如何在现有scala枚举器中有条件地添加新枚举器
我的堆栈是:Scala 2.11.6,ReactiveMongo 0.11.6,Play 2.4.2,Mongo 3.0.4 我有一个到mongo的查询列表,我需要有条件地一个接一个地执行这些查询(如果结果数小于最大可能值) 作为一个解决方案,我正在做以下工作:如何在现有scala枚举器中有条件地添加新枚举器,scala,playframework,enumerator,iterate,Scala,Playframework,Enumerator,Iterate,我的堆栈是:Scala 2.11.6,ReactiveMongo 0.11.6,Play 2.4.2,Mongo 3.0.4 我有一个到mongo的查询列表,我需要有条件地一个接一个地执行这些查询(如果结果数小于最大可能值) 作为一个解决方案,我正在做以下工作: possibleQueries. // find returns Enumerator[JsObject] map(query => searchableCollection.find(query)). /
possibleQueries.
// find returns Enumerator[JsObject]
map(query => searchableCollection.find(query)).
// combine Enumerators with andThen
foldLeft(Enumerator.empty[JsObject])({ (e1, e2) => e1.andThen(e2) }) through
// Take only specified amount of suggestions
take[JsObject](MAX_AMOUNT)
implicit class EnumeratorExtension[E](parent: Enumerator[E]) {
/**
* Create an Enumeratee that combines parent & e if parent was empty.
*/
def andIfEmpty(e: => Enumerator[E]): Enumerator[E] = new Enumerator[E] {
def apply[A](i: Iteratee[E, A]): Future[Iteratee[E, A]] = {
var empty = true
parent.
map(e => {
empty = false
e
})(defaultExecutionContext).
apply(i).
flatMap(r => {
if (empty)
e.apply(r)
else
Future.successful(r)
})(defaultExecutionContext)
}
}
}
如何包装搜索并使其懒惰
class LazyEnumerator[E](e: => Enumerator[E]) extends Enumerator[E] {
lazy _e = e
def apply[A](i: Iteratee[E, A]) = _e.apply(i)
}
possibleQueries.
map(query => new LazyEnumerator(searchableCollection.find(query))).
foldLeft(Enumerator.empty[JsObject])({ (e1, e2) => e1.andThen(e2) }) through
take[JsObject](MAX_AMOUNT)