1.数学函数

有如下表和数组

把num值处于[20,29]之间,改为20

num值处于[30,39]之间的,改为30

UPDATE mian SET num = FLOOR(num/10)*10 WHERE num BETWEEN 20 AND 39;

2.字符串

把good表中商品名为'诺基亚xxxx'的商品,改为'HTCxxxx',

提示:大胆的把列看成变量,参与运算,甚至调用函数来处理 .

substring(),concat()

SELECT goods_id,goods_name,CONCAT('htc',SUBSTRING(goods_name,4)) FROM goods WHERE goods_name LIKE'诺基亚%';

3.where-having-group综合练习题

要求:查询出2门及2门以上不及格者的平均成绩

SELECT * FROMresult;SELECT *,AVG(score) AS pjf FROM result GROUP BY NAME; --显示所有人的平均分

SELECT *,score < 60 FROM result; --比较运算

SELECT NAME,SUM(score < 60) AS gk,AVG(score) AS pj FROM result GROUP BY NAME; --统计每个人的挂科数

--最终的查询语句

SELECT NAME,SUM(score < 60) AS gk,AVG(score) AS pj FROM result GROUP BY NAME HAVING gk>=2;

4.连接查询

根据给出的表结构按要求写出SQL语句。

Match 赛程表

字段名称

字段类型

描述

matchID

int

主键

hostTeamID

int

主队的ID

guestTeamID

int

客队的ID

matchResult

varchar(20)

比赛结果，如(2:0)

matchTime

date

比赛开始时间

Team 参赛队伍表

字段名称

字段类型

描述

teamID

int

主键

teamName

varchar(20)

队伍名称

Match的hostTeamID与guestTeamID都与Team中的teamID关联

查出 2006-6-1 到2006-7-1之间举行的所有比赛，并且用以下形式列出：

拜仁 2：0 不来梅 2006-6-21

mysql> select * from m;

+-----+------+------+------+------------+

| mid | hid | gid | mres | matime |

+-----+------+------+------+------------+

| 1 | 1 | 2 | 2:0 | 2006-05-21 |

| 2 | 2 | 3 | 1:2 | 2006-06-21 |

| 3 | 3 | 1 | 2:5 | 2006-06-25 |

| 4 | 2 | 1 | 3:2 | 2006-07-21 |

+-----+------+------+------+------------+

4 rows in set (0.00 sec)

mysql> select * from t;

+------+----------+

| tid | tname |

+------+----------+

| 1 | 国安 |

| 2 | 申花 |

| 3 | 公益联队 |

SELECT m.mid,t1.`tname` AS hname,m.`mres`,t2.`tname` AS gname, m.`matime` FROM m INNER JOIN t AS t1 ON m.`hid` =t1.`tid`INNER JOIN t AS t2 ON m.`gid` = t2.`tid` WHERE m.`matime` BETWEEN '2006-06-01' AND '2006-07-01';

5.子查询和合并

A表:

+------+------+

| id | num |

+------+------+

| a | 5 |

| b | 10 |

| c | 15 |

| d | 10 |

+------+------+

B表:

+------+------+

| id | num |

+------+------+

| b | 5 |

| c | 15 |

| d | 20 |

| e | 99 |

+------+------+

要求查询出以下效果:

+------+----------+

| id | num |

+------+----------+

| a | 5 |

| b | 15 |

| c | 30 |

| d | 30 |

| e | 99 |

+------+----------+

SELECT id,SUM(num) FROM (SELECT * FROM a UNION ALL SELECT * FROM b) AS temp GROUP BY id;