Question

尝试解析request.queryString，它返回一个Map [String，Seq [String]]

var route = ""
var queryString = "?"
for((k,v) <- request.queryString) {
  if(k == "route"){ route = v.head }
  else {
    queryString += k +"="+ v.head +"&"
  }
}
queryString = queryString.substring(0, queryString.length() -1 );

这很好但很有必要。我确信有一种更实用的方法可以做到这一点。有什么帮助吗？

Answer 1

帮助就在这里！过度评论。

val RouteKey = "route"

val route = request
    .getOrElse(RouteKey, Nil) // will return the route, or empty list
    .headOption               // either Some[head] or None
    .getOrElse("")            // if None, empty string

val queryString = (request - RouteKey)     // remove the route from the request
  .map { case (k, v) =>                    // map each key/value pair
    k + "=" + v.headOption.getOrElse("") } // into key=value strings
  .mkString("?", "&", "")                  // make that list into a single string

您会注意到我使用相同的模式从列表中安全地获取head，以处理空列表。如果您发现自己做了很多，那么您可以将该方法添加到Seq[String]。

implicit def pimpedStringSeq(seq: Seq[String]) = new {
  def headStr = seq.headOption.getOrElse("")
}

val RouteKey = "route"

val route = request.getOrElse(RouteKey, Nil).headStr

val queryString = (request - RouteKey).map { case (k, v) => k + "=" + v.headStr }
  .mkString("?", "&", "")

Answer 2

没有那么好，但你可以使用折叠。

import scalaz._
import Scalaz._

request.queryString.foldLeft(("?", "")) { case ((route, queryString), (k, v)) =>
  if(k == "route")
    (v.head, queryString)
  else
    (route, queryString + k + "=" + v.head + "&")
} :-> (_.init)

那个可爱的微笑操作符（:->）用于转换我们在折叠结束时得到的2元组的第二个元素。它可以理解如下：

t :-> f == (t._1, f(t._2))

您可以看到来源here。

来自控制台的示例：

scala> val requestQueryString = Map("route" -> Seq("a"), "foo" -> Seq("b"), "bar" -> Seq("c"))
requestQueryString: scala.collection.immutable.Map[java.lang.String,Seq[java.lang.String]] = Map(route -> List(a), foo -
> List(b), bar -> List(c))

scala> var route = ""
var queryString = "?"
for((k,v) <- requestQueryString) {
  if(k == "route"){ route = v.head }
  else {
    queryString += k +"="+ v.head +"&"
  }
}
queryString = queryString.substring(0, queryString.length() -1 );
route: java.lang.String = a
queryString: java.lang.String = ?foo=b&bar=c
queryString: java.lang.String = ?foo=b&bar=c

scala> requestQueryString.foldLeft(("?", "")) { case ((queryString, route), (k, v)) =>
  if(k == "route")
    (queryString, v.head)
  else
    (queryString + k + "=" + v.head + "&", route)
}
res8: (java.lang.String, java.lang.String) = (?foo=b&bar=c&,a)

scala> ((_: String).init) <-: res8
res9: (String, java.lang.String) = (?foo=b&bar=c,a)

scala> requestQueryString.foldLeft(("?", "")) { case ((route, queryString), (k, v)) =>
  if(k == "route")
    (v.head, queryString)
  else
    (route, queryString + k + "=" + v.head + "&")
} :-> (_.init)
res10: (java.lang.String, String) = (a,foo=b&bar=c)

什么是scala方式来做到这一点

2 个答案: