JS处理并行请求的四种方式

需求

两个异步请求同时发出，两个请求都返回时再做处理

实现

这里的方法仅提供思路，只做请求成功处理

方法一

使用Promise.all

const startTime = new Date().getTime()
function request(time) {
  return new Promise(resolve => {
    setTimeout(() => {
      resolve(time)
    }, time)
  })
}
let request1 = request(3000)
let request2 = request(2000)
Promise.all([request1, request2]).then(res => {
  console.log(res, new Date() - startTime)	// [ 3000, 2000 ] 3001
})

方法二

自定义状态，在回调中判断返回状态，待2个请求都有返回值时再做处理

const startTime = new Date().getTime()
function request(time) {
  return new Promise(resolve => {
    setTimeout(() => {
      resolve(time)
    }, time)
  })
}
let state = [undefined, undefined]
let request1 = request(3000)
let request2 = request(2000)
request1.then(res => {
  state[0] = res
  process()
})
request2.then(res => {
  state[1] = res
  process()
})
function process() {
  if (state[0] && state[1]) {
    console.log(state, new Date() - startTime) // [ 3000, 2000 ] 3001
  }
}

方法三

generator，yield

const startTime = new Date().getTime()
function ajax(time, cb) {
  setTimeout(() => cb(time), time)
}
function request(time) {
  ajax(time, data => {
    it.next(data);
  })
}
function* main() {
  let request1 = request(3000);
  let request2 = request(2000);
  let res1 = yield request1
  let res2 = yield request2
  console.log(res1, res2, new Date() - startTime) // 2000 3000 3001
}
let it = main();
it.next();

这个地方有点问题，因为request2耗时较短，会先返回，也就是先执行it.next(2000)，导致res1获得了request2的返回值 若使用co函数，则不会存在这个问题，因为co是在promise.then函数中才执行it.next()，相当于it.next()是链式调用

generator使用co函数

const co = require('co')
const startTime = new Date().getTime()
function request (time) {
  return new Promise(resolve => {
    setTimeout(() => {
      resolve(time)
    }, time)
  })
}
co(function* () {
  let request1 = request(3000);
  let request2 = request(2000);
  let res1 = yield request1
  let res2 = yield request2
  console.log(res1, res2, new Date() - startTime) // 3000 2000 3001
})

有了co函数，就不需要生成it和执行next方法了； co的原理其实也简单，就是递归执行next，直到done为true； 如果next返回的value是Promise，则在then函数中执行next，若不是Promise，直接执行next函数 下面是co函数的简版手写实现

function co(func) {
  let it = func()
  let t = it.next()
  next()
  
  function next() {
    if (t.done) return
    if (t.value instanceof Promise) {
      t.value.then(res => {
        t = it.next(res)
        next()
      })
    } else {
      t = it.next(t.value)
      next()
    }
  }
}

方法四

有了generator，很容易想到async/await，毕竟async/await就是由generator实现的

// setTimeout模拟异步请求，time为请求耗时
const startTime = new Date().getTime()
function request (time) {
  return new Promise(resolve => {
    setTimeout(() => {
      resolve(time)
    }, time)
  })
}
(async function () {
  let request1 = request(3000)
  let request2 = request(2000)
  let res1 = await request1
  console.log(res1, new Date() - startTime)	// 3000 3001
  let res2 = await request2
  console.log(res2, new Date() - startTime) // 2000 3005
})()