1

设想：

  1. 用户使用组合成单个流的过滤器
  2. 当过滤器更改时，会触发后端事件以获取“便宜”数据
  3. 当“廉价”数据到达时，另一个具有相同参数的请求被触发到不同的端点，返回“昂贵”数据，这些数据将用于丰富廉价数据。请求应延迟 1 秒，并且仅在用户未更改任何过滤器时触发（否则应等待 1 秒）

我正在为没有中间变量的 3) 选项而苦苦挣扎。

let filterStream = Rx.Observable
.combineLatest(
  filterX,
  filterY,
  (filterX, filterY) => {
    x: filterX,
    y: filterY
  }
 )
 .map((filters) => {
  limit: 100,
  s: filters.x.a,
  f: filters.x.b + filters.y.c,
})
.distinctUntilChanged()


let cheapDataStream = filterStream
.flatMapLatest((filterQuery) =>
Rx.Observable.fromPromise(cheapBackendApiCall(filterQuery)))

// render cheap results
cheapDataStream
.map(result => transformForDisplay(result))
.subscribe(result => { 
  //render
  // how do i invoke expensiveApiCall() with `filterQuery` data here?
  // with a delay, and only if filterQuery has not changed?

});
4

3 回答 3

2

您可以利用隐式转换来避免fromPromise在任何地方显式使用。然后，您可以使用 concat 立即返回廉价数据，然后延迟返回昂贵 + 廉价数据。通过将其嵌套在流中，如果有新查询到达，flatMapLatest还将取消任何未决的。expensiveCalls

var filters = Rx.Observable
.combineLatest(
  filterX,
  filterY,
  (filterX, filterY) => {
    x: filterX,
    y: filterY
  }
 )
 .map((filters) => {
  limit: 100,
  s: filters.x.a,
  f: filters.x.b + filters.y.c,
})
.distinctUntilChanged()
.flatMapLatest(filters => {
  //This kicks off immediately
  var cheapPromise = cheapBackendApiCall(filters);

  //This was added in the latest version 4.1, the function is only called once it is subscribed to, 
  //if you are using earlier you will need to wrap it in a defer instead.
  var expensivePromiseFn = () => expensiveBackendApiCall(filters);

  //For join implicitly calls `fromPromise` so you can pass the same 
  // sort of arguments.
  var cheapAndExpensive = Rx.Observable.forkJoin(
                            cheapPromise, 
                            expensivePromiseFn, 
                            (cheap, expensive) => ({cheap, expensive}));

  //First return the cheap, then wait 1500 millis before subscribing 
  //which will trigger the expensive operation and join it with the result of the cheap one
  //The parent `flatMapLatest` guarantees that this cancels if a new event comes in
  return Rx.Observable.concat(cheap, cheapAndExpensive.delaySubscription(1500));
})
.subscribe(x => /*Render results*/);
于 2016-03-22T22:53:27.153 回答
0

你在寻找去抖吗？
该运算符似乎完全按照您的描述进行。

于 2016-03-22T11:11:32.830 回答
0

我想出的解决方案，虽然不确定 do() 是否合适，而且便宜 + 昂贵的数据合并看起来不太反应。

Rx.Observable
.combineLatest(
  filterX,
  filterY,
  (filterX, filterY) => {
    x: filterX,
    y: filterY
  }
 )
 .map((filters) => {
  limit: 100,
  s: filters.x.a,
  f: filters.x.b + filters.y.c,
})
.distinctUntilChanged()
.flatMapLatest((filterQuery) =>
  Rx
  .Observable
  .fromPromise(cheapBackendApiCall(filterQuery))
  .map((results) => {
    filterQuery: filterQuery,
    results: results
  })
)
.do((filtersAndResults) => {
  // render filtersAndResults.results
})
.debounce(1500)
.flatMapLatest((filtersAndResults) => {
  return Rx
  .Observable
  .fromPromise(expensiveBackendApiCall(filtersAndResults.filterQuery))
  .map(results => {
    expensiveData: results,
    cheapData: filtersAndResults.results
  })
})
.subscribe((result)=> {
  // combine results.cheapData + results.expensiveData with simple .map and .find
  //  and render
})
于 2016-03-22T19:09:34.597 回答