sqlite没有使用类似查询的索引

Question

我一直试图让sqlite使用索引,似乎无济于事.香港专业教育学院尝试整理nocase仍然没有运气.任何人都有任何关于如何让sqlite像命中索引一样的想法.提前致谢

DROP TABLE IF EXISTS "test";
DROP TABLE IF EXISTS "test2";
DROP TABLE IF EXISTS "test3";
create table test(name TEXT COLLATE NOCASE);
create table test2(name TEXT);
create table test3(name TEXT);
create index idx_test_name on test(name);
create index idx_test2_name on test2(name);
create index idx_test3_name on test3(name COLLATE NOCASE);
insert into test(name) values('dan');
insert into test2(name) values('dan');
insert into test3(name) values('dan');
--explain query plan select * from test where name like 'test%' 
-- explain query plan select * from test2 where name like 'test%' 
-- explain query plan select * from test3 where name like 'test%' 

Answer 1

从sqlite邮件列表引用(http://www.mail-archive.com/sqlite-users@sqlite.org/msg27760.html)

LIKE默认情况下不区分大小写.要让它使用您的索引,您需要使索引不区分大小写:

CREATE INDEX test_name ON test(名称COLLATE NOCASE);

或使LIKE区分大小写:

PRAGMA case_sensitive_like = 1;

Answer 2

test在 SQLite 3.6.23.1 中，使用on 索引：

> explain query plan select * from test where name like 'test%';
TABLE test WITH INDEX idx_test_name

> explain query plan select * from test2 where name like 'test%';
TABLE test2

> explain query plan select * from test3 where name like 'test%';
TABLE test3

在 SQLite 3.7.15 的开发版本中，同时使用test's 和's 索引（ on 的索引用于扫描，而不是搜索）：test3test2

> explain query plan select * from test where name like 'test%';
SEARCH TABLE test USING COVERING INDEX idx_test_name (name>? AND name<?) (~31250 rows)

> explain query plan select * from test2 where name like 'test%';
SCAN TABLE test2 USING COVERING INDEX idx_test2_name (~500000 rows)

> explain query plan select * from test3 where name like 'test%';
SEARCH TABLE test3 USING COVERING INDEX idx_test3_name (name>? AND name<?) (~31250 rows)

所以答案是更新 SQLite。