javascript - express.js 异步路由器和错误处理

Question

我有一个异步函数作为路由处理程序，我希望将错误处理为某种中间件。这是我的工作尝试:

router.get(
  "/",
  asyncMiddleware(
    routeProviderMiddleware(
      async ({ y }) => ({
        body: await db.query({x: y})
      })
    )
  )
)

// This is the middleware that catches any errors from the business logic and calls next to render the error page
const asyncMiddleware = fn =>
  (req, res, next) => {
    Promise.resolve(fn(req, res, next))
      .catch(next)
  }

// This is a middleware that provides the route handler with the query and maybe some other services that I don't want the route handler to explicitly access to
const routeProviderMiddleware = routeHandlerFn => async (req, res) => {
  const {status = 200, body = {}} = await routeHandlerFn(req.query)
  res.status(status).json(body)
}

我努力的是一种使路由声明更清晰的方法——我不想要那里的 2 个中间件包装器，理想情况下我只想要那里的业务逻辑功能，并以某种方式声明每条路由都被包装在这些。甚至将这两个中间件组合在一起也很好，但我没有成功。

Answer 1

我使用以下方法:

创建 asyncWrap 作为辅助中间件:

const asyncWrap = fn =>
  function asyncUtilWrap (req, res, next, ...args) {
    const fnReturn = fn(req, res, next, ...args)
    return Promise.resolve(fnReturn).catch(next)
  }

module.exports = asyncWrap

你所有的路由/中间件/ Controller 都应该使用这个asyncWrap来处理错误:

router.get('/', asyncWrap(async (req, res, next) => {
  let result = await db.query({x: y})
  res.send(result)
}));

在app.js，最后一个中间件会收到所有asyncWrap的错误:

// 500 Internal Errors
app.use((err, req, res, next) => {
  res.status(err.status || 500)
  res.send({
    message: err.message,
    errors: err.errors,
  })
})