加载资源失败：服务器响应的状态为404（找不到）-javaScript CSS文件

Question

HTML

doctype html
head
  link(href='https://fonts.googleapis.com/css2?family=Press+Start+2P&display=swap' rel='stylesheet')
  script(src='https://code.jquery.com/jquery-latest.min.js' type='text/javascript')
  script(src='https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js')
  <script type="text/javascript" src="index.js"></script>
  link(rel='stylesheet' type='text/css' href='../css/style.css')
  script(src='/socket.io/socket.io.js')

服务器（serv.js）

我几乎尝试了前面堆栈溢出文章中的所有内容：加载资源失败：服务器以404（未找到）状态响应加载资源失败：服务器以404（未找到）状态响应css

还有其他建议吗？

Answer 1

你的路线错了...

应该是这样的：

app.use(express.static(__dirname + '/public/style.css');
app.use(express.static(__dirname + '/public/index.js');

并且在HTML文件中：

doctype html
head
  link(href='https://fonts.googleapis.com/css2?family=Press+Start+2P&display=swap' rel='stylesheet')
  script(src='https://code.jquery.com/jquery-latest.min.js' type='text/javascript')
  script(src='https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js')
  <script type="text/javascript" src="index.js"></script>
  link(rel='stylesheet' type='text/css' href='/css/style.css')
  script(src='/socket.io/socket.io.js')