我有一个基本的Node.js应用程序，我试图使用Express框架启动。我有一个views文件夹，其中有一个index.html文件。但是我在加载网页时收到以下错误:
Error: Cannot find module 'html'
下面是我的代码。
var express = require('express');
var app = express.createServer();
app.use(express.staticProvider(__dirname + '/public'));
app.get('/', function(req, res) {
res.render('index.html');
});
app.listen(8080, '127.0.0.1')
我错过了什么?
当前回答
index.js
var express = require('express');
var app = express();
app.use(express.static(__dirname + '/public'));
app.get('/', function(req, res) {
res.render('index.html');
});
app.listen(3400, () => {
console.log('Server is running at port 3400');
})
将index.html文件放到公共文件夹中
<!DOCTYPE html>
<html>
<head>
<title>Render index html file</title>
</head>
<body>
<h1> I am from public/index.html </h1>
</body>
</html>
现在在终端上运行以下代码
节点index.js
其他回答
我添加以下2行，它为我工作
app.set('view engine', 'html');
app.engine('html', require('ejs').renderFile);
我想允许对“/”的请求由Express路由处理，而以前它们是由静态中间件处理的。这将允许我渲染index.html的常规版本或加载连接+最小化JS和CSS的版本，这取决于应用程序设置。受到Andrew Homeyer回答的启发，我决定将我的HTML文件(未修改)拖到views文件夹中，并像这样配置Express
app.engine('html', swig.renderFile);
app.set('view engine', 'html');
app.set('views', __dirname + '/views');
创建了一个这样的路由处理器
app.route('/')
.get(function(req, res){
if(config.useConcatendatedFiles){
return res.render('index-dist');
}
res.render('index');
});
结果很好。
我试图用一个RESTful API创建一个angular应用，并多次登陆这个页面，尽管它并没有什么帮助。以下是我发现有效的方法:
app.configure(function() {
app.use(express.static(__dirname + '/public')); // set the static files location
app.use(express.logger('dev')); // log every request to the console
app.use(express.bodyParser()); // pull information from html in POST
app.use(express.methodOverride()); // simulate DELETE and PUT
app.use(express.favicon(__dirname + '/public/img/favicon.ico'));
});
然后在api的回调中，路由看起来像:res.jsonp(users);
您的客户端框架可以处理路由。Express是为API服务的。
我的回家路线是这样的:
app.get('/*', function(req, res) {
res.sendfile('./public/index.html'); // load the single view file (angular will handle the page changes on the front-end)
});
1） 最好的办法是设置静态文件夹。在你的主文件(app.js | server.js | ??):
app.use(express.static(path.join(__dirname, 'public')));
公共/ css / form . html public / css / style . css
然后你从“public”文件夹中获得静态文件:
http://YOUR_DOMAIN/form.html
http://YOUR_DOMAIN/css/style.css
2)
您可以创建您的文件缓存。 使用方法fs.readFileSync
var cache = {};
cache["index.html"] = fs.readFileSync( __dirname + '/public/form.html');
app.get('/', function(req, res){
res.setHeader('Content-Type', 'text/html');
res.send( cache["index.html"] );
};);
It is very sad that it is about 2020 still express hasn't added a way to render an HTML page without using sendFile method of the response object. Using sendFile is not a problem but passing argument to it in the form of path.join(__dirname, 'relative/path/to/file') doesn't feel right. Why should a user join __dirname to the file path? It should be done by default. Why can't the root of the server be by defalut the project directory? Also, installing a templating dependency just to render a static HTML file is again not correct. I don't know the correct way to tackle the issue, but if I had to serve a static HTML, then I would do something like:
const PORT = 8154;
const express = require('express');
const app = express();
app.use(express.static('views'));
app.listen(PORT, () => {
console.log(`Server is listening at port http://localhost:${PORT}`);
});
The above example assumes that the project structure has a views directory and the static HTML files are inside it. For example, let's say, the views directory has two HTML files named index.html and about.html, then to access them, we can visit: localhost:8153/index.html or just localhost:8153/ to load the index.html page and localhost:8153/about.html to load the about.html. We can use a similar approach to serve a react/angular app by storing the artifacts in the views directory or just using the default dist/<project-name> directory and configure it in the server js as follows:
app.use(express.static('dist/<project-name>'));
