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Spark 1.6:java.lang.IllegalArgumentException:spark.sql.execution.id已设置

如何解决《Spark1.6:java.lang.IllegalArgumentException:spark.sql.execution.id已设置》经验,为你挑选了0个好方法。

当我运行以下代码时,我正在使用spark 1.6并遇到上述问题:

// Imports
import org.apache.spark.sql.hive.HiveContext
import org.apache.spark.{SparkConf, SparkContext}
import org.apache.spark.sql.SaveMode
import scala.concurrent.ExecutionContext.Implicits.global
import java.util.Properties
import scala.concurrent.Future

// Set up spark on local with 2 threads
val conf = new SparkConf().setMaster("local[2]").setAppName("app")
val sc = new SparkContext(conf)
val sqlCtx = new HiveContext(sc)

// Create fake dataframe
import sqlCtx.implicits._
var df = sc.parallelize(1 to 50000).map { i => (i, i, i, i, i, i, i) }.toDF("a", "b", "c", "d", "e", "f", "g").repartition(2)
// Write it as a parquet file
df.write.parquet("/tmp/parquet1")
df = sqlCtx.read.parquet("/tmp/parquet1")

// JDBC connection
val url = s"jdbc:postgresql://localhost:5432/tempdb"
val prop = new Properties()
prop.setProperty("user", "admin")
prop.setProperty("password", "")

// 4 futures - at least one of them has been consistently failing for
val x1 = Future { df.write.jdbc(url, "temp1", prop) }
val x2 = Future { df.write.jdbc(url, "temp2", prop) }
val x3 = Future { df.write.jdbc(url, "temp3", prop) }
val x4 = Future { df.write.jdbc(url, "temp4", prop) }

这是github要点:https://gist.github.com/karanveerm/27d852bf311e39f05491

我得到的错误是:at

org.apache.spark.sql.execution.SQLExecution$.withNewExecutionId(SQLExecution.scala:87) ~[org.apache.spark.spark-sql_2.11-1.6.0.jar:1.6.0]
        at org.apache.spark.sql.DataFrame.withNewExecutionId(DataFrame.scala:2125) ~[org.apache.spark.spark-sql_2.11-1.6.0.jar:1.6.0]
        at org.apache.spark.sql.DataFrame.foreachPartition(DataFrame.scala:1482) ~[org.apache.spark.spark-sql_2.11-1.6.0.jar:1.6.0]
        at org.apache.spark.sql.execution.datasources.jdbc.JdbcUtils$.saveTable(JdbcUtils.scala:247) ~[org.apache.spark.spark-sql_2.11-1.6.0.jar:1.6.0]
        at org.apache.spark.sql.DataFrameWriter.jdbc(DataFrameWriter.scala:306) ~[org.apache.spark.spark-sql_2.11-1.6.0.jar:1.6.0]
        at writer.SQLWriter$.writeDf(Writer.scala:75) ~[temple.temple-1.0-sans-externalized.jar:na]
        at writer.Writer$.writeDf(Writer.scala:33) ~[temple.temple-1.0-sans-externalized.jar:na]
        at controllers.Api$$anonfun$downloadTable$1$$anonfun$apply$25.apply(Api.scala:460) ~[temple.temple-1.0-sans-externalized.jar:2.4.6]
        at controllers.Api$$anonfun$downloadTable$1$$anonfun$apply$25.apply(Api.scala:452) ~[temple.temple-1.0-sans-externalized.jar:2.4.6]
        at scala.util.Success$$anonfun$map$1.apply(Try.scala:237) ~[org.scala-lang.scala-library-2.11.7.jar:na]

这是一个火花虫还是我做错了什么/任何变通办法?

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