gulp and running django server with logging - django

Question

I'm trying to run a django server from gulp. Although I've now managed (roughly) to activate the virtual env and then start the server, none of the output/errors from the django server appears. How can I get this to appear when using gulp?
gulpfile.js
var pjson = require('./package.json');
// Run django server
var cmd = 'workon switcher5'
gulp.task('runServer', function() {
exec('~/virtualenvs/' + pjson.name + '/bin/python manage.py runserver', function (err, stdout, stderr) {
console.log(stdout);
console.log(stderr);
});
});
Note although the server runs fine. I am getting the error in the console:
/bin/sh: /Users/User/virtualenvs/switcher5/bin/python: No such file or directory
(Running the django server not via gulp works as normal. )
I tried to spawn a process instead:
gulp.task('runServer', function(){
console.info('starting django server');
var PIPE = {stdio: 'inherit'};
spawn('~/virtualenvs/' + pjson.name + '/bin/python', ['manage.py', 'runserver'], PIPE);
});
However although that ran with
spawn('python', ['manage.py', 'runserver'], PIPE);from the terminal (activating virtualenv before manually running via the above fails.
I'm not clear what's wrong with the path (if anything)
When it ran from the terminal it now gives output, but seems to be only errors from the django server (304's etc...), but no successful requests appear.

Answer 1

I have struggled with this exact concept for a while. The examples that I've seen have everything all in one gulpfile.js, where I have things more modular. Additionally, I had it working with a prior Django project, but then broke it with my next project when I followed the Two Scoops method of breaking up your configuration files.
I finally got it working by creating the files below (Windows-based machine) and then running 'gulp watch' from the bash command line:
Basic Project Folder Structure:
root folder
|_ project folder
|_ config folder
|_ gulp
|_ tasks
runserver.js
watch.js
|_ gulpfile.js
gulpfile.js:
// Load gulp script modules
require('./gulp/tasks/runserver');
require('./gulp/tasks/watch');
runserver.js
// Starts python django server
// Import packages
var gulp = require('gulp'),
spawn = require('child_process').spawn;
// Start the Django runserver
gulp.task('startDjango', function(cb){
var projDir = 'C:\\<<project folder>>'
var envDir = 'C:\\<<virtual environment directory>>'
var cmd = spawn('python', [projDir + '\\manage.py', 'runserver'],
{cwd: envDir, stdio: 'inherit'}
);
cmd.on('close', function(code) {
console.log('startDjango exited with code ' + code);
cb(code);
});
});
watch.js:
// Kicks off Django server, opens and syncs browser upon changes
// Import packages
var gulp = require('gulp'),
watch = require('gulp-watch'),
browserSync = require('browser-sync').create();
// WATCH for changes to files
gulp.task('watch', function(){
// Start python django server
gulp.start('startDjango');
// Sync browsers to python runserver
browserSync.init({
notify: false,
localOnly: true,
proxy: {
target: 'http://localhost:8000/',
},
port: process.env.PORT || 8000,
});
// Refresh the browser whenever any HTML file is changed
watch('./**/*.html', function(){
browserSync.reload();
});
... // Other watch activities here
});
Hope this helps anyone else who's struggling. :)

Answer 2

The key issue in the question was to get gulp to launch a virtualenv then launch the django server while passing the output from it to node via stdout.
The other answer mention unfortunately doesn't achieve this as it doesn't launch a virtualenv at the same time.
The solution below, does work.
(note if you don't use virtualenv in a venv/ folder then just adjust the below to the relevant location
// Activates a virtualenv and runs django server
gulp.task('runServer', function(cb) {
var cmd = spawn('venv/bin/python', ['manage.py', 'runserver'], {stdio: 'inherit'});
cmd.on('close', function(code) {
console.log('runServer exited with code ' + code);
cb(code);
});
});