scala-xml

Scala convert XML to key value map

由 谁都会走 提交于 2020-01-25 07:10:31
问题 Related to This topic Problem is as follows, imagine an XML without any particular schema <persons> <total>2</total> <someguy> <firstname>john</firstname> <name>Snow</name> </someguy> <otherperson> <sex>female</sex> </otherperson> </persons> For processing I want to have this in a Key Value Map: "Persons/total" -> 2 "Persons/someguy/firstname" -> john "Persons/someguy/name" -> Snow "Persons/otherperson/sex" -> female Preferably I have some nice recursive function where I traverse the XML code

error: object xml is not a member of package com.databricks.spark

由 南笙酒味 提交于 2019-12-13 02:54:25
问题 I am trying to read XML file using SBT but i am facing issue when i compile it. build.sbt name:= "First Spark" version:= "1.0" organization := "in.goai" scalaVersion := "2.11.8" libraryDependencies += "org.apache.spark" %% "spark-core" % "2.0.0" libraryDependencies += "org.apache.spark" %% "spark-sql" % "2.0.0" libraryDependencies += "com.databricks" % "spark-avro_2.10" % "2.0.1" libraryDependencies += "org.scala-lang.modules" %% "scala-xml" % "1.0.2" resolvers += Resolver.mavenLocal .scala

Error running introductory example of scala-xml

由 霸气de小男生 提交于 2019-12-11 17:42:00
问题 I'm trying to run the simple example provided on the README of scala-xml, but the code won't run: import org.apache.spark.sql.SQLContext val sqlContext = new SQLContext(sc) val df = sqlContext.read .format("com.databricks.spark.xml") .option("rowTag", "book") .load("books.xml") (copy-pasted from the README; books.xml is indeed in the local directory) This gives me error: Name: Compile Error Message: :1: error: illegal start of definition .format("com.databricks.spark.xml") ^ StackTrace: I'm

ArrayIndexOutofBoundException In Scala Program Usind VTD-XML

由 倖福魔咒の 提交于 2019-12-08 19:33:29
import java.lang._ import com.ximpleware._ object Sample { def main(args :Array[String])= { // println("helloo") try{ var i :Int = -1 val vgen :VTDGen= new VTDGen() val ap :AutoPilot =new AutoPilot() ap.selectXPath("CATALOG/CD/COUNTRY/text()") if(vgen.parseFile("../catalog.xml", false)) { val vnav :VTDNav = vgen.getNav() ap.bind(vnav) while((i=ap.evalXPath)!= -1) { println(vnav.toString(i)) println(vnav.toNormalizedString(vnav.getText())) } ap.resetXPath() } } catch { case e :Exception => println(e) } } } I have imported VTD-XML Library It compiles Well but On Execution prints an Exception

ArrayIndexOutofBoundException In Scala Program Usind VTD-XML

由 ￣綄美尐妖づ 提交于 2019-12-08 07:46:55
问题 import java.lang._ import com.ximpleware._ object Sample { def main(args :Array[String])= { // println("helloo") try{ var i :Int = -1 val vgen :VTDGen= new VTDGen() val ap :AutoPilot =new AutoPilot() ap.selectXPath("CATALOG/CD/COUNTRY/text()") if(vgen.parseFile("../catalog.xml", false)) { val vnav :VTDNav = vgen.getNav() ap.bind(vnav) while((i=ap.evalXPath)!= -1) { println(vnav.toString(i)) println(vnav.toNormalizedString(vnav.getText())) } ap.resetXPath() } } catch { case e :Exception =>

scala String to scala.xml.Elem

由 ε祈祈猫儿з 提交于 2019-12-03 09:04:57
问题 I have to convert a string into XML scala> val string="<play><scala></scala></play>" string: String = <play><scala></scala></play> scala> val xml=(<play><scala></scala></play>) xml: scala.xml.Elem = <play><scala></scala></play> 回答1: scala> scala.xml.XML.loadString(string) res0: scala.xml.Elem = <play><scala/></play> 来源： https://stackoverflow.com/questions/25152738/scala-string-to-scala-xml-elem

scala String to scala.xml.Elem

由 [亡魂溺海] 提交于 2019-12-02 21:56:05
I have to convert a string into XML scala> val string="<play><scala></scala></play>" string: String = <play><scala></scala></play> scala> val xml=(<play><scala></scala></play>) xml: scala.xml.Elem = <play><scala></scala></play> scala> scala.xml.XML.loadString(string) res0: scala.xml.Elem = <play><scala/></play> 来源： https://stackoverflow.com/questions/25152738/scala-string-to-scala-xml-elem
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