环境
我的PostgreSQL（9.2）架构如下所示：
CREATE TABLE first
(
id_first bigint NOT NULL,
first_date timestamp without time zone NOT NULL,
CONSTRAINT first_pkey PRIMARY KEY (id_first)
)
WITH (
OIDS=FALSE
);
CREATE INDEX first_first_date_idx
ON first
USING btree
(first_date);
CREATE TABLE second
(
id_second bigint NOT NULL,
id_first bigint NOT NULL,
CONSTRAINT second_pkey PRIMARY KEY (id_second),
CONSTRAINT fk_first FOREIGN KEY (id_first)
REFERENCES first (id_first) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
WITH (
OIDS=FALSE
);
CREATE INDEX second_id_first_idx
ON second
USING btree
(id_first);
CREATE TABLE third
(
id_third bigint NOT NULL,
id_second bigint NOT NULL,
CONSTRAINT third_pkey PRIMARY KEY (id_third),
CONSTRAINT fk_second FOREIGN KEY (id_second)
REFERENCES second (id_second) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
WITH (
OIDS=FALSE
);
CREATE INDEX third_id_second_idx
ON third
USING btree
(id_second);
所以，我有3张自带PK的桌子。 First的索引位于first_date，Second的FK来自First及其上的索引。 Third作为Second的FK及其上的索引：
First (0 --> n) Second (0 --> n) Third
First表包含约10 000 000条记录。
Second表包含约20 000 000条记录。
Third表包含约18 000 000条记录。
第first_date栏中的日期范围是2016-01-01至今。
random_cost_page设置为2.0。
default_statistics_target设置为100。
所有FK，PK和first_date STATISTICS都设置为5000
要执行的任务
我想计算与Third相关联的所有First行，其中first_date < X
我的查询：
SELECT count(t.id_third) AS count
FROM first f
JOIN second s ON s.id_first = f.id_first
JOIN third t ON t.id_second = s.id_second
WHERE first_date < _my_date
问题描述
_my_date = '2016-01-03' 一切运作良好。查询持续1-2秒。
EXPLAIN ANALYZE：
"Aggregate (cost=8585512.55..8585512.56 rows=1 width=8) (actual time=67.310..67.310 rows=1 loops=1)"
" -> Merge Join (cost=4208477.49..8583088.04 rows=969805 width=8) (actual time=44.277..65.948 rows=17631 loops=1)"
" Merge Cond: (s.id_second = t.id_second)"
" -> Sort (cost=4208477.48..4211121.75 rows=1057709 width=8) (actual time=44.263..46.035 rows=19230 loops=1)"
" Sort Key: s.id_second"
" Sort Method: quicksort Memory: 1670kB"
" -> Nested Loop (cost=0.01..4092310.41 rows=1057709 width=8) (actual time=6.169..39.183 rows=19230 loops=1)"
" -> Index Scan using first_first_date_idx on first f (cost=0.01..483786.81 rows=492376 width=8) (actual time=6.159..12.223 rows=10346 loops=1)"
" Index Cond: (first_date < '2016-01-03 00:00:00'::timestamp without time zone)"
" -> Index Scan using second_id_first_idx on second s (cost=0.00..7.26 rows=7 width=16) (actual time=0.002..0.002 rows=2 loops=10346)"
" Index Cond: (id_first = f.id_first)"
" -> Index Scan using third_id_second_idx on third t (cost=0.00..4316649.89 rows=17193788 width=16) (actual time=0.008..7.293 rows=17632 loops=1)"
"Total runtime: 67.369 ms"
_my_date = '2016-01-11'或更多查询不再使用indexscan - 由seqscan替换，最后3-4分钟...
查询计划：
"Aggregate (cost=8731468.75..8731468.76 rows=1 width=8) (actual time=234411.229..234411.229 rows=1 loops=1)"
" -> Hash Join (cost=4352424.81..8728697.88 rows=1108348 width=8) (actual time=189670.068..234400.540 rows=138246 loops=1)"
" Hash Cond: (t.id_second = o.id_second)"
" -> Seq Scan on third t (cost=0.00..4128080.88 rows=17193788 width=16) (actual time=0.016..124111.453 rows=17570724 loops=1)"
" -> Hash (cost=4332592.69..4332592.69 rows=1208810 width=8) (actual time=98566.740..98566.740 rows=151263 loops=1)"
" Buckets: 16384 Batches: 16 Memory Usage: 378kB"
" -> Hash Join (cost=561918.25..4332592.69 rows=1208810 width=8) (actual time=6535.801..98535.915 rows=151263 loops=1)"
" Hash Cond: (s.id_first = f.id_first)"
" -> Seq Scan on second s (cost=0.00..3432617.48 rows=18752248 width=16) (actual time=6090.771..88891.691 rows=19132869 loops=1)"
" -> Hash (cost=552685.31..552685.31 rows=562715 width=8) (actual time=444.630..444.630 rows=81650 loops=1)"
" -> Index Scan using first_first_date_idx on first f (cost=0.01..552685.31 rows=562715 width=8) (actual time=7.987..421.087 rows=81650 loops=1)"
" Index Cond: (first_date < '2016-01-13 00:00:00'::timestamp without time zone)"
"Total runtime: 234411.303 ms"
出于测试目的，我设置了：
SET enable_seqscan = OFF;
我的查询再次开始使用indexscan并持续1-10秒（取决于范围）。
问题
为什么这样做？如何说服查询规划师使用indexscan？
修改
将random_page_cost缩减为1.1后，我现在可以使用indexscan选择约30天。查询计划有所改变：
"Aggregate (cost=8071389.47..8071389.48 rows=1 width=8) (actual time=4915.196..4915.196 rows=1 loops=1)"
" -> Nested Loop (cost=0.01..8067832.28 rows=1422878 width=8) (actual time=14.402..4866.937 rows=399184 loops=1)"
" -> Nested Loop (cost=0.01..3492321.55 rows=1551849 width=8) (actual time=14.393..3012.617 rows=436794 loops=1)"
" -> Index Scan using first_first_date_idx on first f (cost=0.01..432541.99 rows=722404 width=8) (actual time=14.372..729.233 rows=236007 loops=1)"
" Index Cond: (first_date < '2016-02-01 00:00:00'::timestamp without time zone)"
" -> Index Scan using second_id_first_idx on second s (cost=0.00..4.17 rows=7 width=16) (actual time=0.008..0.009 rows=2 loops=236007)"
" Index Cond: (second = f.id_second)"
" -> Index Scan using third_id_second_idx on third t (cost=0.00..2.94 rows=1 width=16) (actual time=0.004..0.004 rows=1 loops=436794)"
" Index Cond: (id_second = s.id_second)"
"Total runtime: 4915.254 ms"
然而，我仍然不明白为什么要求更多的seqscan ...
有趣的是，当我要求范围超出某种限制时，我会得到这样的查询计划（此处选择40天 - 要求更多会再次生成seqscan）：
"Aggregate (cost=8403399.27..8403399.28 rows=1 width=8) (actual time=138303.216..138303.217 rows=1 loops=1)"
" -> Hash Join (cost=3887619.07..8399467.63 rows=1572656 width=8) (actual time=44056.443..138261.203 rows=512062 loops=1)"
" Hash Cond: (t.id_second = s.id_second)"
" -> Seq Scan on third t (cost=0.00..4128080.88 rows=17193788 width=16) (actual time=0.004..119497.056 rows=17570724 loops=1)"
" -> Hash (cost=3859478.04..3859478.04 rows=1715203 width=8) (actual time=5695.077..5695.077 rows=560503 loops=1)"
" Buckets: 16384 Batches: 16 Memory Usage: 1390kB"
" -> Nested Loop (cost=0.01..3859478.04 rows=1715203 width=8) (actual time=65.250..5533.413 rows=560503 loops=1)"
" -> Index Scan using first_first_date_idx on first f (cost=0.01..477985.28 rows=798447 width=8) (actual time=64.927..1688.341 rows=302663 loops=1)"
" Index Cond: (first_date < '2016-02-11 00:00:00'::timestamp without time zone)"
" -> Index Scan using second_id_first_idx on second s (cost=0.00..4.17 rows=7 width=16) (actual time=0.010..0.012 rows=2 loops=302663)"
" Index Cond: (id_first = f.id_first)"
"Total runtime: 138303.306 ms"
Laurenz Able建议后的更新
在重写查询计划后，Laurenz Able建议：
"Aggregate (cost=9102321.05..9102321.06 rows=1 width=8) (actual time=15237.830..15237.830 rows=1 loops=1)"
" -> Merge Join (cost=4578171.25..9097528.19 rows=1917143 width=8) (actual time=9111.694..15156.092 rows=803657 loops=1)"
" Merge Cond: (third.id_second = s.id_second)"
" -> Index Scan using third_id_second_idx on third (cost=0.00..4270478.19 rows=17193788 width=16) (actual time=23.650..5425.137 rows=803658 loops=1)"
" -> Materialize (cost=4577722.81..4588177.38 rows=2090914 width=8) (actual time=9088.030..9354.326 rows=879283 loops=1)"
" -> Sort (cost=4577722.81..4582950.09 rows=2090914 width=8) (actual time=9088.023..9238.426 rows=879283 loops=1)"
" Sort Key: s.id_second"
" Sort Method: external sort Disk: 15480kB"
" -> Merge Join (cost=673389.38..4341477.37 rows=2090914 width=8) (actual time=3662.239..8485.768 rows=879283 loops=1)"
" Merge Cond: (s.id_first = f.id_first)"
" -> Index Scan using second_id_first_idx on second s (cost=0.00..3587838.88 rows=18752248 width=16) (actual time=0.015..4204.308 rows=879284 loops=1)"
" -> Materialize (cost=672960.82..677827.55 rows=973345 width=8) (actual time=3662.216..3855.667 rows=892988 loops=1)"
" -> Sort (cost=672960.82..675394.19 rows=973345 width=8) (actual time=3662.213..3745.975 rows=476519 loops=1)"
" Sort Key: f.id_first"
" Sort Method: external sort Disk: 8400kB"
" -> Index Scan using first_first_date_idx on first f (cost=0.01..568352.90 rows=973345 width=8) (actual time=126.386..3233.134 rows=476519 loops=1)"
" Index Cond: (first_date < '2016-03-03 00:00:00'::timestamp without time zone)"
"Total runtime: 15244.404 ms"
答案 0 :(得分：3)
首先，看起来有些估计值已经过去了
尝试ANALYZE表，看看是否会更改所选的查询计划。
将random_page_cost降低到略高于1的值可能会有所帮助，看看是否能改善计划。
值得注意的是，快速查询中third_id_second_idx上的索引扫描仅产生17632行而不是超过1700万行，我只能通过假设从该行开始，{{1}的值来解释。 1}}不再匹配id_second和first的联接中的任何行，即合并连接在此之后完成。
您可以尝试使用重写的查询来利用它。尝试
second而不是
JOIN (SELECT id_second, id_third FROM third ORDER BY id_second) t
这可能会导致一个更好的计划，因为PostgreSQL不会优化JOIN third t
，并且计划者可能会决定，因为它必须排序ORDER BY，所以使用合并可能更便宜加入。这样你就会欺骗计划者选择一个它不会被认为理想的计划。通过不同的价值分配，规划者的原始选择可能会更好。