表的創建
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語句還可以復合
3。
select name,birthday, case when birthday>'1983' then 'yong' when name='lee' then 'handsome' else 'just so so ' end from lee;
在這里用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為例
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.#對罰款登記分為三類，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。重點看這里的解決方法 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';