Question

So, I have the grunt file below. I’m wanting to add a task that will start my node app and watch for changes in a directory and restart. I have been using supervisor, node-dev (which are great) but I want to run one command and start my whole app. There has got to be a simple way to do this, but I’m just missing it. It is written in coffeescript as well (not sure if that changes things)…

module.exports = function(grunt) {
  grunt.initConfig({
    /*exec: {
        startApi: {
            command: "npm run-script start-api"
        }
    },*/
    //static server
    server: {
        port: 3333,
        base: './public',
        keepalive: true
    },

    // Coffee to JS compilation
    coffee: {
        compile: {
            files: {
                './public/js/*.js': './src/client/app/**/*.coffee'
            },
            options: {
                //basePath: 'app/scripts'
            }
        }
    },


    mochaTest: {
        all: ['test/**/*.*']
    },


    watch: {
        coffee: {
            files: './src/client/app/**/*.coffee',
            tasks: 'coffee'
        },
        mochaTest: {
            files: 'test/**/*.*',
            tasks: 'mochaTest'
        }
    }
});

grunt.loadNpmTasks('grunt-contrib-coffee');
grunt.loadNpmTasks('grunt-mocha-test');
//grunt.loadNpmTasks('grunt-exec');

grunt.registerTask( 'default', 'server coffee mochaTest watch' );
};

As you can see in the comments, I tries grunt-exec, but the node command stops the execution of the other tasks.

Answer 1

You can set grunt to run default task and the watch task when you start your node app:

in app.js

var cp = require('child_process');
var grunt = cp.spawn('grunt', ['--force', 'default', 'watch'])

grunt.stdout.on('data', function(data) {
    // relay output to console
    console.log("%s", data)
});

Then just run node app as normal!

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