Scala-无法将Scala对象写入Cassandra
我正在尝试使用Spark将Scala case类对象写入Cassandra。但是我在运行代码时遇到了一个异常。我想我无法将case类对象映射到Cassandra行。我的Scala代码如下所示 CassandraPerformerClass.scalaScala-无法将Scala对象写入Cassandra,scala,cassandra,apache-spark,Scala,Cassandra,Apache Spark,我正在尝试使用Spark将Scala case类对象写入Cassandra。但是我在运行代码时遇到了一个异常。我想我无法将case类对象映射到Cassandra行。我的Scala代码如下所示 CassandraPerformerClass.scala object CassandraPerformerClass extends App { override def main(args: Array[String]) { val keyspace = "scalakeys1" val tab
object CassandraPerformerClass extends App
{
override def main(args: Array[String])
{
val keyspace = "scalakeys1"
val tablename = "demotable1"
val conf = new SparkConf().setAppName("CassandraDemo") .setMaster("spark://ct-0015:7077") .setJars(SparkContext.jarOfClass(this.getClass).toSeq)
conf.set("spark.cassandra.connection.host", "192.168.50.103")
conf.set("spark.cassandra.connection.native.port", "9041")
conf.set("spark.cassandra.connection.rpc.port", "9160")
val sc = new SparkContext(conf);
CassandraConnector(conf).withSessionDo
{ session =>
session.execute("DROP KEYSPACE IF EXISTS "+keyspace+" ;");
session.execute("CREATE KEYSPACE "+ keyspace +" WITH replication = {'class': 'SimpleStrategy', 'replication_factor': 3};");
session.execute("CREATE TABLE "+keyspace+"."+tablename+" (keyval bigint, rangef bigint, arrayval text, PRIMARY KEY (rangef, keyval));");
session.execute("CREATE INDEX index_11 ON "+keyspace+"."+tablename+" (keyval) ;");
}
val data = Seq(new Data(1, 10, "string1"), new Data(2, 20, "string2"));
val collection = sc.parallelize(data)
collection.saveToCassandra(keyspace, tablename)
}
case class Data(kv : Long, rf : Long, av : String) extends Serializable
{
private var keyval : Long = kv
private var rangef : Long = rf
private var arrayval : String = av
def setKeyval (kv : Long)
{
keyval = kv
}
def setRangef (rf : Long)
{
rangef = rf
}
def setArrayval (av : String)
{
arrayval = av
}
def getKeyval = keyval
def getRangef = rangef
def getArrayval = arrayval
override def toString = keyval + "," + rangef + "," + arrayval
}
}
例外情况
线程“main”java.lang.IllegalArgumentException中出现异常:RDD中缺少某些主键列或尚未选择:rangef、keyval
位于com.datastax.spark.connector.writer.DefaultRowWriter.checkMissingPrimaryKeyColumns(DefaultRowWriter.scala:44)
位于com.datastax.spark.connector.writer.DefaultRowWriter.(DefaultRowWriter.scala:71)
在com.datastax.spark.connector.writer.DefaultRowWriter$$anon$2.rowWriter上(DefaultRowWriter.scala:109)
在com.datastax.spark.connector.writer.DefaultRowWriter$$anon$2.rowWriter上(DefaultRowWriter.scala:107)
在com.datastax.spark.connector.writer.TableWriter$.apply上(TableWriter.scala:170)
在com.datastax.spark.connector.RDDFunctions.saveToCassandra上(RDDFunctions.scala:23)
在com.cleartrail.spark.scala.cassandra.poc.CassandraPerformerClass$.main上(CassandraPerformerClass.scala:33)
在com.cleartrail.spark.scala.cassandra.poc.CassandraPerformerClass.main(CassandraPerformerClass.scala)
请告诉我如何将案例类对象映射到Cassandra行 Spark的基于Scala的连接器不希望出现一个类似JavaBean的case类,其中包含字段的getter。(无论如何,这是一种糟糕的做法——case类是类似bean的数据容器的不可变替代品,并且具有字段的默认访问器,并且没有变体) 创建一个与Cassandra表具有相同名称和类型的
case类
,只需执行以下操作:
case class Data(keyval: Long, rangef:Long , arrayval: String) extends Serializable
案例类在默认情况下也是可序列化的,因此不需要显式扩展它们。