【发布时间】:2016-01-11 00:46:19
【问题描述】:
我使用的是 spark 1.6,运行以下代码时遇到了上述问题:
// 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) }
这里是 github 要点:https://gist.github.com/karanveerm/27d852bf311e39f05491
我得到的错误是: 在
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]
这是火花错误还是我做错了什么/任何解决方法?
【问题讨论】:
-
请问您在哪台机器上运行此代码?我对 CPU(多少个内核)特别感兴趣?
-
OSX El Capitan 10.11.1 | MacBook Air(13 英寸,2014 年初)| 1.7 GHz 英特尔酷睿 i7 | 8 GB 1600 MHz DDR3(我相信 i7 是 4 核)
-
有趣,我无法在类似的设置(来自 spark shell)上重现这个。这可能是一些讨厌的错误,他们之前在生成 ID 时遇到了问题。您可能想为此创建一个 JIRA。
-
你运行的是哪个版本的 postgres?
-
我在 9.3.5 上运行
标签: scala apache-spark apache-spark-sql spark-dataframe