@Clodoaldo的回答是正確的,因爲它允許你想要的能力,並應該返回正確的結果。不幸的是,它的執行速度很慢。
我建立了包含推文和用戶的實驗數據庫。每個用戶擁有100個推文(1M條推文記錄)。我將PK索引爲u.id,t.id,FK t.user_id和謂詞字段u.username,u.email。
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;
利用一個場作爲謂詞的簡單查詢是非常快:
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
中提出的,其中作爲@Clodoaldo使用情況下的查詢需要近30秒時:
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
觀察那個計劃,我認爲使用union subselect然後過濾其結果以獲得適合於參數化謂詞選擇的id將允許計劃人員爲每個參數使用特定的索引謂語。事實證明我是對的:
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
這回答了OP的問題很好,但我認爲這是有可能產生一個可怕的執行計劃作爲其中的情況下會抑制計劃者選擇的用戶一個很好的指標的能力。 – dbenhur 2012-03-18 00:29:54
@dbenhur是的,我想過,但選項是使用動態查詢,我不知道他是否想聽到。他似乎更關心代碼簡潔性。如果他沒有使用9.1格式化函數將做一個乾淨的動態查詢,它將是必要的連接字符串產生雜亂的代碼,我的印象正是他想避免的。 – 2012-03-18 00:59:00
我很好奇,並建立了一個實驗。事實上,這個案例確實嚴重傷害了這個計劃。查看我的備選答案進行分析,並查詢表單,該表單也可以快速生成,而無需執行動態語句生成。 – dbenhur 2012-03-18 01:01:33