I am trying to follow the instructions mentioned here...
https://www.percona.com/blog/2016/08/17/apache-spark-makes-slow-mysql-queries-10x-faster/
and here...
https://www.percona.com/blog/2015/10/07/using-apache-spark-mysql-data-analysis/
I am using sparkdocker image.
docker run -it -p 8088:8088 -p 8042:8042 -p 4040:4040 -h sandbox sequenceiq/spark:1.6.0 bash
cd /usr/local/spark/
./sbin/start-master.sh
./bin/spark-shell --driver-memory 1G --executor-memory 1g --executor-cores 1 --master local
This works as expected:
scala> sc.parallelize(1 to 1000).count()
But this shows an error:
val jdbcDF = spark.read.format("jdbc").options(
Map("url" -> "jdbc:mysql://1.2.3.4:3306/test?user=dba&password=dba123",
"dbtable" -> "ontime.ontime_part",
"fetchSize" -> "10000",
"partitionColumn" -> "yeard", "lowerBound" -> "1988", "upperBound" -> "2016", "numPartitions" -> "28"
)).load()
And here is the error:
<console>:25: error: not found: value spark
val jdbcDF = spark.read.format("jdbc").options(
How do I connect to MySQL from within spark shell?
With spark 2.0.x,you can use DataFrameReader and DataFrameWriter.
Use SparkSession.read to access DataFrameReader and use Dataset.write to access DataFrameWriter.
Suppose using spark-shell.
read example
val prop=new java.util.Properties()
prop.put("user","username")
prop.put("password","yourpassword")
val url="jdbc:mysql://host:port/db_name"
val df=spark.read.jdbc(url,"table_name",prop)
df.show()
read example 2
val jdbcDF = spark.read
.format("jdbc")
.option("url", "jdbc:mysql:dbserver")
.option("dbtable", “schema.tablename")
.option("user", "username")
.option("password", "password")
.load()
from spark doc
write example
import org.apache.spark.sql.SaveMode
val prop=new java.util.Properties()
prop.put("user","username")
prop.put("password","yourpassword")
val url="jdbc:mysql://host:port/db_name"
//df is a dataframe contains the data which you want to write.
df.write.mode(SaveMode.Append).jdbc(url,"table_name",prop)
Create the spark context first
Make sure you have jdbc jar files in attached to your classpath
if you are trying to read data from jdbc. use dataframe API instead of RDD as dataframes have better performance. refer to the below performance comparsion graph.
here is the syntax for reading from jdbc
SparkConf conf = new SparkConf().setAppName("app"))
.setMaster("local[2]")
.set("spark.serializer",prop.getProperty("spark.serializer"));
JavaSparkContext sc = new JavaSparkContext(conf);
sqlCtx = new SQLContext(sc);
df = sqlCtx.read()
.format("jdbc")
.option("url", "jdbc:mysql://1.2.3.4:3306/test")
.option("driver", "com.mysql.jdbc.Driver")
.option("dbtable","dbtable")
.option("user", "dbuser")
.option("password","dbpwd"))
.load();
It looks like spark is not defined, you should use the SQLContext to connect to the driver like this:
import org.apache.spark.sql.SQLContext
val sqlcontext = new org.apache.spark.sql.SQLContext(sc)
val dataframe_mysql = sqlcontext.read.format("jdbc").option("url", "jdbc:mysql://Public_IP:3306/DB_NAME").option("driver", "com.mysql.jdbc.Driver").option("dbtable", "tblage").option("user", "sqluser").option("password", "sqluser").load()
Later you can user sqlcontext where you used spark (in spark.read etc)
This is a common problem for those migrating to Spark 2.0.0 from the earlier versions. The Spark documentation is not very good. To solve this, you have to define a SparkSession, like this:
import org.apache.spark.sql.SparkSession
val spark = SparkSession
.builder()
.appName("Spark SQL Example")
.config("spark.some.config.option", "some-value")
.getOrCreate()
This solution is hidden in the Spark SQL, Dataframes and Data Sets Guide located here. SparkSession is the new entry point to the DataFrame API and it incorporates both SQLContext and HiveContext and has some additional advantages, so there is no need to define either of those anymore. Further information about this can be found here.
Please accept this as the answer, if you find this useful.
Related
I try to connect to a remote hive cluster using the following code and I get the table data as expected
val spark = SparkSession
.builder()
.appName("adhocattempts")
.config("hive.metastore.uris", "thrift://<remote-host>:9083")
.enableHiveSupport()
.getOrCreate()
val seqdf=sql("select * from anon_seq")
seqdf.show
However, when I try to do this via HiveServer2, I get no data in my dataframe. This table is based on a sequencefile. Is that the issue, since I am actually trying to read this via jdbc?
val sparkJdbc = SparkSession.builder.appName("SparkHiveJob").getOrCreate
val sc = sparkJdbc.sparkContext
val sqlContext = sparkJdbc.sqlContext
val driverName = "org.apache.hive.jdbc.HiveDriver"
Class.forName(driverName)
val df = sparkJdbc.read
.format("jdbc")
.option("url", "jdbc:hive2://<remote-host>:10000/default")
.option("dbtable", "anon_seq")
.load()
df.show()
Can someone help me understand the purpose of using HiveServer2 with jdbc and relevant drivers in Spark2?
I tried to use Sparksession with hive tables.
I had used the following code:
val spark= SparkSession.builder().appName("spark").master("local").enableHiveSupport().getOrCreate()
spark.sql("select * from data").show()
Shows table not found, but the table exists in hive. Please help me with this.
spark.sql("select * from databasename.data").show() - will work
Hello you have to provide the path of warehouse, like:
// warehouseLocation points to the default location for managed databases and tables
val warehouseLocation = new File("spark-warehouse").getAbsolutePath
val spark = SparkSession
.builder()
.appName("Spark Hive Example")
.config("spark.sql.warehouse.dir", warehouseLocation)
.enableHiveSupport()
.getOrCreate()
For more information, You can see here: Hive Tables with Spark
Can someone point me to a working example of saving a csv file to Hbase table using Spark 2.2
Options that I tried and failed (Note: all of them work with Spark 1.6 for me)
phoenix-spark
hbase-spark
it.nerdammer.bigdata : spark-hbase-connector_2.10
All of them finally after fixing everything give similar error to this Spark HBase
Thanks
Add below parameters to your spark job-
spark-submit \
--conf "spark.yarn.stagingDir=/somelocation" \
--conf "spark.hadoop.mapreduce.output.fileoutputformat.outputdir=/somelocation" \
--conf "spark.hadoop.mapred.output.dir=/somelocation"
Phoexin has plugin and jdbc thin client which can connect(read/write) to HBASE, example are in https://phoenix.apache.org/phoenix_spark.html
Option 1 : Connect via zookeeper url - phoenix plugin
import org.apache.spark.SparkContext
import org.apache.spark.sql.SQLContext
import org.apache.phoenix.spark._
val sc = new SparkContext("local", "phoenix-test")
val sqlContext = new SQLContext(sc)
val df = sqlContext.load(
"org.apache.phoenix.spark",
Map("table" -> "TABLE1", "zkUrl" -> "phoenix-server:2181")
)
df
.filter(df("COL1") === "test_row_1" && df("ID") === 1L)
.select(df("ID"))
.show
Option 2 : Use JDBC thin client provied by phoenix query server
more info on https://phoenix.apache.org/server.html
jdbc:phoenix:thin:url=http://localhost:8765;serialization=PROTOBUF
val sql = "select time from table"
val data = sql(sql).map(_.getTimeStamp(0).toString)
In the hive table,time's type is timestamp.when i run this program,it throws NullPointerException.
val data = sql(sql).map(_.get(0).toString)
When I change to the above code,the same Exception be threw.
Is anyone can tell me how to get TimeStamp data in hive using Spark?
Tks.
If you are trying to read data from Hive table you should use, HiveContext instead of SQLContext.
If you are using Spark 2.0, you can try the following.
val spark = SparkSession
.builder()
.appName("Spark Hive Example")
.config("spark.sql.warehouse.dir", warehouseLocation)
.enableHiveSupport()
.getOrCreate()
import spark.implicits._
import spark.sql
val df = sql("select time from table")
df.select($"time").show()
So the read works fine as per documentation:
val cql = new org.apache.spark.sql.cassandra.CassandraSQLContext(sc)
cql.setConf("cluster-src/spark.cassandra.connection.host", "1.1.1.1")
cql.setConf("cluster-dst/spark.cassandra.connection.host", "2.2.2.2")
...
var df = cql.read.format("org.apache.spark.sql.cassandra")
.option("table", "my_table")
.option("keyspace", "my_keyspace")
.option("cluster", "cluster-src")
.load()
But it is not clear how to pass the destination cluster name to the save counterpart. This obviously does not work, it just tries to connect to the local spark host:
df.write
.format("org.apache.spark.sql.cassandra")
.option("table", "my_table")
.option("keyspace", "my_keyspace")
.option("cluster", "cluster-dst")
.save()
Update:
Found a workaround but it is kind of ugly. So instead of:
.option("cluster", "cluster-dst")
use:
.option("spark_cassandra_connection_host", cql.getConf("cluster-dst/spark.cassandra.connection.host")