下面的SQL哪些有可能经过两个阶段（只经过一次shuffle）就计算出结果：（） A: SELECT key1, key2, SUM(value1) FROM (SELECT key1, key2, COUNT(*) FROM dual GROUP BY key1, key2) t1 GROUP BY key2, key1; B: SELECT a.key1, a.key2 FROM table1 a JOIN table2 b ON a.key1 = b.key; C: SELECT *, ROW_NUMBER() OVER(PARTITION BY key ORDER BY value1, value 2) id1, ROW_NUMBER() OVER(PARTITION BY key ORDER BY value1) id2 FROM dual;

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2022-06-08

下面的SQL哪些有可能经过两个阶段（只经过一次shuffle）就计算出结果：（） A: SELECT key1, key2, SUM(value1) FROM (SELECT key1, key2, COUNT() FROM dual GROUP BY key1, key2) t1 GROUP BY key2, key1; B: SELECT a.key1, a.key2 FROM table1 a JOIN table2 b ON a.key1 = b.key; C: SELECT , ROW_NUMBER() OVER(PARTITION BY key ORDER BY value1, value 2) id1, ROW_NUMBER() OVER(PARTITION BY key ORDER BY value1) id2 FROM dual;

下面的SQL哪些有可能经过两个阶段（只经过一次shuffle）就计算出结果：（）
A: SELECT key1, key2, SUM(value1) FROM (SELECT key1, key2, COUNT(*) FROM dual GROUP BY key1, key2) t1 GROUP BY key2, key1;
B: SELECT a.key1, a.key2 FROM table1 a JOIN table2 b ON a.key1 = b.key;
C: SELECT *, ROW_NUMBER() OVER(PARTITION BY key ORDER BY value1, value 2) id1, ROW_NUMBER() OVER(PARTITION BY key ORDER BY value1) id2 FROM dual;

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