When registering a table using the %pyspark interpreter in Zeppelin, I can't access the table in %sql

When registering a table using the %pyspark interpreter in Zeppelin, I can't access the table in %sql

By : sanjeev kumar google
Date : November 26 2020, 09:01 AM
I wish this help you I found the problem to this issue. Prior to 0.6.0 the sqlContext variable is sqlc in %pyspark.
Defect can be found here: https://issues.apache.org/jira/browse/ZEPPELIN-134
code :

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pyspark with HCatalog table in Zeppelin

pyspark with HCatalog table in Zeppelin

By : Hawk
Date : March 29 2020, 07:55 AM
help you fix your problem This an expected behavior. Each table or DataFrame in Spark is bound to a specific SQLContext which has been used to create it and cannot be accessed outside it.
Apache Zeppelin initializes both SparkContext and SQLContext which are shared between interpreters and can be accessed as sc and sqlContext respectively. You should these instances, not create another one, if you want to benefit from interpreter interoperability. In other words don't create custom context by use default one.
zeppelin user Spark interpreter save DataFrame to Hbase Table

zeppelin user Spark interpreter save DataFrame to Hbase Table

By : ncanquest
Date : March 29 2020, 07:55 AM
around this issue hbaseRDD.saveAsNewAPIHadoopDataset(getJob("HWBTEST").getConfiguration)
Solved the problem
Pyspark: Issues With Inserting Into a table in Hive using Zeppelin Notebook

Pyspark: Issues With Inserting Into a table in Hive using Zeppelin Notebook

By : Jovison
Date : March 29 2020, 07:55 AM
may help you . NOTE: This is being written in a Zeppelin Notebook. , You provided the following code.
code :
from pyspark.context import SparkContext

sql = sqlContext.sql("INSERT INTO TABLE kenny_target(`user`, `age`) SELECT 
`user`, COALESCE(`age`, 0L) FROM kenny_source")
frame= sqlContext.createDataFrame(sql).collect()
sql = sqlContext.sql("INSERT INTO TABLE kenny_target('user', 'age') SELECT 'user', COALESCE('age', 0L) FROM kenny_source")
frame = sqlContext.createDataframe(sql)
# Since, you wrote sql = sqlContext.sql('query')
# You could write frame = sql.collect()
frame = sqlContext.createDataFrame(sql).collect()
# SELECT user and age into 'frame' by creating a DataFrame called as 'frame'
frame = sqlContext.sql('''SELECT `user`, COALESCE(`age`, 0L) FROM kenny_source''')

# Write it to the table: kenny_source_test
Zeppelin Notebook %pyspark interpreter vs %python interpreter

Zeppelin Notebook %pyspark interpreter vs %python interpreter

By : Mikhail
Date : March 29 2020, 07:55 AM
I wish did fix the issue. When you run a %pyspark paragraph, zeppelin will create a spark context (spark variable) automatically with the defined parameters (loading spark packages, settings...).* Have a look at the documentation) of the spark-interpreter for some of the possibilities.
In %python paragraph you can create a spark context by your own but it is not done automatically and will not use the defined parameters of the spark interpreter section.
code :
<SparkContext master=local[4] appName=ZeppelinHub>
<pyspark.sql.session.SparkSession at 0x7fe757ca1f60>

Zeppelin - Cannot query with %sql a table I registered with pyspark

Zeppelin - Cannot query with %sql a table I registered with pyspark

By : Sunny
Date : March 29 2020, 07:55 AM
should help you out Zeppelin can create different contexts for different interpreters it is possible that if you executed some code with %spark and some code with %pyspark interpreters your Zeppelin can have two contexts. And when you use %sql it is looking in another context not in %pyspark. Try restart Zeppelin and execute %pyspark code as first statement and than %sql as second.
If you go to 'Interpreters' tab you can add zeppelin.spark.sql.stacktrace there. And after restart Zeppelin you will see full stack trace in a place where you have 'Table not found' now.
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