Try converting that int from decimal to hex and inserting dashes in the appropriate spots - or go the other way.
Also, you are looking at different rows, based upon your selection criteria... ml On Tue, May 24, 2016 at 6:23 AM, Rajesh Radhakrishnan < rajesh.radhakrish...@phe.gov.uk> wrote: > Hi, > > > I got a Cassandra keyspace, but while reading the data(especially UUID) > via Spark SQL using Python is not returning the correct value. > > Cassandra: > -------------- > My table 'SAM'' is described below: > > CREATE table ks.sam (id uuid, dept text, workflow text, type double > primary key (id, dept)) > > SELECT id, workflow FROM sam WHERE dept='blah'; > > The above example CQL gives me the following > id | workflow > --------------------------------------+------------ > 9547v26c-f528-12e5-da8b-001a4q3dac10 | testWK > > > Spark/Python: > ------------------ > from pyspark import SparkConf > from pyspark.sql import SQLContext > import pyspark_cassandra > from pyspark_cassandra import CassandraSparkContext > > .... > conf = > SparkConf().set("spark.cassandra.connection.host",IP_ADDRESS).set("spark.cassandra.connection.native.port",PORT_NUMBER) > sparkContext = CassandraSparkContext(conf = conf) > sqlContext = SQLContext(sparkContext) > > samTable =sparkContext.cassandraTable("ks", "sam").select('id', 'dept',' > workflow') > samTable.cache() > > samdf.registerTempTable("samd") > > sparkSQLl ="SELECT distinct id, dept, workflow FROM samd WHERE workflow=' > testWK' > new_df = sqlContext.sql(sparkSQLl) > results = new_df.collect() > for row in results: > print "dept=",row.dept > print "wk=",row.workflow > print "id=",row.id > ... > The Python code above prints the following: > dept=Biology > wk=testWK > id=293946894141093607334963674332192894528 > > > You can see here that the id (uuid) whose correct value at Cassandra is ' > 9547v26c-f528-12e5-da8b-001a4q3dac10' but via Spark I am getting an int ' > 29394689414109360733496367433219289452'. > What I am doing wrong here? How to get the correct UUID value from > Cassandra via Spark/Python ? Please help me. > > Thank you > Rajesh R > > ************************************************************************** > The information contained in the EMail and any attachments is confidential > and intended solely and for the attention and use of the named > addressee(s). It may not be disclosed to any other person without the > express authority of Public Health England, or the intended recipient, or > both. If you are not the intended recipient, you must not disclose, copy, > distribute or retain this message or any part of it. This footnote also > confirms that this EMail has been swept for computer viruses by > Symantec.Cloud, but please re-sweep any attachments before opening or > saving. http://www.gov.uk/PHE > ************************************************************************** >