[英]how can I render the css files when i use express.js?
I am grouping my files like so:我正在对我的文件进行分组,如下所示:
I've already required those:我已经需要这些:
const express = require('express');
const app = express();
const path = require('path')
what do I do next?我下一步该怎么做?
I assume that you're somewhat of a beginner with Node/Express.我假设您是 Node/Express 的初学者。
I recommend you learn more about how express works before deploying this into an actual app.我建议您在将其部署到实际应用程序之前详细了解 express 的工作原理。
Let's get something straight: I believe that you only have a group of HTML files that you want to show to the user under the file names eg.让我们直截了当:我相信您只有一组 HTML 文件,您想在文件名下显示给用户,例如。 example.com/about.html with the homepage HTML file being called index.html so that express knows what to show where.
example.com/about.html 主页 HTML 文件被称为 index.html 以便快递知道在哪里显示什么。
This is the simplest way I could think i'd achieve this effect.这是我认为可以达到这种效果的最简单方法。
const express = require('express');
const app = express();
const path = require('path');
// This is the port where the application is running on
// Uses the server's enviroment variable named PORT if is defined else
// this will use port 5000
// the page can be seen locally at http://localhost:5000/
const PORT = process.env.PORT || 5000;
// This line makes the app created earlier use the middleware that express provides for "rendering" static files
// inside the express.static method we give a path to the static files
// to create that path we use the path.join method provided by the path module we imported earlier
// this method takes in all the paths that need to be joined.
// the __dirname is the directory the application was launced from (you can use process.cwd() to get the root)
// and ofcourse the structures is the folder which contains all your HTML files
app.use(express.static(path.join(__dirname, "structures")));
// Now we do the same thing we did before but we add the middleware for the styles under the "/styles" URI.
app.use("/styles", express.static(path.join(__dirname, "styles")));
// This will start the server at the PORT which we defined earlier.
app.listen(PORT);
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