使用Scala，从Java DataFrame创建DataFrame或RDD

Question

我不想直接使用spark.read方法创建Dataframe或RDD。我想从java结果集(有5,000,00条记录)形成一个dataframe或RDD。感谢你提供一个勤奋的解决方案。

Answer 1

首先，我们可以使用RowFactory创建行。其次，使用SQLContext.createDataFrame方法可以将所有的行转换成数据格式。希望，这对你也有帮助。)

import java.sql.Connection
import java.sql.ResultSet
import org.apache.spark.sql.RowFactory
import org.apache.spark.sql.DataFrame
import org.apache.spark.sql.Row
import org.apache.spark.sql.SQLContext
import org.apache.spark.sql.types.StringType
import org.apache.spark.sql.types.StructField
import org.apache.spark.sql.types.StructType

var resultSet: ResultSet = null
val rowList = new scala.collection.mutable.MutableList[Row]
var cRow: Row = null

//Resultset is created from traditional Java JDBC.
val resultSet = DbConnection.createStatement().execute("Sql")

//Looping resultset
while (resultSet.next()) {
   //adding two columns into a "Row" object
   cRow = RowFactory.create(resultSet.getObject(1), resultSet.getObject(2))
   //adding each rows into "List" object.
   rowList += (cRow)
}

val sconf = new SparkConf
sconf.setAppName("")
sconf.setMaster("local[*]")
var sContext: SparkContext = new SparkContext(sConf)
var sqlContext: SQLContext = new SQLContext(sContext)

//creates a dataframe
DF = sqlContext.createDataFrame(sContext.parallelize(rowList ,2), getSchema())
DF.show() //show the dataframe.

def getSchema(): StructType = {
    val DecimalType = DataTypes.createDecimalType(38, 10)
    val schema = StructType(
      StructField("COUNT", LongType, false) ::
        StructField("TABLE_NAME", StringType, false) :: Nil)

  //Returning the schema to define dataframe columns.
  schema
}