Express: Is it possible to simplify the path in res.sendFile?

Question

This is my directory structure:
/Users/titi/myproject/app/server/
....app.js
..../public
......../css
......../js
......../view
............index.html
............about.html

---

And my app.js file:

var express = require('express');
var app = express();

app.use(express.static('public'));

app.get('/', function (req, res) {
   res.sendFile('/Users/Titi/myproject/app/server/public/view/index.html');
});

app.get('/about', function (req, res) {
   res.sendFile('/Users/Titi/myproject/app/server/public/view/about.html');
});

app.listen(2000, function () {
  console.log('Example app listening on port 2000!');
});

It works perfectly.
But I would like to know if there is a way to not have to write the whole path (/Users/Titi/myproject/app/server/public/view/about.html).
Is it possible to simplify?

Answer 1

You could specify a root path in the sendFile() options:

var viewOpts = { root: '/Users/Titi/myproject/app/server/public/view' };
app.get('/', function (req, res) {
  res.sendFile('index.html', viewOpts);
});

Or you could use the built-in path to make the absolute path for you:

var path = require('path');

// Assuming this script is inside the `server` portion of the path

app.get('/', function (req, res) {
  res.sendFile(path.resolve('public/view/index.html'));
});