Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
249 views
in Technique[技术] by (71.8m points)

javascript - Running a command with gulp to start Node.js server

So I am using gulp-exec (https://www.npmjs.com/package/gulp-exec) which after reading some of the documentation it mentions that if I want to just run a command I shouldn't use the plugin and make use of the code i've tried using below.

var    exec = require('child_process').exec;

gulp.task('server', function (cb) {
  exec('start server', function (err, stdout, stderr) {
    .pipe(stdin(['node lib/app.js', 'mongod --dbpath ./data']))
    console.log(stdout);
    console.log(stderr);
    cb(err);
  });
})

I'm trying to get gulp to start my Node.js server and MongoDB. This is what i'm trying to accomplish. In my terminal window, its complaining about my

.pipe

However, I'm new to gulp and I thought that is how you pass through commands/tasks. Any help is appreciated, thank you.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)
gulp.task('server', function (cb) {
  exec('node lib/app.js', function (err, stdout, stderr) {
    console.log(stdout);
    console.log(stderr);
    cb(err);
  });
  exec('mongod --dbpath ./data', function (err, stdout, stderr) {
    console.log(stdout);
    console.log(stderr);
    cb(err);
  });
})

For future reference and if anyone else comes across this problem.

The above code fixed my problem. So basically, I found out that the above is its own function and therefore, doesn't need to:

.pipe

I thought that this code:

exec('start server', function (err, stdout, stderr) {

was the name of the task I am running however, it is actually what command I will be running. Therefore, I changed this to point to app.js which runs my server and did the same to point to my MongoDB.

EDIT

As @N1mr0d mentioned below with having no server output a better method to run your server would be to use nodemon. You can simply run nodemon server.js like you would run node server.js.

The below code snippet is what I use in my gulp task to run my server now using nodemon :

// start our server and listen for changes
gulp.task('server', function() {
    // configure nodemon
    nodemon({
        // the script to run the app
        script: 'server.js',
        // this listens to changes in any of these files/routes and restarts the application
        watch: ["server.js", "app.js", "routes/", 'public/*', 'public/*/**'],
        ext: 'js'
        // Below i'm using es6 arrow functions but you can remove the arrow and have it a normal .on('restart', function() { // then place your stuff in here }
    }).on('restart', () => {
    gulp.src('server.js')
      // I've added notify, which displays a message on restart. Was more for me to test so you can remove this
      .pipe(notify('Running the start tasks and stuff'));
  });
});

Link to install Nodemon : https://www.npmjs.com/package/gulp-nodemon


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...