1:创建数据库
$sqlite3 testDB.db
sqlite version 3.7.15.2 2013-01-09 11:53:05
Enter ".help" for instructions
Enter sql statements terminated with a ";"
sqlite>
2.查看数据库
sqlite>.databases
seq name file
--- --------------- ----------------------
0 main /home/sqlite/testDB.db
3.退出
sqlite>.quit
4.dump 命令
$sqlite3 testDB.db .dump > testDB.sql
$sqlite3 testDB.db < testDB.sql
5.创建表
CREATE TABLE testDB.COMPANY(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
6.查看表信息
sqlite>.tables
COMPANY
表的完整信息
sqlite>.schema COMPANY
7.删除表
sqlite>DROP TABLE COMPANY;
8.插入查询
INSERT INTO TABLE_NAME (column1,column2,column3,...columnN)]
VALUES (value1,value2,value3,...valueN);
VALUES (value1,value2,value3,...valueN);
表已经存在直接插入即可
INSERT INTO TABLE_NAME VALUES (value1,value2value3valueN);
或者拷贝其他的表
INSERT INTO first_table_name [(column1,... columnN)]
SELECT column1,...columnN
FROM second_table_name
[WHERE condition];
SELECT column1,...columnN
FROM second_table_name
[WHERE condition];
9.select查询
10.运算符（算数／比较／逻辑／位／）
放在where中的条件
SELECT * FROM COMPANY WHERE SALARY > 50;
11.更新
sqlite> update company set address = 'LISTRON' WHERE ID = '7';
12.删除
sqlite> DELETE FROM COMPANY WHERE ID=7;
13.like子句
sqlite> SELECT * FROM COMPANY WHERE ADDRESS LIKE '%-%';
14.
太多了，没时间整了，以后继续。
