从快速中间件中排除路由 [英] Exclude route from express middleware

问题描述

我有一个节点应用程序，就像防火墙/调度程序一样位于其他微服务前面，它使用如下中间件链:

I have a node app sitting like a firewall/dispatcher in front of other micro services and it uses a middleware chain like below:

...
app.use app_lookup
app.use timestamp_validator
app.use request_body
app.use checksum_validator
app.use rateLimiter
app.use whitelist
app.use proxy
...

但是对于特定的 GET 路由，我想跳过除 rateLimiter 和代理之外的所有路由.他们是一种使用 :except/:only 设置像 Rails before_filter 这样的过滤器的方法吗?

However for a particular GET route I want to skip all of them except rateLimiter and proxy. Is their a way to set a filter like a Rails before_filter using :except/:only?

推荐答案

虽然expressjs没有内置中间件过滤系统，但至少可以通过两种方式实现.

Even though there is no build-in middleware filter system in expressjs, you can achieve this in at least two ways.

第一种方法是挂载所有要跳到正则表达式路径的中间件，而不是包含否定查找:

First method is to mount all middlewares that you want to skip to a regular expression path than includes a negative lookup:

// Skip all middleware except rateLimiter and proxy when route is /example_route
app.use(//((?!example_route).)*/, app_lookup);
app.use(//((?!example_route).)*/, timestamp_validator);
app.use(//((?!example_route).)*/, request_body);
app.use(//((?!example_route).)*/, checksum_validator);
app.use(rateLimiter);
app.use(//((?!example_route).)*/, whitelist);
app.use(proxy);

第二种方法，可能更易读、更简洁，是用一个小的辅助函数包装你的中间件:

Second method, probably more readable and cleaner one, is to wrap your middleware with a small helper function:

var unless = function(path, middleware) {
    return function(req, res, next) {
        if (path === req.path) {
            return next();
        } else {
            return middleware(req, res, next);
        }
    };
};

app.use(unless('/example_route', app_lookup));
app.use(unless('/example_route', timestamp_validator));
app.use(unless('/example_route', request_body));
app.use(unless('/example_route', checksum_validator));
app.use(rateLimiter);
app.use(unless('/example_route', whitelist));
app.use(proxy);

如果你需要比简单的 path === req.path 更强大的路由匹配，你可以使用 path-to-regexp 模块.

If you need more powerfull route matching than simple path === req.path you can use path-to-regexp module that is used internally by Express.

UPDATE :- 在 express 4.17 req.path 只返回 '/'，所以使用 req.baseUrl :

UPDATE :- In express 4.17 req.path returns only '/', so use req.baseUrl :

var unless = function(path, middleware) {
    return function(req, res, next) {
        if (path === req.baseUrl) {
            return next();
        } else {
            return middleware(req, res, next);
        }
    };
};

这篇关于从快速中间件中排除路由的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！