Question

我正在寻找抽象对postgres的数据库访问的方法。在我的例子中，我将在nodejs中使用假设的twitter克隆，但最后它是关于postgres如何处理预准备语句的问题，所以语言和库并不重要：

假设我希望能够通过用户名访问用户的所有推文列表：

name: "tweets by username"
text: "SELECT (SELECT * FROM tweets WHERE tweets.user_id = users.user_id) FROM users WHERE users.username = $1"
values: [username]

这样做很好，但是在实际条款和代码质量方面看起来效率低下，不得不通过电子邮件而不是用户名来处理获取推文的其他功能：

name: "tweets by email"
text: "SELECT (SELECT * FROM tweets WHERE tweets.user_id = users.user_id) FROM users WHERE users.email = $1"
values: [email]

是否可以将字段作为参数包含在预准备语句中？

name: "tweets by user"
text: "SELECT (SELECT * FROM tweets WHERE tweets.user_id = users.user_id) FROM users WHERE users.$1 = $2"
values: [field, value]

虽然在user_id访问推文的情况下这可能效率稍低，但这是我愿意提高代码质量的交易，并希望通过减少查询模板的数量来提高整体效率到1而不是3 +。

Answer 1

@Clodoaldo的回答是正确的，它允许你想要的功能，并应该返回正确的结果。不幸的是，它的执行速度相当慢。

我用推文和用户建立了一个实验数据库。填充10K用户，每个用户有100条推文（1M推文记录）。我将PKs u.id，t.id，FK t.user_id和谓词字段u.username，u.email编入索引。

create table t(id serial PRIMARY KEY, data integer, user_id bignit);
create index t1 t(user_id);
create table u(id serial PRIMARY KEY, name text, email text);
create index u1 on u(name);
create index u2 on u(email);
insert into u(name,email) select i::text, i::text from generate_series(1,10000) i;
insert into t(data,user_id) select i, (i/100)::bigint from generate_series(1,1000000) i;
analyze table t;
analyze table u;

使用一个字段作为谓词的简单查询非常快：

prepare qn as select t.* from t join u on t.user_id = u.id where u.name = $1;

explain analyze execute qn('1111');
 Nested Loop  (cost=0.00..19.81 rows=1 width=16) (actual time=0.030..0.057 rows=100 loops=1)
   ->  Index Scan using u1 on u  (cost=0.00..8.46 rows=1 width=4) (actual time=0.020..0.020 rows=1 loops=1)
         Index Cond: (name = $1)
   ->  Index Scan using t1 on t  (cost=0.00..10.10 rows=100 width=16) (actual time=0.007..0.023 rows=100 loops=1)
         Index Cond: (t.user_id = u.id)
 Total runtime: 0.093 ms

在@Clodoaldo建议的地方使用案例的查询花了将近30秒：

prepare qen as select t.* from t join u on t.user_id = u.id
  where case $2 when 'e' then u.email = $1 when 'n' then u.name = $1 end;

explain analyze execute qen('1111','n');
 Merge Join  (cost=25.61..38402.69 rows=500000 width=16) (actual time=27.771..26345.439 rows=100 loops=1)
   Merge Cond: (t.user_id = u.id)
   ->  Index Scan using t1 on t  (cost=0.00..30457.35 rows=1000000 width=16) (actual time=0.023..17.741 rows=111200 loops=1)
   ->  Index Scan using u_pkey on u  (cost=0.00..42257.36 rows=500000 width=4) (actual time=0.325..26317.384 rows=1 loops=1)
         Filter: CASE $2 WHEN 'e'::text THEN (u.email = $1) WHEN 'n'::text THEN (u.name = $1) ELSE NULL::boolean END
 Total runtime: 26345.535 ms

观察该计划，我认为使用union subselect然后过滤其结果以获得适合参数化谓词选择的id将允许规划器为每个谓词使用特定索引。事实证明我是对的：

prepare qen2 as 
select t.*
from t 
join (
 SELECT id from 
  (
  SELECT 'n' as fld, id from u where u.name = $1
  UNION ALL
  SELECT 'e' as fld, id from u where u.email = $1
  ) poly
 where poly.fld = $2
) uu
on t.user_id = uu.id;

explain analyze execute qen2('1111','n');
 Nested Loop  (cost=0.00..28.31 rows=100 width=16) (actual time=0.058..0.120 rows=100 loops=1)
   ->  Subquery Scan poly  (cost=0.00..16.96 rows=1 width=4) (actual time=0.041..0.073 rows=1 loops=1)
         Filter: (poly.fld = $2)
         ->  Append  (cost=0.00..16.94 rows=2 width=4) (actual time=0.038..0.070 rows=2 loops=1)
               ->  Subquery Scan "*SELECT* 1"  (cost=0.00..8.47 rows=1 width=4) (actual time=0.038..0.038 rows=1 loops=1)
                     ->  Index Scan using u1 on u  (cost=0.00..8.46 rows=1 width=4) (actual time=0.038..0.038 rows=1 loops=1)
                           Index Cond: (name = $1)
               ->  Subquery Scan "*SELECT* 2"  (cost=0.00..8.47 rows=1 width=4) (actual time=0.031..0.032 rows=1 loops=1)
                     ->  Index Scan using u2 on u  (cost=0.00..8.46 rows=1 width=4) (actual time=0.030..0.031 rows=1 loops=1)
                           Index Cond: (email = $1)
   ->  Index Scan using t1 on t  (cost=0.00..10.10 rows=100 width=16) (actual time=0.015..0.028 rows=100 loops=1)
         Index Cond: (t.user_id = poly.id)
 Total runtime: 0.170 ms

Answer 2

SELECT t.* 
FROM tweets t 
inner join users u on t.user_id = u.user_id
WHERE case $2
    when 'username' then u.username = $1
    when 'email' then u.email = $1
    else u.user_id = $1
    end

Postgres准备了不同领域的陈述

2 个答案: