Promise All with Limit of Concurrent N

README.org

The Promise All Problem

in case of processing a very large array e.g. Promise.all(A_VERY_LARGE_ARRAY_OF_XHR_PROMISE)

which would probably blow you browser memory by trying to send all requests at the same time

solution is limit the concurrent of requests, and wrap promise in thunk

Promise.allConcurrent(2)([()=>fetch('BLAH1'), ()=>fetch('BLAH2'),...()=>fetch('BLAHN')])

if set concurrent to 2, it will send request BLAH1 and BLAH2 at the same time

if BLAH1 return response and resolved, will immediatly send request to BLAH3

in this way promise sending at the same time always keep the limit 2 which we’ve just configed before

prelude-test.js

describe('promiseAllStepN', function(){
  describe('3 tasks, and cucurrent is 2', function(){
    let tasks;
    beforeEach(function(){
      tasks = range(3).map(x=>sinon.stub())
    })
    describe('1 is finish', function(){
      it('will kickoff the third task', function(done){
        tasks[0].returns(Promise.resolve(0))
        let task2 = tasks[2]
        tasks[1].returns(new Promise(resolve=>setTimeout(()=>{
          expect(task2.called).to.be.equal(true)
          resolve(1)
          done()
        }, 1000)))
        tasks[2].returns(Promise.resolve(2))

        return Promise.allConcurrent(2)(tasks).then(x=>console.log(x))
      })
    })
  })

  describe('10 tasks, and cucurrent is 3', function(){
    let tasks;
    beforeEach(function(){
      tasks = range(10).map(x=>sinon.stub())
    })
    describe('1st is finish but 2nd stuck', function(){
      it.only('final task will run before 2nd', function(done){
        tasks.forEach((task,index) => task.returns(Promise.resolve(index)))
        let task10 = tasks[9]
        tasks[1].returns(new Promise(resolve=>setTimeout(()=>{
          expect(task10.called).to.be.equal(true)
          resolve(1)
          done()
        }, 1000)))
        return Promise.allConcurrent(2)(tasks).then(x=>console.log(x))
      })
    })
  })

})

prelude.js

function promiseAllStepN(n, list) {
  let tail = list.splice(n)
  let head = list
  let resolved = []
  let processed = 0
  return new Promise(resolve=>{
    head.forEach(x=>{
      let res = x()
      resolved.push(res)
      res.then(y=>{
        runNext()
        return y
      })
    })
    function runNext(){
      if(processed == tail.length){
        resolve(Promise.all(resolved))
      }else{
        resolved.push(tail[processed]().then(x=>{
          runNext()
          return x
        }))
        processed++
      }
    }
  })
}
Promise.allConcurrent = n => list =>  promiseAllStepN(n, list)