update one column in cassandra table

Question

I have a cassandra table person_master (personId: int, customerId: Int, firstName: String, lastName: String, mrids: Set) primaryKey (personId and customerID)

Suppose I have one input RDD of structure [personId, customerId, firstName, lastname, messageType: String, source: String, sourceType: String]

suppose value of RDD:[1001,119,None,None,{abc.xyz} and cassandra row has value [1001,119,Vikash,Singh,{aaa.bbb}]

I want on fetch cassandra row based on RDD value and update the mrids column of cassandra table and using all other column from cassandra row.

e.g. in this I want final RDD value as [1001,119,Vikash,Singh,{aaa.bbb,abc.xyz}] which I will update to cassandra later.

Can anybody give me the solution to do this in Spark using cassandra Connector.

Abhishek Anand Abhishek Anand · Accepted Answer · 2016-04-18T05:41:53

Assuming sc is sparkContext like,

val sparkConf = new SparkConf().setMaster(SPARK_MASTER)
                            .setAppName(SPARK_SCALA_APP_NAME)
                            .setJars(SPARK_SCALA_JAR)
sparkConf.set("spark.cassandra.connection.host", value)
sparkConf.set("spark.cassandra.auth.username", value)
sparkConf.set("spark.cassandra.auth.password", value)
val sc = new SparkContext(sparkConf)

You can use or ignore where clause (where can be used only if its partition key)

val selectedRow = sc.cassandraTable("keyspace", "tableName")
      .select("key", "column2", "column3")
      .where("key IN ?", keys)
      .as((key: String, column2: String, column3: Integer)
          =>(key, column2, column3))

Do filtering and modification on your rdd Then save it like,

selectedRow.saveToCassandra("keyspace",
                           "tableName",
                           SomeColumns("key", "column2", "column3"))

update one column in cassandra table

1 Answers