防止突破/崩溃gulp手表的错误

Question

我正在运行gulp 3.6.2并具有从在线示例设置的以下任务

gulp.task('watch', ['default'], function () {
  gulp.watch([
    'views/**/*.html',        
    'public/**/*.js',
    'public/**/*.css'        
  ], function (event) {
    return gulp.src(event.path)
      .pipe(refresh(lrserver));
  });

  gulp.watch(['./app/**/*.coffee'],['scripts']);
  gulp.watch('./app/**/*.scss',['scss']);
});

任何时候我的CoffeeScript gulp手表都会出现错误停止 - 显然不是我想要的.

正如其他地方所推荐的那样,

gulp.watch(['./app/**/*.coffee'],['scripts']).on('error', swallowError);
gulp.watch('./app/**/*.scss',['scss']).on('error', swallowError);
function swallowError (error) { error.end(); }

但它似乎没有用.

我究竟做错了什么？

---

为了回应@Aperçu的回答,我修改了我的swallowError方法并尝试了以下方法:

gulp.task('scripts', function () {
  gulp.src('./app/script/*.coffee')
    .pipe(coffee({ bare: true }))
    .pipe(gulp.dest('./public/js'))
    .on('error', swallowError);
});

重新启动,然后在我的咖啡文件中创建语法错误.同样的问题:

[gulp] Finished 'scripts' after 306 ?s

stream.js:94
      throw er; // Unhandled stream error in pipe.
            ^
Error: W:\bariokart\app\script\trishell.coffee:5:1: error: unexpected *
*
^
  at Stream.modifyFile (W:\bariokart\node_modules\gulp-coffee\index.js:37:33)
  at Stream.stream.write (W:\bariokart\node_modules\gulp-coffee\node_modules\event-stream\node_modules\through\index.js:26:11)
  at Stream.ondata (stream.js:51:26)
  at Stream.EventEmitter.emit (events.js:95:17)
  at queueData (W:\bariokart\node_modules\gulp\node_modules\vinyl-fs\node_modules\map-stream\index.js:43:21)
  at next (W:\bariokart\node_modules\gulp\node_modules\vinyl-fs\node_modules\map-stream\index.js:71:7)
  at W:\bariokart\node_modules\gulp\node_modules\vinyl-fs\node_modules\map-stream\index.js:85:7
  at W:\bariokart\node_modules\gulp\node_modules\vinyl-fs\lib\src\bufferFile.js:8:5
  at fs.js:266:14
  at W:\bariokart\node_modules\gulp\node_modules\vinyl-fs\node_modules\graceful-fs\graceful-fs.js:104:5
  at Object.oncomplete (fs.js:107:15)

Answer 1

你的swallowError功能应该是这样的:

function swallowError (error) {

  // If you want details of the error in the console
  console.log(error.toString())

  this.emit('end')
}

我认为你必须在error正在watch崩溃的任务事件上绑定这个函数,而不是任务,因为那不是问题所在,你应该在每个可能失败的任务上设置这个错误回调,就像你有的插件一样.错过了一个;或其他东西,以防止watch任务停止.

例子 :

gulp.task('all', function () {
  gulp.src('./app/script/*.coffee')
    .pipe(coffee({ bare: true }))
    .on('error', swallowError)
    .pipe(gulp.dest('./public/js'))

  gulp.src('css/*.scss')
    .pipe(sass({ compass: true }))
    .on('error', swallowError)
    .pipe(cssmin())
    .pipe(gulp.dest('dist'))
})

或者,如果你不介意包含另一个模块,你可以使用gulp-util的日志功能来阻止你在你的:中声明一个额外的函数:gulpfile

.on('error', gutil.log)

但我可能会建议看一下这个令人敬畏的gulp-plumber插件,它用于删除事件的onerror处理程序error,导致流中断.它使用起来非常简单,它可以阻止您捕获可能失败的所有任务.

gulp.src('./app/script/*.coffee')
  .pipe(plumber())
  .pipe(coffee({ bare: true }))
  .pipe(gulp.dest('./public/js'))

有关插件的创建者在本文中有关此内容的更多信息.

Answer 2

以上例子对我不起作用.以下做了:

var plumber = require('gulp-plumber');
var liveReload = require('gulp-livereload');
var gutil = require('gulp-util');
var plumber = require('gulp-plumber');
var compass = require('gulp-compass');
var rename = require('gulp-rename');
var minifycss = require('gulp-minify-css');
var notify = require('gulp-notify');

gulp.task('styles', function () {
    //only process main.scss which imports all other required styles - including vendor files.
    return gulp.src('./assets/scss/main.scss')
            .pipe(plumber(function (error) {
                gutil.log(error.message);
                this.emit('end');
            }))
            .pipe(compass({
                config_file: './config.rb',
                css: './css'
                , sass: './assets/scss'
            }))
            //minify files
            .pipe(rename({suffix: '.min'}))
            .pipe(minifycss())

            //output
            .pipe(gulp.dest('./css'))
            .pipe(notify({message: 'Styles task complete'}));
});

gulp.task('watch', function () {
    liveReload.listen();
    gulp.watch('assets/scss/**/*.scss', ['styles']);
});