Details
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Improvement
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Status: In Progress
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Minor
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Resolution: Unresolved
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3.1.0
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None
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None
Description
The following 2 query produce different plans, as the second one is not recognised as equijoin.
SELECT * FROM t1 FULL OUTER JOIN t2 ON t1.c2 = 2 AND t2.c2 = 2 AND t1.c = t2.c
SortMergeJoin [c#225], [c#236], FullOuter, ((c2#226 = 2) AND (c2#237 = 2))
:- *(2) Sort [c#225 ASC NULLS FIRST], false, 0
: +- Exchange hashpartitioning(c#225, 5), true, [id=#101]
: +- *(1) Project [_1#220 AS c#225, _2#221 AS c2#226]
: +- *(1) LocalTableScan [_1#220, _2#221]
+- *(4) Sort [c#236 ASC NULLS FIRST], false, 0
+- Exchange hashpartitioning(c#236, 5), true, [id=#106]
+- *(3) Project [_1#231 AS c#236, _2#232 AS c2#237]
+- *(3) LocalTableScan [_1#231, _2#232]
SELECT * FROM t1 FULL OUTER JOIN t2 ON t1.c2 = 2 AND t2.c2 = 2
BroadcastNestedLoopJoin BuildRight, FullOuter, ((c2#226 = 2) AND (c2#237 = 2))
:- *(1) Project [_1#220 AS c#225, _2#221 AS c2#226]
: +- *(1) LocalTableScan [_1#220, _2#221]
+- BroadcastExchange IdentityBroadcastMode, [id=#146]
+- *(2) Project [_1#231 AS c#236, _2#232 AS c2#237]
+- *(2) LocalTableScan [_1#231, _2#232]
We could detect the implicit equalities from the join condition.
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