const app = require('express')();
app.get('/', function(req, res) {
res.sendFile(__dirname + "/" + "index.html");
});<link rel="stylesheet" href="style.css">I utilized the Node.js code above to send an HTML file. In order to style the HTML file properly, I also need to send a CSS file called style.css.
My inquiry is: How can I use the sendFile() method to send both the index.html and style.css files together and have them integrated on the client side seamlessly?
To ensure that the browser loads style.css independently, you can set it as a route like this:
app.get('/style.css', function(req, res) {
res.sendFile(__dirname + "/" + "style.css");
});
However, managing multiple files this way can become cumbersome. Fortunately, Express offers a simpler solution for serving static files:
https://expressjs.com/en/starter/static-files.html
const express = require("express");
const app = express();
app.use(express.static(__dirname));
It's important to note that if your server code and index.html are in the same directory, the server code will also be served as static files, which is not ideal.
Instead, consider creating a subdirectory called public and store your index.html, CSS, images, scripts, etc. there. Then use the following code:
app.use(express.static("public"));
By moving your files to the public directory, Express will automatically serve index.html, allowing you to remove the need for app.get("/".
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