Question

我正在尝试使用Spark将Scala案例类对象写入Cassandra。但是我在运行代码时遇到异常。我想我无法将我的case类对象映射到我的Cassandra行。我的Scala代码看起来像这样

CassandraPerformerClass.scala

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
 }
}

异常

线程中的异常＆＃34; main＆＃34; java.lang.IllegalArgumentException：RDD中缺少某些主键列或尚未选择：rangef，keyval 在com.datastax.spark.connector.writer.DefaultRowWriter.checkMissingPrimaryKeyColumns（DefaultRowWriter.scala：44）在com.datastax.spark.connector.writer.DefaultRowWriter。（DefaultRowWriter.scala：71）在com.datastax.spark.connector.writer.DefaultRowWriter $$ anon $ 2.rowWriter（DefaultRowWriter.scala：109）在com.datastax.spark.connector.writer.DefaultRowWriter $$ anon $ 2.rowWriter（DefaultRowWriter.scala：107）在com.datastax.spark.connector.writer.TableWriter $ .apply（TableWriter.scala：170）在com.datastax.spark.connector.RDDFunctions.saveToCassandra（RDDFunctions.scala：23）在com.cleartrail.spark.scala.cassandra.poc.CassandraPerformerClass $ .main（CassandraPerformerClass.scala：33）在com.cleartrail.spark.scala.cassandra.poc.CassandraPerformerClass.main（CassandraPerformerClass.scala）

请告诉我如何将我的case类对象映射到Cassandra行。

Answer 1

用于Spark的基于Scala的连接器不希望类似java-bean的case类具有字段的getter。（无论如何，这是一种不好的做法 - 案例类是类似bean的数据容器的不可替代的替代品，并且具有字段的默认访问器而没有变更器。）

使用与Cassandra表相同的名称和类型创建case class将起作用：

case class Data(keyval: Long, rangef:Long , arrayval: String) extends Serializable

Scala - 无法将Scala对象写入Cassandra

1 个答案: