cloud-fan commented on code in PR #37393:
URL: https://github.com/apache/spark/pull/37393#discussion_r939825908
##########
sql/catalyst/src/main/scala/org/apache/spark/sql/internal/connector/PredicateUtils.scala:
##########
@@ -20,41 +20,28 @@ package org.apache.spark.sql.internal.connector
import org.apache.spark.sql.catalyst.CatalystTypeConverters
import org.apache.spark.sql.connector.expressions.{LiteralValue,
NamedReference}
import org.apache.spark.sql.connector.expressions.filter.{And => V2And, Not =>
V2Not, Or => V2Or, Predicate}
+import org.apache.spark.sql.errors.QueryCompilationErrors
import org.apache.spark.sql.sources.{AlwaysFalse, AlwaysTrue, And,
EqualNullSafe, EqualTo, Filter, GreaterThan, GreaterThanOrEqual, In, IsNotNull,
IsNull, LessThan, LessThanOrEqual, Not, Or, StringContains, StringEndsWith,
StringStartsWith}
import org.apache.spark.sql.types.StringType
private[sql] object PredicateUtils {
def toV1(predicate: Predicate): Option[Filter] = {
- def isValidBinaryPredicate(): Boolean = {
- if (predicate.children().length == 2 &&
- predicate.children()(0).isInstanceOf[NamedReference] &&
- predicate.children()(1).isInstanceOf[LiteralValue[_]]) {
- true
- } else {
- false
- }
- }
-
predicate.name() match {
- case "IN" if predicate.children()(0).isInstanceOf[NamedReference] =>
+ case "IN" if isValidPredicate(predicate) =>
val attribute = predicate.children()(0).toString
val values = predicate.children().drop(1)
if (values.length > 0) {
- if (!values.forall(_.isInstanceOf[LiteralValue[_]])) return None
val dataType = values(0).asInstanceOf[LiteralValue[_]].dataType
- if
(!values.forall(_.asInstanceOf[LiteralValue[_]].dataType.sameType(dataType))) {
- return None
- }
val inValues = values.map(v =>
CatalystTypeConverters.convertToScala(v.asInstanceOf[LiteralValue[_]].value,
dataType))
Some(In(attribute, inValues))
} else {
Some(In(attribute, Array.empty[Any]))
}
- case "=" | "<=>" | ">" | "<" | ">=" | "<=" if isValidBinaryPredicate =>
Review Comment:
+1
--
This is an automated message from the Apache Git Service.
To respond to the message, please log on to GitHub and use the
URL above to go to the specific comment.
To unsubscribe, e-mail: [email protected]
For queries about this service, please contact Infrastructure at:
[email protected]
---------------------------------------------------------------------
To unsubscribe, e-mail: [email protected]
For additional commands, e-mail: [email protected]
