這可以做得更漂亮嗎？匹配字符串到大小寫字段

Question

對不起，如果可以的話，請編輯它以更具描述性！

有沒有辦法用scala來概括這個？我有很多可以過濾的字段，而這只是簡單的醜陋！我碰到的問題是將參數名稱與案例類字段進行匹配，是否可以以更通用的方式完成，而不需要太多的代碼重複？

get("/MostClicked") { request => 
    val res = MongoDbOps.findMostClicked() 
    val res1 = request.params.get("source") match { 
    case None => res 
    case Some(f) => res.filter(_.source == f) 
    } 
    val res2 = request.params.get("category") match { 
    case None => res1 
    case Some(f) => res1.filter(_.category == f) 
    } 
    // more of the same... 

    render.plain { 
    res2.toJson.prettyPrint 
    }.toFuture 
} 

Answer 1

您可以嘗試以下兩種方法中的任意一種。

case class MostClicked(
    source: String, 
    category: String, 
    rating: String) 

    object MongoDbOps { 
    def findMostClicked() = List[MostClicked]() 
    } 

    class Request { 
    val params = Map[String, String]() 
    } 

    def get(path: String)(f: Request => String) = { 
    f(new Request) 
    } 

首先是使用的匹配的列表，然後再應用順序使用foldLeft：

get("/MostClicked") { request => 
    val res = MongoDbOps.findMostClicked() 

    val kfun = List(
     "source" -> ((x: MostClicked, y: String) => x.source == y), 
     "category" -> ((x: MostClicked, y: String) => x.category == y), 
     "rating" -> ((x: MostClicked, y: String) => x.rating == y)) 

    val r = kfun.foldLeft(res) { (x, y) => 
     request.params.get(y._1) 
     .map(f => res.filter(y._2(_, f))) 
     .getOrElse(x) 
    } 

    r.toString 

    // more of the same... 

    render.plain { 
     r.toJson.prettyPrint 
    }.toFuture 

    } 

或者乾脆使其更具可讀性：

get("/MostClicked") { request => 
    val res = MongoDbOps.findMostClicked() 

    val res1 = request.params.get("source") 
     .map(f => res.filter(_.source == f)) 
     .getOrElse(res) 

    val res2 = request.params.get("category") 
     .map(f => res.filter(_.category == f)) 
     .getOrElse(res1) 

    val res3 = request.params.get("rating") 
     .map(f => res.filter(_.rating == f)) 
     .getOrElse(res2) 

    // more of the same... 

    render.plain { 
     res3.toJson.prettyPrint 
    }.toFuture 

    } 

Answer 2

我不得不把它分解對部分和部分進行分類，從而制定出更小的方法。 它在我簡單的實驗中工作，並擺脫它可以重複的地方，但可能不是可讀的？

請注意，除非您進行反思，否則您仍需要創建參數的名稱以及如何過濾它。

這看起來是一樣tuxdna的答案除了與類型，以增加可讀性和可維護性

SETUP

case class Request(params: Map[String, String]) 
case class Result(category: String, source: String) 
type Filterer = (Result, String) => Boolean 
case class FilterInfo(paramName: String, filterer: Filterer) 
type Analyzer = FilterInfo => List[Result] 
val request = Request(Map("source"->"b")) 

提取方法

def reduce(filterInfos: List[FilterInfo], results: List[Result]) = { 
    filterInfos.foldLeft(results) { (currentResult, filterInfo) => 
    request.params.get(filterInfo.paramName) 
     .map(filterVal => currentResult.filter(filterInfo.filterer(_, filterVal))) 
     .getOrElse(currentResult) 
    } 
} 

用法

val filterInfos = List(
       FilterInfo("source", (result, filterVal) => result.source == filterVal), 
       FilterInfo("category", (result, filterVal) => result.category == filterVal)) 
val res = List(Result("a","a"), Result("b", "b")) 

reduce(filterInfos, res) 

用在你的例子會比較像這樣：

get("/MostClicked") { request => 
    val res = MongoDbOps.findMostClicked() 
    val filterInfos = List(
       FilterInfo("source", (result, filterVal) => result.source == filterVal), 
       FilterInfo("category", (result, filterVal) => result.category == filterVal)) 
    val finalResult = reduce(filterInfos, res) 

    render.plain { 
    finalResult.toJson.prettyPrint 
    }.toFuture 
}