I've found people.toDF gives you a data frame (roughly equivalent to the
previous Row RDD),
And you can then call registerTempTable on that DataFrame.
So people.toDF.registerTempTable("people") should work
—
Sent from Mailbox
On Sat, Mar 14, 2015 at 5:33 PM, David Mitchell <jdavidmitch...@gmail.com>
wrote:
> I am pleased with the release of the DataFrame API. However, I started
> playing with it, and neither of the two main examples in the documentation
> work: http://spark.apache.org/docs/1.3.0/sql-programming-guide.html
> Specfically:
> - Inferring the Schema Using Reflection
> - Programmatically Specifying the Schema
> Scala 2.11.6
> Spark 1.3.0 prebuilt for Hadoop 2.4 and later
> *Inferring the Schema Using Reflection*
> scala> people.registerTempTable("people")
> <console>:31: error: value registerTempTable is not a member of
> org.apache.spark
> .rdd.RDD[Person]
> people.registerTempTable("people")
> ^
> *Programmatically Specifying the Schema*
> scala> val peopleDataFrame = sqlContext.createDataFrame(people, schema)
> <console>:41: error: overloaded method value createDataFrame with
> alternatives:
> (rdd: org.apache.spark.api.java.JavaRDD[_],beanClass:
> Class[_])org.apache.spar
> k.sql.DataFrame <and>
> (rdd: org.apache.spark.rdd.RDD[_],beanClass:
> Class[_])org.apache.spark.sql.Dat
> aFrame <and>
> (rowRDD:
> org.apache.spark.api.java.JavaRDD[org.apache.spark.sql.Row],columns:
> java.util.List[String])org.apache.spark.sql.DataFrame <and>
> (rowRDD:
> org.apache.spark.api.java.JavaRDD[org.apache.spark.sql.Row],schema: o
> rg.apache.spark.sql.types.StructType)org.apache.spark.sql.DataFrame <and>
> (rowRDD: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row],schema:
> org.apache
> .spark.sql.types.StructType)org.apache.spark.sql.DataFrame
> cannot be applied to (org.apache.spark.rdd.RDD[String],
> org.apache.spark.sql.ty
> pes.StructType)
> val df = sqlContext.createDataFrame(people, schema)
> Any help would be appreciated.
> David
