sqLite 实例

insert into payIncome(time,inout,money,type) values("2014-05-12","支出",25.5,"衣服");

insert into payIncome(time,inout,money,type) values("2014-05-12","支出",35.5,"饮食");

insert into payIncome(time,inout,money,type) values("2014-05-12","收入",135.5,"打工");

insert into payIncome(time,inout,money,type) values("2014-05-12","收入",245.5,"工资");

insert into payIncome(time,inout,money,type) values("2013-05-12","收入",3245.5,"奖金");

insert into payIncome(time,inout,money,type) values("2014-04-12","支出",500,"学费");

.mode column

.mode list

.mode line

select * from payIncome;

ID time inout money type content

---------- ---------- ---------- ---------- ---------- ----------

1 2014-05-12 支出 25.5 衣服

2 2014-05-12 支出 35.5 饮食

3 2014-05-12 收入 135.5 打工

4 2014-05-12 收入 245.5 工资

5 2013-05-12 收入 3245.5 奖金

6 2014-04-12 支出 500 学费

select inout,type,sum(money) as money frompayIncome where time like "2014-05%"

group by inout,type having sum(money) >100 order by sum(money) desc;

inout type money

---------- ---------- ----------

收入 工资 245.5

收入 打工 135.5

select inout,type,sum(money) as money frompayIncome group by
inout,type having

sum(money) > 100 order by sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

支出 学费 500

收入 工资 245.5

收入 打工 135.5

select inout,type,sum(money) as money frompayIncome group by
type,inout having

sum(money) > 100 order by sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

支出 学费 500

收入 工资 245.5

收入 打工 135.5

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

支出 学费 500

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 衣服 25.5

sqlite> select inout,type,sum(money) asmoney from payIncome group by type,inout

order by sum(money) desc,inout;

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc,inout;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

支出 学费 500

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 衣服 25.5

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc,inout desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

支出 学费 500

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 衣服 25.5

delete from payIncome where money=500 andtype="学费" andinout="收入";

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

支出 学费 500

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 衣服 25.5

delete from payIncome where money=500 andtype="学费";

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 衣服 25.5

update payIncome set type="理发" where money=25.5 andinout="收入";

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 衣服 25.5

update payIncome set type="理发" where money=25.5 andinout="支出";

select inout,type,sum(money) as money frompayIncome group by type,inout order b

y sum(money) desc;

inout type money

---------- ---------- ----------

收入 奖金 3245.5

收入 工资 245.5

收入 打工 135.5

支出 饮食 35.5

支出 理发 25.5

create table user(id integer primary keyautoincrement,name varchar(10),pwd varc

har(10));

sqlite> .tables

.tables

android_metadata payIncome user incomeItem payItem

sqlite> drop table user;

drop table user;

sqlite> .tables

.tables

android_metadata incomeItem payIncome payItem

sqlite>