如何在WebStorm 10中通过Gulpfile调试nodemon启动的Express应用程序？

Question

I've got an Express app that runs via Gulpfile config.

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']);
})

How can I start debugging this application in Webstorm? I've tried 'Edit Configuration' panel, setting up NodeJS configuration and Gulpfile configuration, but still no luck.

I just don't understand how to actually implement this kind of debug process.

Any help would be appreciated. Thanks.

Answer 1

Turns out you have to set bin/www in JavaScript file setting and everything will be working as supposed to.