gulp-notify

Cant make gulp-notify to pop up a error message when gulp-jshint fail

落花浮王杯 提交于 2019-12-09 19:43:53
问题 So, was the whole day working on our new FE workflow, the main idea is to run a few tasks through a watcher, we will have our IDE's and a window with the browsers getting refreshed each time something change (scss, js, html, etc). So, we wont see nothing if all went good: the browser will get reload and we will keep working. But, we want to use gulp notify for errors, so in case we have something wrong, the notifier pop up will appear, and at that point, you can check the console and see the

Calling gulp-notify without using .pipe()

泪湿孤枕 提交于 2019-12-09 04:36:59
问题 In my gulpfile.js I have a task that calls two other tasks. When the previous tasks have run, a notification should be triggered. var notify = require('gulp-notify'); gulp.task('build', ['build:js', 'build:css'], function() { console.log('hello', arguments); gulp.src('gulpfile.js').pipe( notify({ title: 'Production Build', message: 'Done' }) ); }); Is there a way to trigger notify without calling gulp.src('gulpfile.js').pipe( first? While not a real problem, this approach feels unnecessary to

Cant make gulp-notify to pop up a error message when gulp-jshint fail

梦想与她 提交于 2019-12-04 12:08:57
So, was the whole day working on our new FE workflow, the main idea is to run a few tasks through a watcher, we will have our IDE's and a window with the browsers getting refreshed each time something change (scss, js, html, etc). So, we wont see nothing if all went good: the browser will get reload and we will keep working. But, we want to use gulp notify for errors, so in case we have something wrong, the notifier pop up will appear, and at that point, you can check the console and see the error. Example for our sass task when fail: For our styles tasks, everything works fine, but Im not

Calling gulp-notify without using .pipe()

被刻印的时光 ゝ 提交于 2019-12-03 02:50:01
In my gulpfile.js I have a task that calls two other tasks. When the previous tasks have run, a notification should be triggered. var notify = require('gulp-notify'); gulp.task('build', ['build:js', 'build:css'], function() { console.log('hello', arguments); gulp.src('gulpfile.js').pipe( notify({ title: 'Production Build', message: 'Done' }) ); }); Is there a way to trigger notify without calling gulp.src('gulpfile.js').pipe( first? While not a real problem, this approach feels unnecessary to me. It is entirely possible that my question results from an insufficient understanding of gulp. So