mysql面试题练习01

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 * FROM result;

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;