过滤视图时如何利用基础索引？

Question

我有一个非常简单的查询，执行速度非常慢，原因是它在执行 JOIN 之后对视图进行扫描。看到这一点我有点惊讶，因为我希望 Postgres 在加入之前执行过滤，看到其中一个基础表在要过滤的列上有一个 INDEX。

有没有办法我可以以某种方式重新排序查询或提示计划者如何以不同的方式进行查询？

请注意，我确实知道如何通过直接访问基础表来解决问题，但视图隐藏了一些复杂性，最好避免查询。 p>

查询

select * from form where encounter_id= 23728 and type = 'vitals';

解释分析

Subquery Scan on form  (cost=0.57..3439.07 rows=1 width=622) (actual time=8.187..8.187 rows=0 loops=1)
  Filter: ((form.encounter_id = 23728) AND (form.type = 'vitals'::text))
  Rows Removed by Filter: 12000
  ->  Unique  (cost=0.57..3259.07 rows=12000 width=626) (actual time=0.008..7.612 rows=12000 loops=1)
        ->  Merge Join  (cost=0.57..3229.07 rows=12000 width=626) (actual time=0.007..5.485 rows=12000 loops=1)
              Merge Cond: (fd.form_id = f.id)
              ->  Index Scan using _idx_form_details on _form_details fd  (cost=0.29..2636.78 rows=12000 width=603) (actual time=0.003..1.918 rows=12000 loops=1)
              ->  Index Scan using pk_form on _form f  (cost=0.29..412.29 rows=12000 width=27) (actual time=0.002..1.214 rows=12000 loops=1)
Planning time: 0.170 ms
Execution time: 8.212 ms

TABLE 和 VIEW 定义

CREATE TABLE _form (
  id INT NOT NULL,
  encounter_id INT REFERENCES _encounter (id)            NOT NULL,
  type         TEXT                                      NOT NULL,
  CONSTRAINT pk_form PRIMARY KEY (id),
  FOREIGN KEY (cid) REFERENCES _user_in_role (id)
);

CREATE INDEX encounter_id ON _form (encounter_id, type);

CREATE TABLE _form_details (
  id INT NOT NULL,
  form_id   INT REFERENCES _form (id) NOT NULL,
  archived  BOOLEAN                   NOT NULL DEFAULT FALSE,
  CONSTRAINT pk_form_details PRIMARY KEY (id),
  FOREIGN KEY (cid) REFERENCES _user_in_role (id)
);

CREATE VIEW form AS
  SELECT DISTINCT ON (f.id)
    f.id,
    f.encounter_id,
    f.type,
    fd.archived,
    f.cid
  FROM _form f 
       JOIN _form_details fd 
         ON (f.id = fd.form_id)
  ORDER BY f.id, fd.id DESC;

编辑： 有人发布了一个答案（随后被删除），其中包含一个重要的信息：即使基础表中的encounter_id 列已被索引，VIEW 中的ORDER BY 操作也无法实现其目的。不幸的是，我们无法摆脱ORDER BY，因为DISTINCT ON 需要它才能工作。

Answer 1

• DISTINCT ON ... ORDER BY 是性能杀手（不能分解子查询）
• form_id INT REFERENCES _form (id) FK 缺少索引
• aNOT EXISTS() antijoin 或row_number() 可用于避免DISTINCT 子查询

---

SET search_path=tmp;
/***/
\i tmp.sql

CREATE TABLE tform (
  id INT NOT NULL
  , encounter_id INT NOT NULL -- REFERENCES tencounter (id)
  , ztype         TEXT                                      NOT NULL
  , CONSTRAINT pk_form PRIMARY KEY (id)
  -- FOREIGN KEY (cid) REFERENCES _user_in_role (id)
);

CREATE TABLE tform_details (
  id INT NOT NULL
  , form_id   INT REFERENCES tform (id) NOT NULL
  , archived  BOOLEAN                   NOT NULL DEFAULT FALSE
  , CONSTRAINT pk_form_details PRIMARY KEY (id)
  -- , FOREIGN KEY (cid) REFERENCES _user_in_role (id)
);

-- ALTER TABLE tform ADD FOREIGN KEY(encounter_id) REFERENCES tencounter (id) ;
CREATE INDEX encounter_id ON tform (encounter_id, ztype);

INSERT INTO tform (id, encounter_id, ztype)
SELECT gs, 23720+gs%29, 'ztype_' || gs::text
FROM generate_series(1,10000) gs
        ;

INSERT INTO tform_details (id, form_id, archived)
SELECT 10000*gs+tf.id, tf.id, (random() > 0.3) ::boolean
FROM tform tf
CROSS JOIN generate_series(0,22) gs
        ;


UPDATE tform
SET ztype = 'vitals'
WHERE random() < 0.2;

/***/
DROP INDEX xxxx ;
CREATE UNIQUE INDEX xxxx ON tform_details (form_id, id);

VACUUM ANALYZE tform;
VACUUM ANALYZE tform_details;

\d tform;
\d tform_details;

select COUNT(*) FROM tform;
select COUNT(*) FROM tform_details;

DROP VIEW form ;
CREATE VIEW form AS
  SELECT DISTINCT ON (f.id)
    f.id
    , f.encounter_id
    , f.ztype
    , fd.archived
    -- , f.cid
  FROM tform f
       JOIN tform_details fd ON f.id = fd.form_id
  ORDER BY f.id, fd.id DESC
        ;

DROP VIEW form2 ;
CREATE VIEW form2 AS
  SELECT f.id
    , f.encounter_id
    , f.ztype
    , fd.archived
  FROM tform f
  JOIN tform_details fd
    ON f.id = fd.form_id
  WHERE NOT EXISTS ( SELECT *
        FROM tform_details nx
        WHERE nx.form_id = fd.form_id
        AND nx.id > fd.id
        )
        ;

DROP VIEW form3 ;
CREATE VIEW form3 AS
  SELECT f.id
    , f.encounter_id
    , f.ztype
    , fd.archived
  FROM tform f
  JOIN ( select xx.form_id, xx.archived
        , row_number() OVER (PARTITION BY xx.form_id ORDER BY xx.id DESC) AS rn
         FROM tform_details xx
        ) fd ON f.id = fd.form_id AND fd.rn = 1
        ;

\echo burn-in
EXPLAIN ANALYZE
select * from form where encounter_id= 23728 and ztype = 'vitals' ;

\echo plain
EXPLAIN ANALYZE
select * from form where encounter_id= 23728 and ztype = 'vitals' ;

EXPLAIN ANALYZE
select * from form2 where encounter_id= 23728 and ztype = 'vitals' ;

EXPLAIN ANALYZE
select * from form3 where encounter_id= 23728 and ztype = 'vitals' ;

\echo no_hash
SET enable_hashjoin = False;

EXPLAIN ANALYZE
select * from form where encounter_id= 23728 and ztype = 'vitals' ;

EXPLAIN ANALYZE
select * from form2 where encounter_id= 23728 and ztype = 'vitals' ;

EXPLAIN ANALYZE
select * from form3 where encounter_id= 23728 and ztype = 'vitals' ;

Answer 2

@a_horse_with_no_name 在聊天中给了我迄今为止最快的解决方案，但从未提供过答案。所以作为参考，这是他的解决方案，使用横向连接来创建视图。

CREATE VIEW form AS
    SELECT f.id, 
        f.encounter_id, 
        f.type, 
        fd.archived, 
        f.cid 
    FROM _form f 
    JOIN LATERAL ( 
        SELECT form_id, archived 
        FROM _form_details _fd 
        WHERE _fd.form_id = f.id 
        ORDER BY _fd.id DESC 
        LIMIT 1 
    ) AS fd ON TRUE; 

这比任何其他解决方案快约 10 倍。如果创建为form4，基于与@wildplasser 相同的表，它的执行方式如下：

select * from form where encounter_id= 23728 and ztype = 'vitals' ;
Time: 181.065 ms
select * from form2 where encounter_id= 23728 and ztype = 'vitals' ;
Time: 12.395 ms
select * from form3 where encounter_id= 23728 and ztype = 'vitals' ;
Time: 122.305 ms
select * from form4 where encounter_id= 23728 and ztype = 'vitals' ;
Time: 1.305 ms

Postgres 9.3 中引入的关于横向连接的一些很好的指导：

• https://medium.com/kkempin/postgresqls-lateral-join-bfd6bd0199df
• https://heapanalytics.com/blog/engineering/postgresqls-powerful-new-join-type-lateral
• What is the difference between LATERAL and a subquery in PostgreSQL?