在Express.js中投放资产时遇到问题

时间:2018-09-22 22:03:32

标签: javascript node.js express

我具有以下文件结构:

 (http://localhost:3000/Chart.js:10917:4)
dashboard:178 Uncaught TypeError: Cannot read property 'style' of null
    at ChartElement.customTooltips (dashboard:178)
    at ChartElement.update (Chart.js:8660)
    at ChartElement.handleEvent (Chart.js:8952)
    at Chart.eventHandler (Chart.js:4679)
    at listener (Chart.js:4609)
    at HTMLCanvasElement.proxies.(anonymous)

然后在我的- server.js - controllers - [...] - public - utils - views - home - index.html - js - index.js - css - index.css 内部,在应用程序启动时执行此操作:

server.js

这是我“回家”路线中的逻辑。

app.use(express.static( path.join(__dirname, "public") ));

当我导航到app.get("/", (req, res) => { const publicViews = path.join(__dirname, "public", "views"); res.sendFile("home/index.html", { root: publicViews }); }); index.html呈现时,服务器未找到并返回localhost:{port}/index.js

这就是我在index.css中引用它们的方式

index.html

<link rel="stylesheet" href="./css/index.css">

我在做什么错了?

2 个答案:

答案 0 :(得分:1)

You can add one more static folder

app.use(logger('dev'));
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: false }));
app.use(cookieParser());
app.use(express.static(path.join(__dirname, 'public')));

app.use(express.static(path.join(__dirname, 'views', 'home')));

app.use('/', index);
app.use('/users', users);

Then it will be possible to use your templates like this

var express = require('express');
var router = express.Router();

/* GET home page. */
router.get('/', function(req, res, next) {
  res.render('home/index', { title: 'Express' });
});

module.exports = router;

And your assets will be accessible

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>
    </title>
</head>
<body>
    <script src="/js/index.js"></script>
</body>

答案 1 :(得分:1)

Feel free to assign method specific routing instead of using static middleware for serving resources from /public/utils/.

app.use('/utils', express.static(path.join(__dirname, 'public', 'utils')))
app.use('/', express.static(path.join(__dirname, 'public', 'views', 'home')))
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