CREATE TABLE `lee` (
`id` int(10) NOT NULL AUTO_INCREMENT,
`name` char(20) DEFAULT NULL,
`birthday` datetime DEFAULT NULL,
PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8
數據插入：
insert into lee(name,birthday) values ('sam','1990-01-01');
insert into lee(name,birthday) values ('lee','1980-01-01');
insert into lee(name,birthday) values ('john','1985-01-01');
使用case when語句
1。
select name,
case
when birthday<'1981' then 'old'
when birthday>'1988' then 'yong'
else 'ok' END YORN
from lee;
2。
select NAME,
case name
when 'sam' then 'yong'
when 'lee' then 'handsome'
else 'good' end
from lee;
當然了case when語句還可以複合
3。
select name,birthday,
case
when birthday>'1983' then 'yong'
when name='lee' then 'handsome'
else 'just so so ' end
from lee;
在這裏用sql語句進行日期比較的話，需要對年加引號。要不然可能結果可能和預期的結果會不同。我的mysql版本5.1
當然也可以用year函數來實現，以第一個sql爲例
select NAME,
CASE
when year(birthday)>1988 then 'yong'
when year(birthday)<1980 then 'old'
else 'ok' END
from lee;
create table penalties
(
paymentno INTEGER not NULL,
payment_date DATE not null,
amount DECIMAL(7,2) not null,
primary key(paymentno)
)
insert into penalties values(1,'2008-01-01',3.45);
insert into penalties values(2,'2009-01-01',50.45);
insert into penalties values(3,'2008-07-01',80.45);
1.#對罰款登記分爲三類，第一類low，包括大於0小於等於40的罰款，第二類moderate大於40
#到80之間的罰款，第三類high包含所有大於80的罰款。
2.#統計出屬於low的罰款編號。
第一道題的解法與上面的相同
select paymentno,amount,
case
when amount>0 and amount<=40 then 'low'
when amount>40 and amount<=80 then 'moderate'
when amount>80 then 'high'
else 'incorrect' end lvl
from `penalties`
2.#統計出屬於low的罰款編號。重點看這裏的解決方法
方法1.
select paymentno,amount
from `penalties`
where case
when amount>0 and amount<=40 then 'low'
when amount>40 and amount<=80 then 'moderate'
when amount>80 then 'high'
else 'incorrect' end ='low';
方法2
select *
from (select paymentno,amount,
case
when amount>0 and amount<=40 then 'low'
when amount>40 and amount<=80 then 'moderate'
when amount>80 then 'high'
else 'incorrect' end lvl
from `penalties`) as p
where p.lvl='low';