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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