I run a daily job to write data to BigQuery using Databricks Pyspark. There was a recent update of configuration for Databricks (https://docs.databricks.com/data/data-sources/google/bigquery.html) which caused the job to fail. I followed all the steps in the docs. Reading data works again but writing throws the following error: java.lang.RuntimeException: java.lang.ClassNotFoundException: Class com.google.cloud.hadoop.fs.gcs.GoogleHadoopFS not found
I tried adding configuration also right in the code (as advised for similar errors in Spark) but it did not help:
spark._jsc.hadoopConfiguration().set('fs.gs.impl', 'com.google.cloud.hadoop.fs.gcs.GoogleHadoopFileSystem')
spark._jsc.hadoopConfiguration().set('fs.gs.auth.service.account.enable', 'true')
spark._jsc.hadoopConfiguration().set('google.cloud.auth.service.account.json.keyfile', "<path-to-key.json>")
My code is:
upload_table_dataset = 'testing_dataset'
upload_table_name = 'testing_table'
upload_table = upload_table_dataset + '.' + upload_table_name
(import_df.write.format('bigquery')
.mode('overwrite')
.option('project', 'xxxxx-test-project')
.option('parentProject', 'xxxxx-test-project')
.option('temporaryGcsBucket', 'xxxxx-testing-bucket')
.option('table', upload_table)
.save()
)
You need to install the GCS connector on your cluster first
Related
I'm trying to use glue streaming and write data to AWS TimestreamDB but I'm having a hard time in configuring the JDBC connection.
Steps I’m following are below and the documentation link: https://docs.aws.amazon.com/timestream/latest/developerguide/JDBC.configuring.html
I’m uploading the jar to S3. There are multiple jars here and I tried with each one of it. https://github.com/awslabs/amazon-timestream-driver-jdbc/releases
In the glue job I’m pointing the jar lib path to the above s3 location
In the job script I’m trying to read from timestream using both spark/ glue with the below code but its not working. Can someone explain what I'm doing wrong here
This is my code:
url = jdbc:timestream://AccessKeyId=<myAccessKeyId>;SecretAccessKey=<mySecretAccessKey>;SessionToken=<mySessionToken>;Region=us-east-1
source_df = sparkSession.read.format("jdbc").option("url",url).option("dbtable","IoT").option("driver","software.amazon.timestream.jdbc.TimestreamDriver").load()
datasink1 = glueContext.write_dynamic_frame.from_options(frame = applymapping0, connection_type = "jdbc", connection_options = {"url":url,"driver":"software.amazon.timestream.jdbc.TimestreamDriver", database = "CovidTestDb", dbtable = "CovidTestTable"}, transformation_ctx = "datasink1")
To this date (April 2022) there is not support for write operations using timestream's jdbc driver (reviewed the code and saw a bunch of no write support exceptions). It is possible to read data from timestream using glue though. Following steps worked for me:
Upload timestream-query and timestream-jdbc to an S3 bucket that you can reference in your glue script
Ensure that the IAM role for the script has access to read operations to the timestream database and table
You don't need to use the access key and secret parameters in the jdbc url, using something like jdbc:timestream://Region=<timestream-db-region> should be enough
Specify the driver and fetchsize options option("driver","software.amazon.timestream.jdbc.TimestreamDriver")
option("fetchsize", "100") (tweak the fetchsize according to your needs)
Following is a complete example of reading a dataframe from timestream:
val df = sparkSession.read.format("jdbc")
.option("url", "jdbc:timestream://Region=us-east-1")
.option("driver","software.amazon.timestream.jdbc.TimestreamDriver")
// optionally add a query to narrow the data to fetch
.option("query", "select * from db.tbl where time between ago(15m) and now()")
.option("fetchsize", "100")
.load()
df.write.format("console").save()
Hope this helps
Here is my spark sql code, where I am trying to read a presto table based on this guide; https://spark.apache.org/docs/latest/sql-data-sources-jdbc.html
val df = spark.read
.format("jdbc")
.option("driver", "com.facebook.presto.jdbc.PrestoDriver")
.option("url", "jdbc:presto://localhost:8889/mycatalog")
.option("query", "select * from mydb.mytable limit 1")
.option("user", "myuserid")
.load()
I am getting the following exception, unrecognized connection property 'url'
Exception in thread "main" java.sql.SQLException: Unrecognized connection property 'url'
at com.facebook.presto.jdbc.PrestoDriverUri.validateConnectionProperties(PrestoDriverUri.java:345)
at com.facebook.presto.jdbc.PrestoDriverUri.<init>(PrestoDriverUri.java:102)
at com.facebook.presto.jdbc.PrestoDriverUri.<init>(PrestoDriverUri.java:92)
at com.facebook.presto.jdbc.PrestoDriver.connect(PrestoDriver.java:87)
at org.apache.spark.sql.execution.datasources.jdbc.connection.BasicConnectionProvider.getConnection(BasicConnectionProvider.scala:49)
at org.apache.spark.sql.execution.datasources.jdbc.connection.ConnectionProvider$.create(ConnectionProvider.scala:68)
at org.apache.spark.sql.execution.datasources.jdbc.JdbcUtils$.$anonfun$createConnectionFactory$1(JdbcUtils.scala:62)
at org.apache.spark.sql.execution.datasources.jdbc.JDBCRDD$.resolveTable(JDBCRDD.scala:56)
at org.apache.spark.sql.execution.datasources.jdbc.JDBCRelation$.getSchema(JDBCRelation.scala:226)
at org.apache.spark.sql.execution.datasources.jdbc.JdbcRelationProvider.createRelation(JdbcRelationProvider.scala:35)
at org.apache.spark.sql.execution.datasources.DataSource.resolveRelation(DataSource.scala:354)
at org.apache.spark.sql.DataFrameReader.loadV1Source(DataFrameReader.scala:326)
at org.apache.spark.sql.DataFrameReader.$anonfun$load$3(DataFrameReader.scala:308)
at scala.Option.getOrElse(Option.scala:189)
at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:308)
at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:226)
at org.apache.spark.sql.DataFrameReader.jdbc(DataFrameReader.scala:341)
Seems like this issue is related to https://github.com/prestodb/presto/issues/9254 where the property url is not a recognized property in Presto and looks like the fix needs to be done on the Spark side? Are there any other workaround for this issue?
PS:
Spark Version: 3.1.1
presto-jdbc version: 0.245
looks like a spark bug fixed 3.3
https://issues.apache.org/jira/browse/SPARK-36163
There is no issue with spark or presto JDBC driver. I don't think URL which you specified will work.
You should change that to below format.
jdbc:presto://localhost:8889/mycatalog
UPDATE
Not sure how it's working with spark version < 3. As an workaround you can use another jar where strict config check has been removed as specified here.
#odonnry is correct that the issue was fixed in spark 3.3.x, but if anyone cannot upgrade to Spark 3.3.x and is trying to use Trino, I created a workaround below according to the Jira issue linked by #Mohana
https://github.com/amitferman/trino
I have a Spark project with AWS Glue implementation running locally.
I listen to a Kinesis stream so when Data is arrived in JSON format, I can storage to S3 correctly.
I want to store in AWS RDS instead of storing in S3.
I have tried to use:
dataFrame.write
.format("jdbc")
.option("url","jdbc:mysql://aurora.cluster.region.rds.amazonaws.com:3306/database")
.option("user","user")
.option("password","password")
.option("dbtable","test-table")
.option("driver","com.mysql.jdbc.Driver")
.save()
Spark project get data from a Kinesis stream using AWS glue job.
I want to add the data to Aurora database.
It fails with error
Caused by: java.sql.SQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL
server version for the right syntax to use near '-glue-table (`label2` TEXT , `customerid` TEXT , `sales` TEXT , `name` TEXT )' a
t line 1
This is the test dataFrame Im using, dataFrame.show():
+------+----------+-----+--------------------+
|label2|customerid|sales| name|
+------+----------+-----+--------------------+
| test6| test| test|streamingtesttest...|
+------+----------+-----+--------------------+
Using Spark DynamicFrame instead of DataFrame and using the glueContext sink to publish to Aurora:
So the final code could be:
lazy val mysqlJsonOption = jsonOptions(MYSQL_AURORA_URI)
//Write to Aurora
val dynamicFrame = DynamicFrame(joined, glueContext)
glueContext.getSink("mysql", mysqlJsonOption).writeDynamicFrame(dynamicFrame)
I am trying to integrate Cassandra with Spark and facing the below issue.
Issue:
com.datastax.spark.connector.util.ConfigCheck$ConnectorConfigurationException: Invalid Config Variables
Only known spark.cassandra.* variables are allowed when using the Spark Cassandra Connector.
spark.cassandra.keyspace is not a valid Spark Cassandra Connector variable.
Possible matches:
spark.cassandra.sql.keyspace
spark.cassandra.output.batch.grouping.key
at com.datastax.spark.connector.util.ConfigCheck$.checkConfig(ConfigCheck.scala:50)
at com.datastax.spark.connector.cql.CassandraConnectorConf$.apply(CassandraConnectorConf.scala:253)
at org.apache.spark.sql.cassandra.CassandraSourceRelation$.apply(CassandraSourceRelation.scala:263)
at org.apache.spark.sql.cassandra.CassandraCatalog.org$apache$spark$sql$cassandra$CassandraCatalog$$buildRelation(CasandraCatalog.scala:41)
at org.apache.spark.sql.cassandra.CassandraCatalog$$anon$1.load(CassandraCatalog.scala:26)
at org.apache.spark.sql.cassandra.CassandraCatalog$$anon$1.load(CassandraCatalog.scala:23)
Please find the below versions of spark Cassandra and connector I am using.
Spark : 1.6.0
Cassandra : 2.1.17
Connector Used : spark-cassandra-connector_2.10-1.6.0-M1.jar
Below is the code snippet I am using to connect Cassandra from spark.
val conf: org.apache.spark.SparkConf = new SparkConf(true) \
.setAppName("Spark Cassandra") \
.set"spark.cassandra.connection.host", "abc.efg.lkh") \
.set("spark.cassandra.auth.username", "xyz") \
.set("spark.cassandra.auth.password", "1234") \
.set("spark.cassandra.keyspace","abcded")
val sc = new SparkContext("local[*]", "Spark Cassandra",conf)
val csc = new CassandraSQLContext(sc)
csc.setKeyspace("abcded")
val my_df = csc.sql("select * from table")
Here when I try to create DF, I am getting above posted error. I tried without passing schema in conf but it is trying to access in default schema where mentioned user doesn't have access.
Already a JIRA was opened and closed.
https://datastax-oss.atlassian.net/browse/SPARKC-102
yet I am getting this issue. Please let me know whether I need to use lastest connector to resolve this issue.
Thanks in advance.
The important information is in the error message you posted [formatted for readability]:
Invalid Config Variables
Only known spark.cassandra.* variables are allowed when using the Spark Cassandra Connector.
spark.cassandra.keyspace is not a valid Spark Cassandra Connector variable.
Possible matches: spark.cassandra.sql.keyspace
spark.cassandra.keyspace is not an available property for the connector. A full list of the available properties can be found here: https://github.com/datastax/spark-cassandra-connector/blob/master/doc/reference.md
You may have some luck using the suggested spark.cassandra.sql.keyspace; otherwise you may just need to explicitly specify the keyspace for every Cassandra interaction you perform using the connector.
I have written a spark code -
customDF.registerTempTable("customTable")
var query = "select date, id ts,errorode from customTable"
val finalCustomDF = hiveContext.sql(query)
finalCustomDF.write.format("com.databricks.spark.csv").save("/user/oozie/data")
When i run this code using spark submit, it runs fine but when i run it using oozie coordinator. I get following exception.
User class threw exception: org.apache.spark.sql.AnalysisException: character '<EOF>' not supported here; line 1 pos 111
I have tried reading data from existing hive table it works but issue is with customTable.