Question

我想在用于创建案例类的动态列表中转义字符。

case class Profile(biography: String,
                  userid: String,
                  creationdate: String) extends Serializable

object Profile {

  val cs = this.getClass.getConstructors
  def createFromList(params: List[Any]) = params match {
    case List(biography: Any,
              userid: Any,
              creationdate: Any) => Profile(StringEscapeUtils.escapeJava(biography.asInstanceOf[String]),
                                            StringEscapeUtils.escapeJava(creationdate.asInstanceOf[String]),
                                            StringEscapeUtils.escapeJava(userid.asInstanceOf[String]))
  }
}

JSON.parseFull("""{"biography":"An avid learner","userid":"165774c2-a0e7-4a24-8f79-0f52bf3e2cda", "creationdate":"2015-07-13T07:48:47.924Z"}""")
  .map(_.get.asInstanceOf[scala.collection.immutable.Map[String, Any]])
  .map { 
    m => Profile.createFromList(m.values.to[collection.immutable.List])
   } saveToCassandra("testks", "profiles", SomeColumns("biography", "userid", "creationdate"))

我收到此错误：

scala.MatchError: List(An avid learner, 165774c2-a0e7-4a24-8f79-0f52bf3e2cda, 2015-07-13T07:48:47.925Z) (of class scala.collection.immutable.$colon$colon)

有什么想法吗？

Answer 1

使用与scala.util.parsing.json不同的（外部）JSON库可能更简单（自Scala 2.11以来已弃用）。

lot of good Scala Json libraries下面有一个使用json4s的示例。

import org.json4s._
import org.json4s.native.JsonMethods._

case class Profile(biography: String, userid: String, creationdate: String)

val json = """{
|  "biography":"An avid learner",
|  "userid":"165774c2-a0e7-4a24-8f79-0f52bf3e2cda", 
|  "creationdate":"2015-07-13T07:48:47.924Z"
|}""".stripMargin

parse(json).extract[Profile]
// Profile(An avid learner,165774c2-a0e7-4a24-8f79-0f52bf3e2cda,2015-07-13T07:48:47.924Z)

转义动态列表中的字符

1 个答案: