如何在WebStorm 10中通过Gulpfiledebugging由nodemon启动的快速应用程序?

我有一个通过Gulpfileconfiguration运行的应用程序。

gulpfile.js

'use strict'; var gulp = require('gulp'); var sass = require('gulp-sass'); var prefix = require('gulp-autoprefixer'); var browserSync = require('browser-sync'); var nodemon = require('gulp-nodemon'); var reload = browserSync.reload; // we'd need a slight delay to reload browsers // connected to browser-sync after restarting nodemon var BROWSER_SYNC_RELOAD_DELAY = 500; gulp.task('nodemon', function (cb) { var called = false; return nodemon({ // nodemon our expressjs server script: './bin/www', // watch core server file(s) that require server restart on change watch: ['app.js'] }) .on('start', function onStart() { // ensure start only got called once if (!called) { cb(); } called = true; }) .on('restart', function onRestart() { // reload connected browsers after a slight delay setTimeout(function reload() { reload({ stream: false }); }, BROWSER_SYNC_RELOAD_DELAY); }); }); gulp.task('browser-sync', ['nodemon'], function () { // for more browser-sync config options: http://www.browsersync.io/docs/options/ browserSync({ // informs browser-sync to proxy our expressjs app which would run at the following location proxy: 'http://localhost:3000', // informs browser-sync to use the following port for the proxied app // notice that the default port is 3000, which would clash with our expressjs port: 4000, // open the proxied app in chrome browser: ['google-chrome'] }); }); gulp.task('js', function () { return gulp.src('public/**/*.js') // do stuff to JavaScript files //.pipe(uglify()) //.pipe(gulp.dest('...')); }); gulp.task('sass', function () { return gulp.src('./public/scss/*.scss') .pipe(sass({outputStyle: 'compressed', sourceComments: 'map'}, {errLogToConsole: true})) .pipe(prefix("last 2 versions", "> 1%", "ie 8", "Android 2", "Firefox ESR")) .pipe(gulp.dest('./public/stylesheets')) .pipe(reload({stream:true})); }); gulp.task('css', function () { return gulp.src('public/**/*.css') .pipe(reload({ stream: true })); }) gulp.task('bs-reload', function () { reload(); }); gulp.task('default', ['browser-sync'], function () { gulp.watch('public/**/*.js', ['js', reload()]); gulp.watch('public/scss/*.scss', ['sass']); gulp.watch('public/**/*.html', ['bs-reload']); }) 

我如何开始在Webstorm中debugging这个应用程序? 我已经尝试了“编辑configuration”面板,设置NodeJSconfiguration和Gulpfileconfiguration,但仍然没有运气。

我只是不明白如何实际执行这种debugging过程。

任何帮助,将不胜感激。 谢谢。

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