OM COMPANY WHERE ID IN (SELECT ID FROM COMPANY WHERE SALARY > 45000);
INSERT语句中的子查询使用，基本语法：
INSERT INTO table_name [ (column1 [, column2 ]) ]
SELECT [ *|column1 [, column2 ]
FROM table1 [, table2 ]
[ WHERE VALUE OPERATOR ]
实例如下：
sqlite> INSERT INTO COMPANY_BKP
SELECT * FROM COMPANY
WHERE ID IN (SELECT ID
FROM COMPANY) ;
UPDATE语句中的子查询使用，基本语法如下：
UPDATE table
SET column_name = new_value
[ WHERE OPERATOR [ VALUE ]
(SELECT COLUMN_NAME
FROM TABLE_NAME)
[ WHERE) ]
实例如下：
sqlite> UPDATE COMPANY
SET SALARY = SALARY * 0.50
WHERE AGE IN (SELECT AGE FROM COMPANY_BKP
WHERE AGE >= 27 );
DELETE语句中的子查询使用，语法如下：
DELETE FROM TABLE_NAME
[ WHERE OPERATOR [ VALUE ]
(SELECT COLUMN_NAME
FROM TABLE_NAME)
[ WHERE) ]
实例如下：
sqlite> DELETE FROM COMPANY
WHERE AGE IN (SELECT AGE FROM COMPANY_BKP
WHERE AGE > 27 );
8，EXPLAIN分析
没有建立索引之前，分析都是表扫描：
sqlite> EXPLAIN SELECT * FROM COMPANY WHERE Salary < 20000;
addr opcode p1 p2 p3 p4 p5 comment
---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
0 Trace 0 0 0 00
1 Integer 20000 1 0 00
2 Goto 0 16 0 00
3 OpenRead 0 2 0 5 00
4 Rewind 0 14 0 00
5 Column 0 4 2 00
6 Ge 1 13 2 collseq(BI 6b
7 Column 0 0 4 00
8 Column 0 1 5 00
9 Column 0 2 6 00
10 Column 0 3 7 00
11 Column 0 4 8 00
12 ResultRow 4 5 0 00
13 Next 0 5 0 01
14 Close 0 0 0 00
15 Halt 0 0 0 00
16 Transactio 0 0 0 00
17 VerifyCook 0 1 0 00
18 TableLock 0 2 0 COMPANY 00
19 Goto 0 3 0 00
sqlite> EXPLAIN QUERY PLAN SELECT * FROM COMPANY WHERE Salary < 20000;
order from detail
---------- ---------- -------------
0 0 TABLE COMPANY
sqlite>
建立索引，再进行EXPLAIN分析查看结果，走了idx_sal索引扫描：
sqlite> CREATE INDEX idx_sal ON COMPANY(SALARY);
sqlite> EXPLAIN SELECT * FROM COMPANY WHERE Salary < 20000;
addr opcode p1 p2 p3 p4 p5 comment
---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
0 Trace 0 0 0 00
1 Integer 20000 1 0 00
2 Goto 0 25 0 00
3 OpenRead 0 2 0 5 00
4 OpenRead 1 3 0 keyinfo(1, 00
5 Affinity 2 0 0 cb 00
6 Rewind 1 22 2 0 00
7 SCopy 1 2 0 00
8 IsNull 2 22 0 00
9 Affinity 2 1 0 cb 00
10 IdxGE 1 22 2 1 00
11 Column 1 0 3 00
12 IsNull 3 21 0 00
13 IdxRowid 1 3 0 00
14 Seek 0 3 0 00
15 Column 0 0 4 00
16 Column 0 1 5 00
17 Column 0 2 6 00
18 Column 0 3 7 00
19 Column 1 0 8 00
20 ResultRow 4 5 0 00
21 Next 1 10 0 00
22 Close 0 0 0 00
23 Close 1 0 0 00
24 Halt 0 0 0 00
25 Transactio 0 0 0 00
26 VerifyCook 0 2 0 00
27 TableLock 0 2 0 COMPANY 00
28 Goto 0 3 0 00
sqlite> EXPLAIN QUERY PLAN SELECT * FROM COMPANY WHERE Salary < 20000;
order from detail
---------- ---------- --------------------------------
0 0 TABLE COMPANY WITH INDEX idx_sal
sqlite>
9，删除重复数据并且保留最新一条记录
录入测试数据
sqlite> .dump
PRAGMA foreign_keys=OFF;
BEGIN TRANSACTION;
CREATE TABLE COMPANY(ID INT NOT NULL, NAME VARCHAR(20),AGE INT,ADDRESS VARCHAR(20),SALARY DECIMAL(7,2));
INSERT INTO "COMPANY" VALUES(2,'Allen',25,'Texas',15000);
INSERT INTO "COMPANY" VALUES(3,'Teddy',23,'Norway',20000);
INSERT INTO "COMPANY" VALUES(4,'Mark',25,'Rich-Mond',65000);
INSERT INTO "COMPANY" VALUES(5,'David',27,'Texas',85000);
INSERT INTO "COMPANY" VALUES(6,'Kim',22,'South-Hall',45000);
INSERT INTO "COMPANY" VALUES(7,'James',24,'Houston',10000);
INSERT INTO "COMPANY" VALUES(7,'James',28,'Houston',20000);
INSERT INTO "COMPANY" VALUES(4,'Mark',29,'Rich-Mond',95000);
COMMIT;
sqlite>?
查看重复记录数
sqlite> select * from company order by name;
ID NAME AGE ADDRESS SALARY
---------- ---------- ---------- ---------- ----------
2 Allen 25 Texas 15000
5 David 27 Texas 85000
7 James 24 Houston 10000
7 James 28 Houston 20000
6 Kim 22 South-Hall 4