Re: Kryo not registered class

2017-11-20 Thread Vadim Semenov 
Try:

Class.forName("[Lorg.apache.spark.sql.execution.datasources.PartitioningAwareFileIndex$SerializableFileStatus$SerializableBlockLocation;")

On Sun, Nov 19, 2017 at 3:24 PM, Angel Francisco Orta <
angel.francisco.o...@gmail.com> wrote:

> Hello, I'm with spark 2.1.0 with scala and I'm registering all classes
> with kryo, and I have a  problem registering this class,
>
> org.apache.spark.sql.execution.datasources.PartitioningAwareFileIndex$
> SerializableFileStatus$SerializableBlockLocation[]
>
> I can't register with classOf[Array[Class.forName("org.apache.spark.sql.
> execution.datasources.PartitioningAwareFileIndex$SerializableFileStatus$
> SerializableBlockLocation").type]]
>
>
> I have tried as well creating a java class like register and registering
> the class as org.apache.spark.sql.execution.datasources.
> PartitioningAwareFileIndex$SerializableFileStatus$
> SerializableBlockLocation[].class;
>
> Any clue is appreciatted,
>
> Thanks.
>
>

Kryo not registered class

2017-11-19 Thread Angel Francisco Orta 
Hello, I'm with spark 2.1.0 with scala and I'm registering all classes with
kryo, and I have a  problem registering this class,

org.apache.spark.sql.execution.datasources.PartitioningAwareFileIndex$SerializableFileStatus$SerializableBlockLocation[]

I can't register with
classOf[Array[Class.forName("org.apache.spark.sql.execution.datasources.PartitioningAwareFileIndex$SerializableFileStatus$SerializableBlockLocation").type]]


I have tried as well creating a java class like register and registering
the class as
org.apache.spark.sql.execution.datasources.PartitioningAwareFileIndex$SerializableFileStatus$SerializableBlockLocation[].class;

Any clue is appreciatted,

Thanks.