javascript - 如何处理 Koa 2 中的 404？

Question

我有一个 404.jade 文件，每当有无效的 GET 请求时我都想渲染它。

这是我当前的代码:

app.js

import Koa from 'koa'
import views from 'koa-views'
import serve from 'koa-static'
import rootRoutes from './routes/index'
import userRoutes from './routes/user'

const app = new Koa()

app.use(views(`${__dirname}/views`, { extension: 'jade' }))
app.use(serve(`${__dirname}/public`))
app.use(rootRoutes.routes())
app.use(userRoutes.routes())

app.listen(3000, () => {
  console.log('Server running at http://localhost:3000')
})

export default app

routes/index.js

import Router from 'koa-router'
const router = new Router()

router.get('/', async ctx => {
  await ctx.render('index')
})

router.get('/about', async ctx => {
  await ctx.render('about')
})

export default router

routes/user.js

import Router from 'koa-router'
const router = new Router({ prefix: '/user' })

router.get('/:name', async ctx => {
  const user = ctx.params.name
  await ctx.render('user', { user })
})

export default router

我如何处理任何类型的无效 GET 请求并在其发生时以某种方式使用 await ctx.render('404')？

Answer 1

您可以在app.js 文件中添加自定义中间件。

import Koa from 'koa'
import views from 'koa-views'
import serve from 'koa-static'
import rootRoutes from './routes/index'
import userRoutes from './routes/user'

const app = new Koa()

app.use(async(ctx, next) => {
  try {
    await next()
    const status = ctx.status || 404
    if (status === 404) {
        ctx.throw(404)
    }
  } catch (err) {
    ctx.status = err.status || 500
    if (ctx.status === 404) {
      //Your 404.jade
      await ctx.render('404')
    } else {
      //other_error jade
      await ctx.render('other_error')
    }
  }
})

app.use(views(`${__dirname}/views`, { extension: 'jade' }))
app.use(serve(`${__dirname}/public`))
app.use(rootRoutes.routes())
app.use(userRoutes.routes())

app.listen(3000, () => {
  console.log('Server running at http://localhost:3000')
})

export default app