CODEWITHSUNDEEP
pythonapache-sparkpyspark
solarenqu
solarenqu

Reputation: 814

Spark - PySpark sql error

I have a simple pyspark code but i can't run it. I try to run it on Ubuntu system and I use PyCharm IDE. I would like to connect to Oracle XE Database and I want to print my test table.

Here comes my spark python code: 

from pyspark import SparkContext
from pyspark.sql import SQLContext

sc = SparkContext()

sqlContext = SQLContext(sc)

demoDf = sqlContext.read.format("jdbc").options(
    url="jdbc:oracle:thin:@10.10.10.10:1521:XE",
    driver="oracle.jdbc.driver.OracleDriver",
    table="tst_table",
    user="xxx",
    password="xxx").load()

demoDf.show()

And this is my trace:

Traceback (most recent call last):
  File "/home/kebodev/PycharmProjects/spark_tst/cucc_spark.py", line 13, in <module>
    password="xxx").load()
  File "/home/kebodev/spark-2.0.1/python/pyspark/sql/readwriter.py", line 153, in load
    return self._df(self._jreader.load())
  File "/home/kebodev/spark-2.0.1/python/lib/py4j-0.10.3-src.zip/py4j/java_gateway.py", line 1133, in __call__
  File "/home/kebodev/spark-2.0.1/python/pyspark/sql/utils.py", line 63, in deco
    return f(*a, **kw)
  File "/home/kebodev/spark-2.0.1/python/lib/py4j-0.10.3-src.zip/py4j/protocol.py", line 319, in get_return_value
py4j.protocol.Py4JJavaError: An error occurred while calling o27.load.
: java.lang.RuntimeException: Option 'dbtable' not specified
    at scala.sys.package$.error(package.scala:27)
    at org.apache.spark.sql.execution.datasources.jdbc.JDBCOptions$$anonfun$2.apply(JDBCOptions.scala:30)
    at org.apache.spark.sql.execution.datasources.jdbc.JDBCOptions$$anonfun$2.apply(JDBCOptions.scala:30)
    at scala.collection.MapLike$class.getOrElse(MapLike.scala:128)
    at org.apache.spark.sql.execution.datasources.CaseInsensitiveMap.getOrElse(ddl.scala:117)
    at org.apache.spark.sql.execution.datasources.jdbc.JDBCOptions.<init>(JDBCOptions.scala:30)
    at org.apache.spark.sql.execution.datasources.jdbc.JdbcRelationProvider.createRelation(JdbcRelationProvider.scala:33)
    at org.apache.spark.sql.execution.datasources.DataSource.resolveRelation(DataSource.scala:330)
    at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:149)
    at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:122)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:498)
    at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:237)
    at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:357)
    at py4j.Gateway.invoke(Gateway.java:280)
    at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:132)
    at py4j.commands.CallCommand.execute(CallCommand.java:79)
    at py4j.GatewayConnection.run(GatewayConnection.java:214)
    at java.lang.Thread.run(Thread.java:745)


Process finished with exit code 1

Can Anybody help me?

Upvotes: 1

Views: 3771

Answers (2)

sau
sau

Reputation: 1356

Try something like this:

def testQuery(query):
        df = sqlContext.read.format("jdbc").options(
             url="jdbc:oracle:thin:@10.10.10.10:1521:XE",
             driver="oracle.jdbc.driver.OracleDriver",
             dbtable="( " + query + " ) as temp",
             user="xxx",
             password="xxx").load()
        return df

Upvotes: 0

scarecrow
scarecrow

Reputation: 1615

Change to dbtable from table like this,

demoDf = sqlContext.read.format("jdbc").options(
    url="jdbc:oracle:thin:@10.10.10.10:1521:XE",
    driver="oracle.jdbc.driver.OracleDriver",
    dbtable="tst_table",
    user="xxx",
    password="xxx").load()

Upvotes: 1

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