javascript - 为什么以下处理程序被识别为404处理程序

Question

我正在查看express生成器生成的app.js，有以下代码:

app.use('/', index);
app.use('/users', users);

// catch 404 and forward to error handler
app.use(function (req, res, next) {
    var err = new Error('Not Found');
    err.status = 404;
    next(err);
});

我的问题是为什么最后一个中间件函数被标识为应该返回 not find 错误时执行的函数？

是否基于这样的假设:如果调用此函数，则意味着没有其他中间件/路由器函数使用 res.send() 完成了请求的处理，因此对该请求不感兴趣，所以可能没有该请求的处理程序？如果是这样，那么这个 404 处理函数应该始终添加到最后，对吗？

Answer 1

正如你所说，如 http://expressjs.com/en/starter/faq.html 中所述

How do I handle 404 responses? In Express, 404 responses are not the result of an error, so the error-handler middleware will not capture them. This behavior is because a 404 response simply indicates the absence of additional work to do; in other words, Express has executed all middleware functions and routes, and found that none of them responded. All you need to do is add a middleware function at the very bottom of the stack (below all other functions) to handle a 404 response:

app.use(function (req, res, next) {
    res.status(404).send("Sorry can't find that!")
})