11

我有一个404.jade文件，只要有无效的 GET 请求，我就想渲染它。

这是我当前的代码：

应用程序.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

路线/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

路线/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

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

4

2 回答 2

15

您可以middleware在app.js文件中添加自定义。

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
于 2016-05-10T02:39:50.907 回答
2

ctx.response.status 的默认值为 404

application.js 第 125 行：

callback() {
    const fn = compose(this.middleware);

    if (!this.listeners('error').length) this.on('error', this.onerror);

    const handleRequest = (req, res) => {
      res.statusCode = 404;  // defaul
      const ctx = this.createContext(req, res);
      const onerror = err => ctx.onerror(err);
      const handleResponse = () => respond(ctx);
      onFinished(res, onerror);
      return fn(ctx).then(handleResponse).catch(onerror);
    };

    return handleRequest;
}

如果你打电话：

this.render('index',{});  
this.send();  
this.body='';

状态码会自动改变。
所以我们可以使用这个：

app.use(async (ctx, next) => {
    if(parseInt(ctx.status) === 404){
       ctx.status = 404
       ctx.body = {msg:'emmmmmmm, seems 404'};
    }
})

此处警告，如果您使用koa-router，请确保使用app.use( app = new Koa() ) 调用上述函数，而不是router.use

于 2017-11-15T06:32:42.513 回答