I am trying to read a simple csv file Azure Data Lake Storage V2 with Spark 2.4 on my IntelliJ-IDE on mac
Code Below
package com.example
import org.apache.spark.SparkConf
import org.apache.spark.sql._
object Test extends App {
val appName: String = "DataExtract"
val master: String = "local[*]"
val sparkConf: SparkConf = new SparkConf()
.setAppName(appName)
.setMaster(master)
.set("spark.scheduler.mode", "FAIR")
.set("spark.sql.session.timeZone", "UTC")
.set("spark.sql.shuffle.partitions", "32")
.set("fs.defaultFS", "abfs://development#xyz.dfs.core.windows.net/")
.set("fs.azure.account.key.xyz.dfs.core.windows.net", "~~key~~")
val spark: SparkSession = SparkSession
.builder()
.config(sparkConf)
.getOrCreate()
spark.time(run(spark))
def run(spark: SparkSession): Unit = {
val df = spark.read.csv("abfs://development#xyz.dfs.core.windows.net/development/sales.csv")
df.show(10)
}
}
It's able to read, and throwing security exception
Exception in thread "main" java.lang.NullPointerException
at org.wildfly.openssl.CipherSuiteConverter.toJava(CipherSuiteConverter.java:284)
at org.wildfly.openssl.OpenSSLEngine.toJavaCipherSuite(OpenSSLEngine.java:1094)
at org.wildfly.openssl.OpenSSLEngine.getEnabledCipherSuites(OpenSSLEngine.java:729)
at org.wildfly.openssl.OpenSSLContextSPI.getCiphers(OpenSSLContextSPI.java:333)
at org.wildfly.openssl.OpenSSLContextSPI$1.getSupportedCipherSuites(OpenSSLContextSPI.java:365)
at org.apache.hadoop.fs.azurebfs.utils.SSLSocketFactoryEx.<init>(SSLSocketFactoryEx.java:105)
at org.apache.hadoop.fs.azurebfs.utils.SSLSocketFactoryEx.initializeDefaultFactory(SSLSocketFactoryEx.java:72)
at org.apache.hadoop.fs.azurebfs.services.AbfsClient.<init>(AbfsClient.java:79)
at org.apache.hadoop.fs.azurebfs.AzureBlobFileSystemStore.initializeClient(AzureBlobFileSystemStore.java:817)
at org.apache.hadoop.fs.azurebfs.AzureBlobFileSystemStore.<init>(AzureBlobFileSystemStore.java:149)
at org.apache.hadoop.fs.azurebfs.AzureBlobFileSystem.initialize(AzureBlobFileSystem.java:108)
Can anyone help me, what is the mistake?
As per my research, you will receive this error message when you have incompatible jar with the hadoop version.
I would request you to kindly go through the below issues:
http://mail-archives.apache.org/mod_mbox/spark-issues/201907.mbox/%3CJIRA.13243325.1562321895000.591499.1562323440292#Atlassian.JIRA%3E
https://issues.apache.org/jira/browse/HADOOP-16410
Had the same issue, resolved by adding wildfly version 1.0.7, as the docs shared by #cheekatlapradeep-msft mentioned
<dependency>
<groupId>org.wildfly.openssl</groupId>
<artifactId>wildfly-openssl</artifactId>
<version>1.0.7.Final</version>
</dependency>
Related
intellij(spark)--->Hive(Remote)---storage on S3(orc format)
Not able to read remote Hive table through spark/scala.
was able to read table schema but not able to read table.
Error -Exception in thread "main" java.lang.IllegalArgumentException:
AWS Access Key ID and Secret Access Key must be specified as the
username or password (respectively) of a s3 URL, or by setting the
fs.s3.awsAccessKeyId or fs.s3.awsSecretAccessKey properties
(respectively).
import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.sql.{Encoders, SparkSession}
import org.apache.spark.sql.hive.HiveContext
import org.apache.spark.sql.hive.orc._
import org.apache.spark.sql.types.StructType
object mainclas {
def main(args: Array[String]): Unit = {
val spark = SparkSession.builder
.master("local[*]")
.appName("hivetable")
.config("hive.metastore.uris", "thrift://10.20.30.40:9083")
.config("access-key","PQHFFDEGGDDVDVV")
.config("secret-key","FFGSGHhjhhhdjhJHJHHJGJHGjHH")
.config("format", "orc")
.enableHiveSupport()
.getOrCreate()
val res = spark.sqlContext.sql("show tables").show()
val res1 =spark.sql("select *from ace.visit limit 5").show()
}
}`
Try this:
val spark = SparkSession.builder
.master("local[*]")
.appName("hivetable")
.config("hive.metastore.uris", "thrift://10.20.30.40:9083")
.config("fs.s3n.awsAccessKeyId","PQHFFDEGGDDVDVV")
.config("fs.s3n.awsSecretAccessKey","FFGSGHhjhhhdjhJHJHHJGJHGjHH")
.config("format", "orc")
.enableHiveSupport()
.getOrCreate()
you need to prefix all the fs. options with spark.hadoop if you are setting them in the spark config. And as noted: use s3a over s3n if you can.
we are reading the date/table from AWS(hive) through Spark/scala using Intellij(witch is on local machine). we can able to see the schema of table. but not able to read data.
please find below flow to get better understanding
Intellij(spark/scala)------> hive:9083(remote)------> s3(orc)
Note: Here Intellij is present on local and hive and S3 present on AWS
Please find below code for the same:
import org.apache.spark.SparkConf
import org.apache.spark.sql.SparkSession
//import org.apache.spark.sql.hive.HiveContext
object hiveconnect {
def main(args: Array[String]): Unit = {
val spark = SparkSession
.builder()
.appName("SparkHiveExample")
.config("hive.metastore.uris", "thrift://10.20.30.40:9083")
.master("local[*]")
.config("spark.sql.warehouse.dir", "s3://abc/test/main")
.config("spark.driver.allowMultipleContexts", "true")
.config("access-key","key")
.config("secret-key","key")
.enableHiveSupport()
.getOrCreate()
println("Start of SQL Session--------------------")
spark.sql("show databases").show()
spark.sql("select *from ace.visit limit 5").show()
}
}
Error: Exception in thread "main" java.lang.IllegalArgumentException: AWS Access Key ID and Secret Access Key must be specified as the username or password (respectively) of a s3 URL, or by setting the fs.s3.awsAccessKeyId or fs.s3.awsSecretAccessKey properties (respectively).
I want to write and update by Kudu API.
This is the maven dependency:
<dependency>
<groupId>org.apache.kudu</groupId>
<artifactId>kudu-client</artifactId>
<version>1.1.0</version>
</dependency>
<dependency>
<groupId>org.apache.kudu</groupId>
<artifactId>kudu-spark2_2.11</artifactId>
<version>1.1.0</version>
</dependency>
In the following code, I have no idea about KuduContext parameter.
My code in spark2-shell:
val kuduContext = new KuduContext("master:7051")
Also the same error in Spark 2.1 streaming:
import org.apache.kudu.spark.kudu._
import org.apache.kudu.client._
val sparkConf = new SparkConf().setAppName("DirectKafka").setMaster("local[*]")
val ssc = new StreamingContext(sparkConf, Seconds(2))
val messages = KafkaUtils.createDirectStream("")
messages.foreachRDD(rdd => {
val spark = SparkSession.builder.config(rdd.sparkContext.getConf).getOrCreate()
import spark.implicits._
val bb = spark.read.options(Map("kudu.master" -> "master:7051","kudu.table" -> "table")).kudu //good
val kuduContext = new KuduContext("master:7051") //error
})
Then the error:
org.apache.spark.SparkException: Only one SparkContext may be running
in this JVM (see SPARK-2243). To ignore this error, set
spark.driver.allowMultipleContexts = true. The currently running
SparkContext was created at:
org.apache.spark.sql.SparkSession$Builder.getOrCreate(SparkSession.scala:860)
Update your version of Kudu to the latest one (currently 1.5.0). The KuduContext takes the SparkContext as an input parameter in later versions and that should prevent this problem.
Also, do the initial Spark initialization outside of the foreachRDD. In the code you provided, move both the spark and kuduContext out of the foreach. Also, you do not need to create a separate sparkConf, you can use the newer SparkSession only.
val spark = SparkSession.builder.appName("DirectKafka").master("local[*]").getOrCreate()
import spark.implicits._
val kuduContext = new KuduContext("master:7051", spark.sparkContext)
val bb = spark.read.options(Map("kudu.master" -> "master:7051", "kudu.table" -> "table")).kudu
val messages = KafkaUtils.createDirectStream("")
messages.foreachRDD(rdd => {
// do something with the bb table and messages
})
Spark version 2.0.2.6
Scala version 2.11.11
Using DataStax 5.0
import org.apache.log4j.{Level, Logger}
import java.util.Calendar
import org.apache.spark.sql.functions._
import org.apache.spark.SparkContext
import org.apache.spark.SparkConf
import com.datastax.spark.connector._
import org.apache.spark.sql._
object csvtocassandra {
def main(args: Array[String]): Unit = {
val key_space = scala.io.StdIn.readLine("Please enter cassandra Key Space Name: ")
val table_name = scala.io.StdIn.readLine("Please enter cassandra Table Name: ")
// Cassandra Part
val conf = new SparkConf().setAppName("Sample1").setMaster("local[*]")
val sc = new SparkContext(conf)
sc.setLogLevel("ERROR")
println(Calendar.getInstance.getTime)
// Scala Read CSV Part
val spark1 = org.apache.spark.sql.SparkSession.builder().master("local").config("spark.cassandra.connection.host", "127.0.0.1")
.appName("Spark SQL basic example").getOrCreate()
val csv_input = scala.io.StdIn.readLine("Please enter csv file location: ")
val df_csv = spark1.read.format("csv").option("header", "true").option("inferschema", "true").load(csv_input)
df_csv.printSchema()
}
}
Why am I not able to run this program as a Job trying to submit it to spark. When I run this program using IntelliJ it works.
But When I create a JAR and run it I am getting following Error.
Command:
> dse spark-submit --class "csvtospark" /Users/del/target/scala-2.11/csvtospark_2.11-1.0.jar
I am getting following Error:
ERROR 2017-11-02 11:46:10,245 org.apache.spark.deploy.DseSparkSubmitBootstrapper: Failed to start or submit Spark application
org.apache.spark.sql.AnalysisException: Path does not exist: dsefs://127.0.0.1/Users/Desktop/csv/example.csv;
Why is it appending dsefs://127.0.0.1 part even though I am giving just the path /Users/Desktop/csv/example.csv when asked.
I tried giving --mater option as well. How ever I am getting the same error. I am running DataStax Spark in Local Machine. No Cluster.
Please correct me where I am doing things wrong.
Got it. Never mind. Sorry about that.
input should be file:///file_name
I would like to use spark SQL in an Intellij IDEA SBT project.
Even though I have imported the library the code does not seem to import it.
Spark Core seems to be working however.
You can't create a DataFrame from a scala List[A]. You need first to create an RDD[A], and then transform that to a DataFrame. You also need an SQLContext:
val conf = new SparkConf()
.setMaster("local[*]")
.setAppName("test")
val sc = new SparkContext(conf)
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
val test = sc.parallelize(List(1,2,3,4)).toDF
For reference this is how the Spark 2.0 boilerplate with spark sql should look like:
import org.apache.spark.sql.SparkSession
object Test {
def main(args: Array[String]) {
val spark = SparkSession.builder()
.master("local")
.appName("some name")
.getOrCreate()
import spark.sqlContext.implicits._
}
}