RXSwift flatmap從異步完成處理程序返回

Question

我想創建一個搜索欄是通過MKLocalSearch與請求搜索地址，並使用RXSwift並結合RXCocoa

到目前爲止，我已經做了以下 1.過濾器和反跳，以避免過多的請求：

let searchRes = searchBar.rx.text 
    .orEmpty 
    .filter { query in 
     return query.characters.count > 4 
    } 
    .debounce(1, scheduler: MainScheduler.instance) 

所得串查詢是圖：ED創建MKLocationSearch然後flatMapp：ED返回MKMapItems的陣列是能夠將數組綁定到tableView的項目：

searchRes.map{query -> MKLocalSearch in 
    let request = MKLocalSearchRequest() 
    request.naturalLanguageQuery = query 
    request.region = self.mapView.region 
    return MKLocalSearch(request: request) 
}.flatMapLatest{search -> Observable<[MKMapItem]> in 
    search.start(completionHandler:{(response, error) in 
     let items: Variable<[MKMapItem]> = Variable([]) 
     if let resp = response { 
      //need to return the result form this 
      // i.e. items.value = response.mapItems 
     } 
     //can not return from here since the request is async 
    } 
}.bindTo //continue to bind to tableview 

怎麼可以這樣使用RXSwift做的，我找不到任何相關的例子

Answer 1

你必須創建一個可觀察與ansync請求。你可以定義下面的方法來做到這一點：

func mapItems(for searchRequest: MKLocalSearch) -> Observable<[MKMapItem]> { 
    return Observable.create { observer in 
     searchRequest.start(completionHandler: { (response, error) in 
      if let error = error { 
       observer.onError(error) 
      } else { 
       let items = response?.mapItems ?? [] 
       observer.onNext(items) 
       observer.onCompleted() 
      } 
     }) 

     return Disposables.create { 
      searchRequest.cancel() 
     } 
    } 
} 

然後，你可以使用flatMapLatest與方法：

searchRes 
    .map { query -> MKLocalSearch in 
     let request = MKLocalSearchRequest() 
     request.naturalLanguageQuery = query 
     request.region = self.mapView.region 
     return MKLocalSearch(request: request) 
    } 
    .flatMapLatest{ [unowned self] search -> Observable<[MKMapItem]> in 
     self.mapItems(for: search) 
    } 
    .bindTo(....