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Posted to dev@phoenix.apache.org by "Fengdong Yu (JIRA)" <ji...@apache.org> on 2015/09/24 10:56:04 UTC
[jira] [Created] (PHOENIX-2290) Spark Phoenix cannot recognize
Phoenix view fields
Fengdong Yu created PHOENIX-2290:
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Summary: Spark Phoenix cannot recognize Phoenix view fields
Key: PHOENIX-2290
URL: https://issues.apache.org/jira/browse/PHOENIX-2290
Project: Phoenix
Issue Type: Bug
Affects Versions: 4.5.1
Reporter: Fengdong Yu
I created base table in base shell:
{code}
create 'test_table', {NAME => 'cf1', VERSIONS => 1}
put 'test_table', 'row_key_1', 'cf1:col_1', '200'
{code}
This is a very simple table. then create phoenix view in Phoenix shell.
{code}
create view "test_table" (pk varchar primary key, "cf1"."col_1" archer)
{code}
then do following in Spark shell:
val df = sqlContext.load("org.apache.phoenix.spark", Map("table" -> "\"test_table\"", "zkUrl" -> "localhost:2181"))
df.registerTempTable("temp")
sqlContext.sql("select * from temp where col_1='200')
java.lang.RuntimeException: org.apache.phoenix.schema.ColumnNotFoundException: ERROR 504 (42703): Undefined column. columnName=col_1
at org.apache.phoenix.mapreduce.PhoenixInputFormat.getQueryPlan(PhoenixInputFormat.java:125)
at org.apache.phoenix.mapreduce.PhoenixInputFormat.getSplits(PhoenixInputFormat.java:80)
at org.apache.spark.rdd.NewHadoopRDD.getPartitions(NewHadoopRDD.scala:95)
at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:219)
at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:217)
at scala.Option.getOrElse(Option.scala:120)
at org.apache.spark.rdd.RDD.partitions(RDD.scala:217)
at org.apache.phoenix.spark.PhoenixRDD.getPartitions(PhoenixRDD.scala:47)
at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:219)
at org.apache.spark.rdd.RDD$$anonfun$partitions$2.apply(RDD.scala:217)
at scala.Option.getOrElse(Option.scala:120)
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