I have a server.js
file in the same directory with gulpfile.js
.
In gulpfile.js
I require my server.js
file:
var express = require('./server.js')
I want to run it in the default
task:
gulp.task('default' , [ 'build','watch','connect'] , function(){
gulp.run('express');
});
I tried to run it like that but it didn't work. I suppose you can run only tasks in that way. How can I run it in default
task?
The server.js
file includes this code:
var express = require('express');
var app = express();
app.get('/api/test', function(req,res){
res.send('Hello world!')
});
app.listen(3000, function(){
console.log('Example app listening on port 3000!');
})
The solution to that would be just replacing my code with the following :
gulp.task('default' , [ 'build','watch','connect'] , function(){
express;
});