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我有一个查询，运行时间太长。我正在使用 PostgreSQL 10.3。在我参与此查询的表中，每个表中有大约 350 万条记录。查询是：

SELECT thf.attr1, thf.attr2, thf.attr3, thf.attr4
FROM tb_one AS thf
INNER JOIN tb_two AS ths
ON ths.tb_hit_hitid = thf.tb_hit_hitid
WHERE ths.source IN ('source1', 'source2')

在这些表中，我有索引：

CREATE INDEX tb_two_idx_1 on tb_two (Source ASC, attr5 ASC);
CREATE INDEX tb_one_idx_1 on tb_one USING btree (attr1 ASC,attr2 ASC,attr3 ASC,attr4 ASC);
CREATE INDEX tb_one_idx_2 on tb_hit_feature (tb_hit_HitId ASC);
CREATE INDEX tb_two_idx_2 on tb_hit_source (tb_hit_HitId ASC);

这是查询计划（explain (analyse, buffers)）：

    QUERY PLAN                                                                             
-------------------------------------------------------------------------------------------------------------------------------------------------------------------
 Merge Join  (cost=3.85..287880.35 rows=1771004 width=44) (actual time=0.091..3894.024 rows=1726970 loops=1)
   Merge Cond: (thf.tb_hit_hitid = ths.tb_hit_hitid)
   Buffers: shared hit=354821
   ->  Index Scan using tb_one_idx_2 on tb_one thf  (cost=0.43..124322.43 rows=3230800 width=52) (actual time=0.014..655.036 rows=1726946 loops=1)
         Buffers: shared hit=27201
   ->  Index Scan using tb_two_idx_2 on tb_two ths  (cost=0.43..139531.97 rows=1771004 width=8) (actual time=0.069..1604.789 rows=1726973 loops=1)
         Filter: ((source)::text = ANY ('{source1,source2}'::text[]))
         Rows Removed by Filter: 1651946
         Buffers: shared hit=327620
 Planning time: 2.737 ms
 Execution time: 4117.573 ms
(11 rows)
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1 回答 1

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对于此查询：

SELECT thf.attr1, thf.attr2, thf.attr3, thf.attr4
FROM tb_one thf INNER JOIN
     tb_two ths
     ON ths.tb_hit_hitid = thf.tb_hit_hitid
WHERE ths.source IN ('source1', 'source2');

你想要索引tb_two(source, tb_hit_hitid)和tb_one(tb_hit_hitid)。这可能是最好的索引。

如果查询返回重复项（由于连接），我可能建议将其写为：

SELECT thf.attr1, thf.attr2, thf.attr3, thf.attr4
FROM tb_one thf
WHERE EXISTS (SELECT 1
              FROM tb_two ths
              WHERE ths.tb_hit_hitid = thf.tb_hit_hitid AND
                    ths.source IN ('source1', 'source2')
             );

对于此版本，您希望索引为tb_two(tb_hit_hitid, source).

于 2018-04-16T20:32:48.573 回答