Question

我尝试使用不同的排序顺序为我的电影列表创建输出（取决于我输入的参数），但参数不起作用。错误：值排序不是Scala.Movie的成员

def printResult(list: List[Movie], sorting : String) = {
  val movieList = list.filter(element => element.isInstanceOf[Movie])
  if (movieList.length > 0) 
  {
    val resultSize = chooseResultSize()
    val sizedList = movieList.filter(element => movieList.indexOf(element) < resultSize)
    val formtLength = sizedList.map(_.title.length).max + 9
    val printPattern = "%-" + formtLength + "s"

    println(s"\n$printPattern %-9s %-9s %-9s".format("Title", "Year", "Votes", "Rating"))

    for (movie <- ((for (film <- sizedList) yield film.asInstanceOf[Movie]).sortBy(_.sorting).reverse)) println(s"$printPattern %-9s %-9d %-9.1f".format(movie.title.trim, movie.year, movie.votes, movie.rating))
  } 
  else println("Sorry - no matches found!")

}

Answer 1

sortBy函数会等待函数(Scala.Movie) => B。因此，您不能只传递一个String值来指示您要排序的字段，您必须传递一个函数。

val movies = for (movie <- for (film <- sizedList) yield film.asInstanceOf[Movie]) yield movie
val sortedMovies = (sorting match {
    case "byYear" => movies.sortBy(_.year)
    case "byVotes" => movies.sortBy(_.votes)
    case _ => movies.sortBy(_.title)
}).reverse
sortedMovies foreach(movie => println(s"$printPattern %-9s %-9d %-9.1f".format(movie.title.trim, movie.year, movie.votes, movie.rating)))

使用参数更改排序顺序

1 个答案: