com/questions/22453381/gulp-watch-terminates-immediately