Github user cloud-fan commented on a diff in the pull request:
https://github.com/apache/spark/pull/16233#discussion_r95306084
--- Diff:
sql/hive/src/main/scala/org/apache/spark/sql/hive/HiveMetastoreCatalog.scala ---
@@ -125,11 +132,16 @@ private[hive] class
HiveMetastoreCatalog(sparkSession: SparkSession) extends Log
// Otherwise, wrap the table with a Subquery using the table name.
alias.map(a => SubqueryAlias(a, qualifiedTable,
None)).getOrElse(qualifiedTable)
} else if (table.tableType == CatalogTableType.VIEW) {
+ val tableIdentifier = table.identifier
val viewText = table.viewText.getOrElse(sys.error("Invalid view
without text."))
- SubqueryAlias(
- alias.getOrElse(table.identifier.table),
- sparkSession.sessionState.sqlParser.parsePlan(viewText),
- Option(table.identifier))
+ // The relation is a view, so we wrap the relation by:
+ // 1. Add a [[View]] operator over the relation to keep track of the
view desc;
+ // 2. Wrap the logical plan in a [[SubqueryAlias]] which tracks the
name of the view.
+ val child = View(
+ desc = table,
+ output = table.schema.toAttributes,
+ child = sparkSession.sessionState.sqlParser.parsePlan(viewText))
--- End diff --
sorry I may have asked this question before, but I can't recall your
answer. Why do we need the `output` field of `View`? can we do something like
```
val child = sparkSession.sessionState.sqlParser.parsePlan(viewText)
val projectList = schema.map { field =>
Alias(UnresolvedAttribute(Seq(field.name)), field.name)(explicitMetadata
= Some(field.metadata))
}
View(
desc = table,
child = Project(projectList, child))
case class View(...) {
def output = child.output
}
```
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