为所有以/api开头的路由运行Express中间件? [英] Run express middleware for all routes except that are starting with /api?

问题描述

我有一个Express中间件，可以像这样在所有其他路由之后将React呈现在服务器上:

I have an Express middleware that renders React on the server like so, placed at the end, after all other routes:

app.use(function(req, res) {
  Router.run(routes, req.path, function(Handler) {
    var markup = React.renderToString(<Handler />);
    res.send(swig.renderFile('views/index.html', { html: markup }));
  });
});

由于未指定任何路由，因此该中间件将对所有请求执行.我想从中排除所有/api路线.问题是，即使我在此中间件之前创建了所有API路由，也总是有可能在/api/nonexistantendpoint路径上执行该中间件.

Since no route is specified this middleware will execute for all requests. I would like to exclude all /api routes from it. The problem is even if I create all API routes before this middleware, there is always a possibility for this middleware to be executed on /api/nonexistantendpoint path.

更确切地说，我正在寻找一个正则表达式，使该中间件对除/api开头的任何路径之外的所有路径都执行.

To be more precise, I am looking for a regex that would make this middleware execute for all paths except any path starting with /api.

谢谢！

我已经查看过 这样的帖子，但找不到对我有用的东西.

I have already looked at this SO post but couldn't find anything useful to me.

此帖子包含很多变通办法，这也不是什么我在寻找.

And this post contains a lot of workarounds which is also not what I am looking for.

此帖子要求您有条件地检查URL在中间件中.

This post asks you to conditionally check for URL inside the middleware.

推荐答案

尝试一下:

app.use(/^\/(?!api).*/, function(req, res) {
  Router.run(routes, req.path, function(Handler) {
    var markup = React.renderToString(<Handler />);
    res.send(swig.renderFile('views/index.html', { html: markup }));
  });
});

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