Github user dongjoon-hyun commented on the issue:
https://github.com/apache/spark/pull/14132
In database, the purpose of view is **hiding the underneath tables**. We
should not support to specify the table inside the view. However, you can
specify the hint on the view itself. Let me give you example. I assumed that
you asked the following scenario.
```scala
scala> sql("create temporary view view_u as select * from u")
```
We know that `u` is inside the view. But any table names inside the view
should be ignored because they are not is unrecognized table name in this
context.
```scala
scala> sql("SELECT /*+ MAPJOIN(u) */ * FROM t JOIN view_u ON t.id =
view_u.id").explain(true)
== Physical Plan ==
*SortMergeJoin [id#0L], [id#4L], Inner
:- *Sort [id#0L ASC], false, 0
: +- Exchange hashpartitioning(id#0L, 200)
: +- *Range (0, 1000000000, splits=8)
+- *Sort [id#4L ASC], false, 0
+- ReusedExchange [id#4L], Exchange hashpartitioning(id#0L, 200)
```
However, if you give the hint on view **view_u**, it can be propagated.
```scala
scala> sql("SELECT /*+ MAPJOIN(view_u) */ * FROM t JOIN view_u ON t.id =
view_u.id").explain
== Physical Plan ==
*BroadcastHashJoin [id#0L], [id#4L], Inner, BuildRight
:- *Range (0, 1000000000, splits=8)
+- BroadcastExchange HashedRelationBroadcastMode(List(input[0, bigint,
false]))
+- *Range (0, 1000000000, splits=8)
```
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