Postgres按外键性能排序？答案

【问题标题】：Postgres order by foreign key performance?Postgres按外键性能排序？
【发布时间】：2011-11-07 21:08:18
【问题描述】：

我在 Postgres 中按外键排序时遇到了奇怪的 (?) 问题。这是第二个表和查询，使用 order by 比不使用要花费更长的时间。

EXPLAIN ANALYZE SELECT "spoleczniak_zdjecia"."id", "spoleczniak_zdjecia"."postac_id", "spoleczniak_zdjecia"."zdjecie", "spoleczniak_zdjecia"."opis", "spoleczniak_zdjecia"."data", "spoleczniak_zdjecia"."avatar", "spoleczniak_zdjecia"."tagi", "postac_postacie"."id", "postac_postacie"."user_id", "postac_postacie"."avatar", "postac_postacie"."ikonka", "postac_postacie"."imie", "postac_postacie"."nazwisko", "postac_postacie"."pseudonim", "postac_postacie"."plec", "postac_postacie"."wzrost", "postac_postacie"."waga", "postac_postacie"."ur_tydz", "postac_postacie"."ur_rok", "postac_postacie"."ur_miasto_id", "postac_postacie"."akt_miasto_id", "postac_postacie"."kasa", "postac_postacie"."punkty", "postac_postacie"."zmeczenie", "postac_postacie"."zdrowie", "postac_postacie"."kariera" FROM "spoleczniak_zdjecia" INNER JOIN "taggit_taggeditem" ON ("spoleczniak_zdjecia"."id" = "taggit_taggeditem"."object_id") INNER JOIN "taggit_tag" ON ("taggit_taggeditem"."tag_id" = "taggit_tag"."id") INNER JOIN "postac_postacie" ON ("spoleczniak_zdjecia"."postac_id" = "postac_postacie"."id") WHERE ("taggit_tag"."slug" = 'ja' AND "taggit_taggeditem"."content_type_id" = 922 ) ORDER BY "spoleczniak_zdjecia"."id" DESC LIMIT 28;
                                                                                QUERY PLAN
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 Limit  (cost=27.88..27.89 rows=7 width=198) (actual time=2984.689..2984.697 rows=28 loops=1)
   ->  Sort  (cost=27.88..27.89 rows=7 width=198) (actual time=2984.688..2984.692 rows=28 loops=1)
         Sort Key: spoleczniak_zdjecia.id
         Sort Method: top-N heapsort  Memory: 32kB
         ->  Nested Loop  (cost=2.31..27.78 rows=7 width=198) (actual time=1.063..2974.901 rows=9091 loops=1)
               ->  Nested Loop  (cost=2.31..22.02 rows=7 width=109) (actual time=1.057..2899.010 rows=9091 loops=1)
                     ->  Nested Loop  (cost=2.31..19.92 rows=7 width=4) (actual time=1.046..2848.853 rows=9103 loops=1)
                           ->  Index Scan using taggit_tag_slug on taggit_tag  (cost=0.00..4.27 rows=1 width=4) (actual time=0.025..0.027 rows=1 loops=1)
                                 Index Cond: ((slug)::text = 'ja'::text)
                           ->  Bitmap Heap Scan on taggit_taggeditem  (cost=2.31..15.56 rows=7 width=8) (actual time=1.019..2847.244 rows=9103 loops=1)
                                 Recheck Cond: (tag_id = taggit_tag.id)
                                 Filter: (content_type_id = 922)
                                 ->  Bitmap Index Scan on taggit_taggeditem_tag_id  (cost=0.00..2.31 rows=7 width=0) (actual time=0.954..0.954 rows=9103 loops=1)
                                       Index Cond: (tag_id = taggit_tag.id)
                     ->  Index Scan using spoleczniak_zdjecia_pkey on spoleczniak_zdjecia  (cost=0.00..0.29 rows=1 width=109) (actual time=0.005..0.005 rows=1 loops=9103)
                           Index Cond: (id = taggit_taggeditem.object_id)
               ->  Index Scan using postac_postacie_pkey on postac_postacie  (cost=0.00..0.81 rows=1 width=89) (actual time=0.007..0.007 rows=1 loops=9091)
                     Index Cond: (id = spoleczniak_zdjecia.postac_id)
 Total runtime: 2984.760 ms

这里没有订单：

EXPLAIN ANALYZE SELECT "spoleczniak_zdjecia"."id", "spoleczniak_zdjecia"."postac_id", "spoleczniak_zdjecia"."zdjecie", "spoleczniak_zdjecia"."opis", "spoleczniak_zdjecia"."data", "spoleczniak_zdjecia"."avatar", "spoleczniak_zdjecia"."tagi", "postac_postacie"."id", "postac_postacie"."user_id", "postac_postacie"."avatar", "postac_postacie"."ikonka", "postac_postacie"."imie", "postac_postacie"."nazwisko", "postac_postacie"."pseudonim", "postac_postacie"."plec", "postac_postacie"."wzrost", "postac_postacie"."waga", "postac_postacie"."ur_tydz", "postac_postacie"."ur_rok", "postac_postacie"."ur_miasto_id", "postac_postacie"."akt_miasto_id", "postac_postacie"."kasa", "postac_postacie"."punkty", "postac_postacie"."zmeczenie", "postac_postacie"."zdrowie", "postac_postacie"."kariera" FROM "spoleczniak_zdjecia" INNER JOIN "taggit_taggeditem" ON ("spoleczniak_zdjecia"."id" = "taggit_taggeditem"."object_id") INNER JOIN "taggit_tag" ON ("taggit_taggeditem"."tag_id" = "taggit_tag"."id") INNER JOIN "postac_postacie" ON ("spoleczniak_zdjecia"."postac_id" = "postac_postacie"."id") WHERE ("taggit_tag"."slug" = 'ja' AND "taggit_taggeditem"."content_type_id" = 922 ) LIMIT 28;
                                                                            QUERY PLAN
-------------------------------------------------------------------------------------------------------------------------------------------------------------------
 Limit  (cost=2.31..27.78 rows=7 width=198) (actual time=1.113..1.482 rows=28 loops=1)
   ->  Nested Loop  (cost=2.31..27.78 rows=7 width=198) (actual time=1.112..1.477 rows=28 loops=1)
         ->  Nested Loop  (cost=2.31..22.02 rows=7 width=109) (actual time=1.102..1.292 rows=28 loops=1)
               ->  Nested Loop  (cost=2.31..19.92 rows=7 width=4) (actual time=1.092..1.145 rows=28 loops=1)
                     ->  Index Scan using taggit_tag_slug on taggit_tag  (cost=0.00..4.27 rows=1 width=4) (actual time=0.017..0.017 rows=1 loops=1)
                           Index Cond: ((slug)::text = 'ja'::text)
                     ->  Bitmap Heap Scan on taggit_taggeditem  (cost=2.31..15.56 rows=7 width=8) (actual time=1.072..1.118 rows=28 loops=1)
                           Recheck Cond: (tag_id = taggit_tag.id)
                           Filter: (content_type_id = 922)
                           ->  Bitmap Index Scan on taggit_taggeditem_tag_id  (cost=0.00..2.31 rows=7 width=0) (actual time=0.989..0.989 rows=9103 loops=1)
                                 Index Cond: (tag_id = taggit_tag.id)
               ->  Index Scan using spoleczniak_zdjecia_pkey on spoleczniak_zdjecia  (cost=0.00..0.29 rows=1 width=109) (actual time=0.004..0.005 rows=1 loops=28)
                     Index Cond: (id = taggit_taggeditem.object_id)
         ->  Index Scan using postac_postacie_pkey on postac_postacie  (cost=0.00..0.81 rows=1 width=89) (actual time=0.005..0.005 rows=1 loops=28)
               Index Cond: (id = spoleczniak_zdjecia.postac_id)
 Total runtime: 1.562 ms

什么会导致问题？是查询？配置？我应该检查任何特定的配置吗？在我的最后一个问题中，有更复杂的查询，但该查询根本不复杂。有什么建议吗？

顺便说一句。该查询由 Django 生成（准确地说是 django-taggit）。顺便说一句。第二部分，硬件一点也不差（i7、16 GB RAM、用于操作系统和数据的 RAID 10 3x2 + 用于 WAL 的 2 个 RAID1 磁盘、512 MB RAID 缓存 + BBU）

纯文本查询：

SELECT "spoleczniak_zdjecia"."id", "spoleczniak_zdjecia"."postac_id", "spoleczniak_zdjecia"."zdjecie", "spoleczniak_zdjecia"."opis", "spoleczniak_zdjecia"."data", "spoleczniak_zdjecia"."avatar" ", "spoleczniak_zdjecia"."tagi", "postac_postacie"."id", "postac_postacie"."user_id", "postac_postacie"."avatar", "postac_postacie"."ikonka", "postac_postacie"."imie", “postac_postacie”。“nazwisko”，“postac_postacie”。“pseudonim”，“postac_postacie”。“plec”，“postac_postacie”。“wzrost”，“postac_postacie”。“waga”，“postac_postacie”。“ur_tydz”，“postac_postacie” "."ur_rok", "postac_postacie"."ur_miasto_id", "postac_postacie"."akt_miasto_id", "postac_postacie"."kasa", "postac_postacie"."punkty", "postac_postacie"."zmeczenie", "postac_postacie"。 “zdrowie”，“postac_postacie”。“kariera” FROM “spoleczniak_zdjecia” INNER JOIN “taggit_taggeditem” ON（“spoleczniak_zdjecia”。“id” = “taggit_taggeditem”。“object_id”）INNER JOIN “taggit_tag” ON（“taggit_taggeditem”。” tag_id" = "taggit_tag"."id") INNER JOIN "postac_postaci e" ON ("spoleczniak_zdjecia"."postac_id" = "postac_postacie"."id") WHERE ("taggit_tag"."slug" = 'ja' AND "taggit_taggeditem"."content_type_id" = 922 ) 由 "spoleczniak_zdjecia" 订购。 "id" DESC 限制 28;

【问题讨论】：

能否再添加查询的明文？水平滚动不是我的专长 ;-)
将查询简化为骨架：SELECT "spoleczniak_zdjecia"."id" -- PK -- , "spoleczniak_zdjecia"."postac_id" -- FK FROM "spoleczniak_zdjecia" INNER JOIN "taggit_taggeditem" ON ( "spoleczniak_zdjecia"."id" = "taggit_taggeditem"."object_id") -- PK == PK INNER JOIN "taggit_tag" ON ("taggit_taggeditem"."tag_id" = "taggit_tag"."id") -- FK -> PK INNER JOIN "postac_postacie" ON ("spoleczniak_zdjecia"."postac_id" = "postac_postacie"."id") -- FK -> PK WHERE "taggit_tag"."slug" = 'ja' AND "taggit_taggeditem"."content_type_id" = 922 ORDER BY "spoleczniak_zdjecia"."id" DESC LIMIT 28;

标签： database django postgresql rdbms

【解决方案1】：

第二个返回你找到的前 28 条记录，不管顺序。

首先您必须对结果进行排序，然后返回 28 条第一条记录。

如果数据没有被修改，ORDER BY的查询每次都会返回相同的28条记录。

但是第二个查询可以在每次执行时返回 28 条不同的记录。结果无法保证。

【讨论】：

【解决方案2】：

区别就在 EXPLAIN 输出的第二行：

->  Sort  (cost=27.88..27.89 rows=7 width=198) (actual time=2984.688..2984.692 rows=28 loops=1)

请注意，“实际时间”几乎是查询的整个时间。排序不仅需要一堆比较（即排序任何东西的成本），还需要额外的数据管理，服务器需要将一些数据（行或指向行的指针）复制到临时位置，以便可以在不干扰其他任何内容的情况下对其进行排序.

除非您很幸运，并且您的排序与磁盘上的顺序匹配并且优化器可以注意到它们匹配，否则任何查询都将花费更长的时间进行排序。

【讨论】：