小编the*_*ker的帖子

我正在尝试将列表转换为地图并收到下面给出的错误，任何帮助将不胜感激。

我的输入是这样的

list=(ageis=21,nameis="xyz",profession="Tester",city="cuba")

我想要这样的输出

Map(ageis->21,nameis->"xyz",profession->"Tester",city->"cuba")

我正在尝试读取属性文件，但遇到了下面给出的错误。我编写了一个 Scala 包，我正在尝试读取属性文件并调用 abc.scala 程序。任何帮助将不胜感激。

文件：- xyz.properties

driver = "oracle.jdbc.driver.OracleDriver"
url = "jdbc:oracle:thin:@xxxx:1521/xxxx.xxxx"
username = "xxx"
password = "xxx"
input_file = "C:\\Users\\xxx\\test\\src\\main\\resources\\xxxx.xlsx"

构建.sbt

name := "xxx.xxxx.xxxxx"

scalaVersion := "2.10.6"

ivyScala := ivyScala.value map{ _.copy(overrideScalaVersion = true) }


libraryDependencies ++= Seq(
"org.apache.spark" %% "spark-core" % "2.1.0",
"com.databricks" %% "spark-csv" % "1.5.0",
"org.apache.commons" % "commons-configuration2" % "2.1.1",
"commons-beanutils" % "commons-beanutils" % "1.9.3",
"org.apache.spark" %% "spark-sql" % "2.1.0",
"org.scala-lang" % "scala-xml" % "2.11.0-M4"  )

包裹

package com.xxx.zzzz.xxx1

        import java.io.File
        import org.apache.commons.configuration2.builder.fluent.{Configurations, Parameters}


        object …