Node.js 中间件原理

Question

const app = {
  middleware: [],

  callback (ctx) {
    console.log(ctx)
  },

  use (fn) {
    this.middleware.push(fn)
  },

  go (ctx) {
    this.middleware.reduceRight((next, fn, i) => {
      return () => {
        console.log('next: ', next)
        console.log('fn: ', fn)
        fn(ctx, next)
      }
    }, this.callback.bind(this, ctx))()
  }
}

app.use((ctx, next) => {
  ctx.name = 'Lucy'
  next()
})

app.use((ctx, next) => {
  ctx.age = 12
  next()
})

app.use((ctx, next) => {
  console.log(`${ctx.name} is ${ctx.age} years old.`)
  next()
})

app.go({})

上面代码的执行结果是

next:  () => {
  console.log('next: ', next)
  console.log('fn: ', fn)
  fn(ctx, next)
}
fn:  (ctx, next) => {
  ctx.name = 'Lucy'
  next()
}
next:  () => {
  console.log('next: ', next)
  console.log('fn: ', fn)
  fn(ctx, next)
}
fn:  (ctx, next) => {
  ctx.age = 12
  next()
}
next:  ƒ callback(ctx) {
  console.log(ctx)
}
fn:  (ctx, next) => {
  console.log(`${ctx.name} is ${ctx.age} years old.`)
  next()
}
Lucy is 12 years old.
{name: "Lucy", age: 12}

有没有大佬能解释一下执行结果为啥是这个

Answer 1

可以看一下reduceRight
第一个参数是上次调用回调的返回值也就是next
第二个参数是当前元素值
reduceRight从右往左执行，执行完毕将最后一步也就是数组的第一位的回调返回并将第二位的回调以参数next的形式传入，然后以此类推，最后调用callbakc方法