表达静态服务索引.html如何否认它

Question

当我从节点快速提供我的角度 5 应用程序时，我遇到了这个问题作为快速静态服务索引.html当我访问 domain.com 但我访问时 domain.com/something 它工作正常，我可以提供帮助如何解决这个问题

app.use(express.static(path.join(__dirname, 'dist'))); //This line serves index.html how to deny it
app.use("/api", routes);
app.get('*', function(req, res) { //The reason it needs to be * so that it works with angular routes or it wont work
//This is never called when i visit domain.com. but it does when i visit domain.com/somthing
   console.log("i was serverd"); 
   res.sendFile(path.join(__dirname, 'dist/index.html'));
});

提前致谢:)

Answer 1

找到解决方案

var options = {
  index: false
}
app.use(express.static(path.join(__dirname, 'dist'), options));

索引

：发送指定的目录索引文件。设置为 false 以禁用目录索引。

参考： http://expressjs.com/en/api.html

Answer 2

我已经为您的prooblem找到了解决方法

 app.use('/',(req,res,next)=>{
  if(req.path!='/')
    next();
  console.log('ii');
  res.sendFile(path.join(__dirname,'dist/index.html'));
},express.static(path.join(__dirname, 'dist')));
app.get('*',(req,res,next)=>{
  console.log('ii');
  res.sendFile(path.join(__dirname,'dist/index.html'));
});

这将满足您的所有要求。但是通过这种方式，您必须在 2 条路由中编写相同的代码

Answer 3

将*替换为 / 。

这应该在app.use(express.static调用之前注册。(谢谢jfriend00(

前任：

 app.get('/', function(req, res) { //see, no asterix
    //This is never called when i visit domain.com. but it does when i visit domain.com/somthing
       console.log("i was serverd"); 
       res.sendFile(path.join(__dirname, 'dist/index.html'));
    });
app.use(express.static(path.join(__dirname, 'dist')));