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Sparksql error after select command

Sparksql error after select command

New Contributor

I try to request my dataset register table and i get this error.

Can you help know please?

org.apache.spark.sql.AnalysisException: cannot recognize input near '/' '*' 'Nombre'; line 1 pos 0
	at org.apache.spark.sql.hive.HiveQl$.createPlan(HiveQl.scala:318)
	at org.apache.spark.sql.hive.ExtendedHiveQlParser$$anonfun$hiveQl$1.apply(ExtendedHiveQlParser.scala:41)
	at org.apache.spark.sql.hive.ExtendedHiveQlParser$$anonfun$hiveQl$1.apply(ExtendedHiveQlParser.scala:40)
	at scala.util.parsing.combinator.Parsers$Success.map(Parsers.scala:136)
	at scala.util.parsing.combinator.Parsers$Success.map(Parsers.scala:135)
	at scala.util.parsing.combinator.Parsers$Parser$$anonfun$map$1.apply(Parsers.scala:242)
	at scala.util.parsing.combinator.Parsers$Parser$$anonfun$map$1.apply(Parsers.scala:242)
	at scala.util.parsing.combinator.Parsers$$anon$3.apply(Parsers.scala:222)
	at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append$1$$anonfun$apply$2.apply(Parsers.scala:254)
	at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append$1$$anonfun$apply$2.apply(Parsers.scala:254)
	at scala.util.parsing.combinator.Parsers$Failure.append(Parsers.scala:202)
	at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append$1.apply(Parsers.scala:254)
	at scala.util.parsing.combinator.Parsers$Parser$$anonfun$append$1.apply(Parsers.scala:254)
	at scala.util.parsing.combinator.Parsers$$anon$3.apply(Parsers.scala:222)
	at scala.util.parsing.combinator.Parsers$$anon$2$$anonfun$apply$14.apply(Parsers.scala:891)
	at scala.util.parsing.combinator.Parsers$$anon$2$$anonfun$apply$14.apply(Parsers.scala:891)
	at scala.util.DynamicVariable.withValue(DynamicVariable.scala:57)
	at scala.util.parsing.combinator.Parsers$$anon$2.apply(Parsers.scala:890)
	at scala.util.parsing.combinator.PackratParsers$$anon$1.apply(PackratParsers.scala:110)
	at org.apache.spark.sql.catalyst.AbstractSparkSQLParser.parse(AbstractSparkSQLParser.scala:34)
	at org.apache.spark.sql.hive.HiveQl$.parseSql(HiveQl.scala:295)
1 REPLY 1

Re: Sparksql error after select command

Can you please post your query

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