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.
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!