The issue is cased by the data, and indeed a type miss match between Hive
schema and Spark. Now it is fixed.
Without that kind of data, the problem won't be trigged in some brands.
Thanks taking a look of this problem.
Yong
________________________________
From: ayan guha <guha.a...@gmail.com>
Sent: Tuesday, June 13, 2017 1:54 AM
To: Angel Francisco Orta
Cc: Yong Zhang; user@spark.apache.org
Subject: Re: Parquet file generated by Spark, but not compatible read by Hive
Try setting following Param:
conf.set("spark.sql.hive.convertMetastoreParquet","false")
On Tue, Jun 13, 2017 at 3:34 PM, Angel Francisco Orta
<angel.francisco.o...@gmail.com<mailto:angel.francisco.o...@gmail.com>> wrote:
Hello,
Do you use df.write or you make with hivecontext.sql(" insert into ...")?
Angel.
El 12 jun. 2017 11:07 p. m., "Yong Zhang"
<java8...@hotmail.com<mailto:java8...@hotmail.com>> escribió:
We are using Spark 1.6.2 as ETL to generate parquet file for one dataset, and
partitioned by "brand" (which is a string to represent brand in this dataset).
After the partition files generated in HDFS like "brand=a" folder, we add the
partitions in the Hive.
The hive version is 1.2.1 (In fact, we are using HDP 2.5.0).
Now the problem is that for 2 brand partitions, we cannot query the data
generated in Spark, but it works fine for the rest of partitions.
Below is the error in the Hive CLI and hive.log I got if I query the bad
partitions like "select * from tablename where brand='BrandA' limit 3;"
Failed with exception
java.io.IOException:org.apache.hadoop.hive.ql.metadata.HiveException:
java.lang.UnsupportedOperationException: Cannot inspect
org.apache.hadoop.io<http://org.apache.hadoop.io>.LongWritable
Caused by: java.lang.UnsupportedOperationException: Cannot inspect
org.apache.hadoop.io<http://org.apache.hadoop.io>.LongWritable
at
org.apache.hadoop.hive.ql.io<http://org.apache.hadoop.hive.ql.io>.parquet.serde.primitive.ParquetStringInspector.getPrimitiveWritableObject(ParquetStringInspector.java:52)
at
org.apache.hadoop.hive.serde2.lazy.LazyUtils.writePrimitiveUTF8(LazyUtils.java:222)
at
org.apache.hadoop.hive.serde2.lazy.LazySimpleSerDe.serialize(LazySimpleSerDe.java:307)
at
org.apache.hadoop.hive.serde2.lazy.LazySimpleSerDe.serializeField(LazySimpleSerDe.java:262)
at
org.apache.hadoop.hive.serde2.DelimitedJSONSerDe.serializeField(DelimitedJSONSerDe.java:72)
at
org.apache.hadoop.hive.serde2.lazy.LazySimpleSerDe.doSerialize(LazySimpleSerDe.java:246)
at
org.apache.hadoop.hive.serde2.AbstractEncodingAwareSerDe.serialize(AbstractEncodingAwareSerDe.java:50)
at
org.apache.hadoop.hive.ql.exec.DefaultFetchFormatter.convert(DefaultFetchFormatter.java:71)
at
org.apache.hadoop.hive.ql.exec.DefaultFetchFormatter.convert(DefaultFetchFormatter.java:40)
at
org.apache.hadoop.hive.ql.exec.ListSinkOperator.process(ListSinkOperator.java:90)
... 22 more
There are not too much I can find by googling this error message, but it points
to that the schema in Hive is different as in parquet file.
But this is a very strange case, as the same schema works fine for other
brands, which defined as a partition column, and share the whole Hive schema as
the above.
If I query like: "select * from tablename where brand='BrandB' limit 3:",
everything works fine.
So is this really caused by the Hive schema mismatch with parquet file
generated by Spark, or by the data within different partitioned keys, or really
a compatible issue between Spark/Hive?
Thanks
Yong
--
Best Regards,
Ayan Guha
