使用不同参数的ExpressJS路由

Question

I have two routes : app.use serves the static files.

 app.use('/test', earlyAccess(), express.static(path.join(__dirname, staticFolder)))

app.get("/test", callback);
app.get("/test/:id", callback);

// Here is the callback
var fileStream = fs.createReadStream(staticFolder + '/test.html');
fileStream.on('open', function () {
  fileStream.pipe(res);
});

From the browser, if I call localhost:80/test/1 - works fine. but if I call localhost:80/test - it gets redirected to main page. In the server console, I got 304 warning.

How to use routes based on parameter in expressjs?

Answer 1

尝试使用res.sendFile代替fs.createReadStream ，类似于：

res.sendFile('test.html');

Answer 2

You can check for request parameters within the callback,

app.get('/test', callback);
app.get('/test/123', callback);

// callback
if(req.params.second) {
  // code for route with param. - /test/123
}
else {
// code for route without param. - /test
}

Parameters in the request are stored in req.params as

{first:'NameOfRoute', second:'NextParam'} ...

So here,

{first:'test', second:'123'}

Answer 3

you can try it like this:

app.use("/test/*", (req, res, next) => {
   // your callback(req, res);
   next();
});
app.get("/test/:id", callback);

app.use , mean you can write a middleware, and it will call next();