Start a shell process in Gulp with callback when process started

问题

I'm trying to run some tasks in Gulp, sequentially. One of those tasks is a shell script that performs a simple $ node app.js. How can I fire the callback so I can tell Gulp that the server has started?

tl;dr

So here's the bigger picture of what I'm trying to accomplish:

I'm using gulp run-sequence to fire up a number of tasks sequentially, which specifies a couple ways you should be writing your tasks in order for them to be run in sequence.

Each gulp.task() must either:

return the stream or
call a callback on the task

My setup:

gulp.task("clean", ..); // returns the stream, all OK
gulp.task("compile", ..); // returns the stream, all OK
gulp.task("spin-server", ..); // calls the callback with a hack
gulp.task("init-browser-sync", ..); // last task

Here's my spin-server task:

gulp.task("spin-server", function(cb) {
  exec("sh shell-utilities/spin-server");

  // @HACK allow time for the server to start before `runSequence`
  // runs the next task.
  setTimeout(function() {
    cb();
  }, 2500);
});

And here's the spin-server.sh shell script:

## Start Node server ##

node server/app.js

#######
# EOF #
#######

The problem

Right now I'm using a setTimeout hack to make sure my Node/Express server has fired up before proceeding to run the init-browser-sync task.

How can I eliminate that setTimeout hack and call cb() when my Express server actually fires up?

回答1:

If you'd like to spawn a process but listen to it's output, you can start it with exec and then attach listeners to the stdout of the process.

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

gulp.task("spin-server", function() {

  var child = exec("sh shell-utilities/spin-server");

  child.stdout.on('data', function(data) {
    console.log('stdout: ' + data);
  });
  child.stderr.on('data', function(data) {
    console.log('stderr: ' + data);
  });
  child.on('close', function(code) {
    console.log('closing code: ' + code);
  });
});

回答2:

Use spawn instead of exec (require('child_process').spawn) like this :

var srv = spawn("sh shell-utilities/spin-server");

gulp.task("spin-server", function(cb) {
  var srv = spawn("sh shell-utilities/spin-server");

  srv.stdout.on('data', data => {
    if(data.includes('server listening')) { // Or whatever your server outputs when it's done initializing
      console.log('Server initialization completed');
      return cb();
    }
  });
});

When the string server listening is found in the output of the spawned process, the cb() is called and the server is guaranteed to be initialized at that point.

回答3:

gulp.task("spin-server", function(cb) {
  const spawn = require('child_process').spawn;
  const process = spawn('bash', ['', 'setup.sh']);

  process.stdout.on('data', (data) => {
      if(data = "SCRIPT CHECKPOINT")
          cb();
      });
});

Use spawn for non blocking async exec.

Let your setup.sh file print "SCRIPT CHECKPOINT" or something similar to inform that the server is up and running".

This will trigger the cb().

来源：https://stackoverflow.com/questions/38190673/start-a-shell-process-in-gulp-with-callback-when-process-started

标签

javascript

node.js

gulp