16

我正在使用 spark 1.6 并在运行以下代码时遇到上述问题：

// 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：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？

4

3 回答 3

3

在尝试了几件事后，我发现全局创建的线程之一ForkJoinPool将其spark.sql.execution.id属性设置为随机值。我无法确定实际执行此操作的过程，但我可以使用自己的ExecutionContext.

import java.util.concurrent.Executors
import concurrent.ExecutionContext
val executorService = Executors.newFixedThreadPool(4)
implicit val ec = ExecutionContext.fromExecutorService(executorService)

我使用了来自http://danielwestheide.com/blog/2013/01/16/the-neophytes-guide-to-scala-part-9-promises-and-futures-in-practice.html的代码。ForkJoinPool在创建新线程属性时，可能会克隆线程属性，如果在 SQL 执行的上下文中发生这种情况，它将获得其非空值，而 aFixedThreadPool将在实例化时创建线程。

于 2016-01-13T05:59:27.037 回答
1

请检查SPARK-13​​747

如果适用于您的环境，请考虑使用 Spark 2.2.0 或更高版本。

于 2017-12-21T09:08:44.243 回答
0

测试 1：如果您以串行方式而不是并行方式运行每个 df.write 操作是否有帮助？

测试 2：如果您持久化数据帧，然后并行执行所有 df.write 操作并在所有完成后进行序列化以取消持久化以查看这是否有帮助，是否有帮助？

于 2016-01-13T05:51:54.817 回答