Mysql常见的面试总结

第一题

CREATE TABLE `table1` (
  `name` VARCHAR(20) DEFAULT NULL,
  `kecheng` VARCHAR(20) DEFAULT NULL,
  `fenshu` INT(10) DEFAULT NULL
);


INSERT INTO table1(NAME,kecheng,fenshu) VALUES('张三','语文',81);
INSERT INTO table1(NAME,kecheng,fenshu) VALUES('张三','数学',75);
INSERT INTO table1(NAME,kecheng,fenshu) VALUES('李四','语文',76);
INSERT INTO table1(NAME,kecheng,fenshu) VALUES('李四','数学',76);
INSERT INTO table1(NAME,kecheng,fenshu) VALUES('王五','语文',81);
INSERT INTO table1(NAME,kecheng,fenshu) VALUES('王五','数学',100);

#求出每门课都大于80分的学生姓名

第一种方法:

SELECT NAME FROM table1 GROUP BY NAME 
HAVING MIN(fenshu)>80

第二种方法:

SELECT NAME FROM table1 GROUP BY NAME 
HAVING COUNT(1)=SUM(CASE WHEN fenshu>80 THEN 1 ELSE 0 END)

第三种方法:

SELECT NAME FROM table1 GROUP BY NAME 
HAVING NAME NOT IN (SELECT NAME FROM table1 WHERE fenshu<=80)

第二题

CREATE TABLE `table2` (
  `语文` INT(10) DEFAULT NULL,
  `数学` INT(10) DEFAULT NULL,
  `英语` INT(10) DEFAULT NULL
);
INSERT INTO table2(语文,数学,英语) VALUES(58,70,80);

有一张表,3个字段是语文,数学,英语, 有1条记录分别表示语文68,数学70,英语80, 得出结果分数变等级(>=80分是优秀, >=60分是及格, <60是不及格)

SELECT CASE WHEN 语文>=80 THEN '优秀' WHEN 语文>=60 THEN '及格' WHEN 语文<60 THEN '不及格' END 语文,
CASE WHEN 数学>=80 THEN '优秀' WHEN 数学>=60 THEN '及格' WHEN 数学<60 THEN '不及格' END 数学,
CASE WHEN 英语>=80 THEN '优秀' WHEN 英语>=60 THEN '及格' WHEN 英语<60 THEN '不及格' END 英语 
FROM table2

第三题

CREATE TABLE `table3` (
  `date` VARCHAR(20) DEFAULT NULL,
  `result` VARCHAR(10) DEFAULT NULL
)

#查询出来的结果：
#date 胜 负
#2011-02-01 2 1
#2011-02-02 1 1

SELECT DATE, 
(SELECT COUNT(1) FROM table3 WHERE DATE=t.date AND result='胜') AS 胜, 
(SELECT COUNT(1) FROM table3 WHERE DATE=t.date AND result='负') AS 负
FROM table3 t GROUP BY DATE

第四题

#1 2005001 张三 0001 数学 69
#2 2005002 李四 0001 数学 89
#3 2005001 张三 0001 数学 69

CREATE TABLE IF NOT EXISTS `student_test` (
	id INT UNSIGNED AUTO_INCREMENT,
	stu_id VARCHAR(10) NULL,
	NAME VARCHAR(10) NULL,
	couser_id VARCHAR(10) NULL,
	couser VARCHAR(10) NULL,
	score DECIMAL(10,2) NULL,
	PRIMARY KEY (id) 
)
INSERT INTO student_test(id,stu_id,NAME,couser_id,couser,score) VALUES(NULL,'2005001','张三','0001','数学',69);
INSERT INTO student_test(id,stu_id,NAME,couser_id,couser,score) VALUES(NULL,'2005002','李四','0001','数学',89);
INSERT INTO student_test(id,stu_id,NAME,couser_id,couser,score) VALUES(NULL,'2005001','张三','0001','数学',69);

#删除冗余字段, 保留ID最小的那个

DELETE FROM student_test WHERE id NOT IN (
 SELECT * FROM (
  SELECT MIN(id) FROM student_test GROUP BY stu_id,NAME,couser_id,couser,score
 ) s
)

第五题

CREATE TABLE IF NOT EXISTS `product` (
	id INT UNSIGNED AUTO_INCREMENT,
	NAME VARCHAR(10) NULL,
	STATUS VARCHAR(10) NULL,
	PRIMARY KEY (id)
)
INSERT INTO product(id,NAME,STATUS) VALUES(NULL,'苹果','好吃');
INSERT INTO product(id,NAME,STATUS) VALUES(NULL,'梨','难吃');
INSERT INTO product(id,NAME,STATUS) VALUES(NULL,'橘子','好吃');
INSERT INTO product(id,NAME,STATUS) VALUES(NULL,'葡萄',NULL);

#现在需要把status的‘好吃’更新为‘0’，‘难吃’更新为‘1’

UPDATE product SET STATUS=(CASE WHEN STATUS='好吃' THEN '0' WHEN STATUS='难吃' THEN '1' END)
1

第六题

1) 创建一张学生表，包含以下信息，学号，姓名，年龄，性别，家庭住址，联系电话

CREATE TABLE IF NOT EXISTS `t_student` (
	stu_id INT UNSIGNED AUTO_INCREMENT,
	NAME VARCHAR(10) NULL,
	age INT NULL,
	gender VARCHAR(2) NULL,
	address VARCHAR(50) NULL,
	phone VARCHAR(11) NULL,
	PRIMARY KEY (stu_id)
);

#2) 修改学生表的结构，添加一列信息，学历

ALTER TABLE t_student ADD COLUMN `education` VARCHAR(20) NOT NULL;

#3) 修改学生表的结构，删除一列信息，家庭住址

ALTER TABLE t_student DROP COLUMN address;

#4) 向学生表添加如下信息：

INSERT INTO t_student(stu_id,NAME,age,gender,phone,education) VALUES(NULL,'A',22,'男','123456','小学');
INSERT INTO t_student(stu_id,NAME,age,gender,phone,education) VALUES(NULL,'B',21,'男','119','中学');
INSERT INTO t_student(stu_id,NAME,age,gender,phone,education) VALUES(NULL,'C',23,'男','110','高中');
INSERT INTO t_student(stu_id,NAME,age,gender,phone,education) VALUES(NULL,'D',18,'女','114','大学');

#5) 修改学生表的数据，将电话号码以11开头的学员的学历改为“大专”

UPDATE t_student SET education='大专' WHERE phone LIKE '11%';

#6) 删除学生表的数据，姓名以C开头，性别为‘男’的记录删除

DELETE FROM t_student WHERE NAME LIKE 'C%' AND gender='男';

#7) 查询学生表的数据，将所有年龄小于22岁的，学历为“大专”的，学生的姓名和学号示出来

SELECT stu_id,NAME FROM t_student WHERE age<22 AND education='大专'

#8) 查询学生表的数据，查询所有信息，列出前25%的记录

SELECT * FROM t_student WHERE stu_id<=(SELECT COUNT(*) FROM t_student)*0.25;

#9) 查询出所有学生的姓名，性别，年龄降序排列

SELECT NAME,gender,age FROM t_student ORDER BY age DESC;

#10) 按照性别分组查询所有的平均年龄

SELECT gender,AVG(age) FROM t_student GROUP BY gender;

第七题

CREATE TABLE IF NOT EXISTS test1(
	a DECIMAL(10,2) NOT NULL,
	b DECIMAL(10,2) NOT NULL,
	c DECIMAL(10,2) NOT NULL,
	d VARCHAR(10) NOT NULL
)
INSERT INTO test1(a,b,c,d) VALUES(0.5,1.5,2.0,'A1');
INSERT INTO test1(a,b,c,d) VALUES(1.5,0.5,0.5,'A1');
INSERT INTO test1(a,b,c,d) VALUES(0.5,0.5,1.5,'A1');
INSERT INTO test1(a,b,c,d) VALUES(1.5,1.5,1.5,'B1');
INSERT INTO test1(a,b,c,d) VALUES(0.5,2.0,2.0,'B1');
INSERT INTO test1(a,b,c,d) VALUES(1.5,2.0,0.5,'B1');

统计a,b,c三列大于1的个数

SELECT 
SUM(CASE WHEN a > 1 THEN 1 ELSE 0 END) a, 
SUM(CASE WHEN b > 1 THEN 1 ELSE 0 END) b, 
SUM(CASE WHEN c > 1 THEN 1 ELSE 0 END) c 
FROM test1;

按a分组, 用b排序取top1
其实用hive或者Oracle很简单了, 直接用row_number即可
方法一

SELECT test1.* 
FROM test1 
JOIN (SELECT a,MAX(b) maxb FROM test1 GROUP BY a) t 
ON test1.b=t.maxb 
GROUP BY a

方法二

SELECT a, SUBSTRING_INDEX(GROUP_CONCAT(b ORDER BY b DESC),',',1) b, 
SUBSTRING_INDEX(GROUP_CONCAT(c ORDER BY b DESC),',',1) c, 
SUBSTRING_INDEX(GROUP_CONCAT(d ORDER BY b DESC),',',1) d 
FROM test1 GROUP BY a;

用到了下面的两个函数:

group_concat(v)

先看看它的语法:

group_concat([DISTINCT] 要连接的字段 [ORDER BY ASC/DESC 排序字段] [SEPARATOR
‘分隔符’])

举例:

select GROUP_CONCAT(c ORDER BY b DESC) from test1 group a
#解析: 按a进行分组, 把c字段的值打印在一起, 逗号分割(默认), 并且c字段按b排序

substring_index(v1, v2, v3) – 将v1按照v2进行分割取下标为v3的值

从表中随机取出3条数据
下面这种方法虽然可以取出随机记录, 但是不推荐, 数据量大的话会很慢很慢

SELECT * FROM test1 ORDER BY RAND() LIMIT 3

可以使用下面这种方法

SELECT * 
FROM (SELECT test1.*, RAND() r FROM test1) t 
ORDER BY t.r LIMIT 3

第八题
Table表有三个字段ID,Name,Location, 请以两种方式写出SQL语句删除表中ID以及Name都重复的记录.

CREATE TABLE table4(
  ID INT NOT NULL,
  NAME VARCHAR(10) NOT NULL,
  Location VARCHAR(12) NOT NULL
);

INSERT INTO table4(ID,NAME,Location) VALUES(1,'zhangsan','bj');
INSERT INTO table4(ID,NAME,Location) VALUES(1,'zhangsan','bj');
INSERT INTO table4(ID,NAME,Location) VALUES(2,'lisi','bj');
INSERT INTO table4(ID,NAME,Location) VALUES(3,'wangwu','sh');

第一种方法

DELETE FROM table4 WHERE (id,NAME) IN (
SELECT * FROM (SELECT id,NAME FROM table4 GROUP BY ID,NAME HAVING COUNT(1)>1) t)

第二种方法

DELETE FROM table4 WHERE (id,NAME) IN (
SELECT * FROM(
SELECT id,NAME FROM table4 t1 WHERE (
  (SELECT COUNT(1) FROM table4 t2 WHERE t1.ID=t2.ID AND t1.Name=t2.Name)>1
)
) tmp)

第九题

现有BOOKS表, 查询所有图书中价格小于60元的作者有多少本图书在售

CREATE TABLE books(
  bno INT NOT NULL,
  bname VARCHAR(10) NOT NULL,
  author VARCHAR(10) NOT NULL,
  price DECIMAL(10,2) NOT NULL
)
INSERT INTO books VALUES(1,'倚天屠龙记','金庸',12.99);
INSERT INTO books VALUES(2,'雪山飞狐','金庸',64.99);
INSERT INTO books VALUES(3,'东方不败','金庸',12.99);
INSERT INTO books VALUES(4,'仙逆','耳根',72.99);
INSERT INTO books VALUES(5,'求魔','耳根',12.99);

答案:
使用子查询查出图书价格小于60的作者有哪些, 然后按照作者分组, 查询这些作者的图书数量.

SELECT author,COUNT(1) FROM books GROUP BY author HAVING author IN 
(SELECT DISTINCT(author) FROM books WHERE price<60)

第十题

表中有A,B,C三列, 用SQL语句实现, 依次取A,B,C中第一个不为空的值

CREATE TABLE table_null(
  A VARCHAR(10),
  B VARCHAR(10),
  C VARCHAR(10)
)

INSERT INTO table_null VALUES(NULL,NULL,NULL);
INSERT INTO table_null VALUES('abc',NULL,NULL);
INSERT INTO table_null VALUES(NULL,'ccc',NULL);
INSERT INTO table_null VALUES(NULL,NULL,'ddd');

答案:
分别按照每一个字段分组, 所有为NUll的分为一组, 然后取不为null的第一条即可.

SELECT * FROM 
(SELECT A FROM table_null GROUP BY A LIMIT 1,1) t1 JOIN
(SELECT B FROM table_null GROUP BY B LIMIT 1,1) t2 JOIN
(SELECT C FROM table_null GROUP BY C LIMIT 1,1) t3

第十一题

有一张表stu_score, 所有课程都>=90则特优, 所有课程都>=80则优秀, 都>=60则及格, 否则挂科.

CREATE TABLE stu_score(
  kecheng VARCHAR(10),
  stu_id INT,
  score INT
)
INSERT INTO stu_score VALUES('001',1,90);
INSERT INTO stu_score VALUES('002',1,92);
INSERT INTO stu_score VALUES('001',2,80);
INSERT INTO stu_score VALUES('002',2,92);
INSERT INTO stu_score VALUES('001',3,76);
INSERT INTO stu_score VALUES('002',3,92);
INSERT INTO stu_score VALUES('001',4,50);
INSERT INTO stu_score VALUES('002',4,92);

答案:
按照学号分组, 使用 CASE WHEN 判断最小分数 MIN(score) 是否满足条件.

SELECT 
   stu_id, (CASE WHEN MIN(score)>=90 THEN '特优' WHEN MIN(score)>=80 THEN '优秀' WHEN MIN(score)>=60 THEN '及格' ELSE '挂科' END) stat
FROM stu_score GROUP BY stu_id

第十二题

运用SQL实现数据转换
转换前:

转换后:

CREATE TABLE test33(
  grade_id INT,
  subject_name VARCHAR(10),
  max_score INT
)
INSERT INTO test33 VALUES(1,'语文',98);
INSERT INTO test33 VALUES(2,'数学',95);
INSERT INTO test33 VALUES(2,'政治',87);
INSERT INTO test33 VALUES(5,'语文',97);
INSERT INTO test33 VALUES(5,'数学',100);
INSERT INTO test33 VALUES(5,'政治',92);

答案:

SELECT 年级,
SUM(CASE WHEN  语文 IS NULL THEN 0 ELSE 语文 END) 语文,
SUM(CASE WHEN  数学 IS NULL THEN 0 ELSE 数学 END) 数学,
SUM(CASE WHEN  政治 IS NULL THEN 0 ELSE 政治 END) 政治
FROM
(
(SELECT 
(CASE grade_id WHEN 1 THEN '一年级' WHEN 2 THEN '二年级' WHEN 5 THEN '五年级' END) 年级,
(CASE subject_name WHEN '语文' THEN max_score ELSE NULL END) 语文,
(CASE subject_name WHEN '数学' THEN max_score ELSE NULL END) 数学,
(CASE subject_name WHEN '政治' THEN max_score ELSE NULL END) 政治
FROM test33
)  UNION (
SELECT 
(CASE grade_id WHEN 1 THEN '一年级' WHEN 2 THEN '二年级' WHEN 5 THEN '五年级' END) 年级,
(CASE subject_name WHEN '语文' THEN max_score ELSE NULL END) 语文,
(CASE subject_name WHEN '数学' THEN max_score ELSE NULL END) 数学,
(CASE subject_name WHEN '政治' THEN max_score ELSE NULL END) 政治
FROM test33
)  UNION (
SELECT 
(CASE grade_id WHEN 1 THEN '一年级' WHEN 2 THEN '二年级' WHEN 5 THEN '五年级' END) 年级,
(CASE subject_name WHEN '语文' THEN max_score ELSE NULL END) 语文,
(CASE subject_name WHEN '数学' THEN max_score ELSE NULL END) 数学,
(CASE subject_name WHEN '政治' THEN max_score ELSE NULL END) 政治
FROM test33
)
) t1 GROUP BY 年级

第十三题

A表与B表结构相同, 现在想用B表的数据更新A表的用户姓名, 存在就更新, 不存在就添加.
表A

CREATE TABLE A(
  user_id VARCHAR(10),
  user_name VARCHAR(20)
)
CREATE TABLE B(
  user_id VARCHAR(10),
  user_name VARCHAR(20)
)

答案:
mysql有这样的语法: ON DUPLICATE KEY UPDATE 对唯一索引或主键索引可以实现更新.

#给B表添加唯一索引

ALTER TABLE B ADD UNIQUE KEY user_name(user_name)

#存在相同的user_name就执行UPDATE

INSERT INTO B (user_id,user_name) SELECT user_id,user_name FROM A 
ON DUPLICATE KEY UPDATE user_id=VALUES(user_id),user_name=VALUES(user_name)

下面列几个常见的Hive必考HQL
#1. 级联求和问题 #######################

select a.username, a.month, max(a.salSum), sum(b.salSum) as salSum
from 
  (select username, month, sum(salary) as salSum from t_salary_detail group by username,month) a
  join
  (select username, month, sum(salary) as salSum from t_salary_detail group by username,month) b
  on a.username=b.username
where a.month>=b.month
group by a.username,a.month;

#2. 连续3个月都有销售额的商家 #######################

select shopid,count(flag) as cnt
from 
  (select shopid,dt,sale,rn,date_sub(to_date(dt),rn) as flag 
   from 
     (select shopid,dt,sale,
        row_number() over(partition by shopid order by dt) as rn 
      from t_jd
     ) tmp
  ) tmp2
group by shopid,flag having cnt>=3

#3. 求TOPN问题 #######################

--统计每日最热门页面的top10
select month,day,request,request_count, od 
from 
 (select 
 	month,day,request,request_count,
    row_number() over(distribute by concat(month,day) sort by request_count desc) as od
  from 
    (select month,day,request,count(1) as request_count 
     from ods_weblog_detail
     where datestr='20130918' 
     group by request,month,day having request is not null
    ) b
 ) c where od<=10

#4. 行转列, 列转行问题 #######################

–当前有用户人生阶段表lifeStage, 有用户唯一ID字段uid,用户人生阶段字段stage,其中stage字段内容为各个人生阶段标签按照英文逗号分割的拼接内容,
–如:计划买车,已买房, 并且每个用户的内容不同, 请使用hive sql统计每个人生阶段的用户量.
–考察点: lateral view使用, explode函数

select stage_someone,count(distinct uid) as uids
from lifeStage
  lateral view explode(split(stage,',')) lifeStage_tmp as stage_someone
group by stage_someone;

–和上一题相同的数据场景, 但是lifeStage中每行数据存储一个用户的人生阶段数据
–如: 一行数据uid是43, stage内容为计划买车, 另一行数据uid字段为43,stage字段为已买房, 请输出类似于uid为43,stage字段为计划买车,已买房这样的新整合数据.
–考察点: collect_set函数, concat_ws函数

select uid,concat_ws(',',collect_set(stage_someone)) as stage
from lifestage_multline
group by uid

#5. 多表关联的基础上, 行转列与列转行问题 #######################

–权限组表(g)中记录了组id和组名称，用户表(u)记录了用户id和用户名称，权限表(gu)是权限组表和用户表的关系，记录了每一个权限组与用户的对应关系
–题目是根据gt和ut表将gu表中的所有id转换为名称

create table gu_trans as
select t.gname as gname,concat_ws(',',collect_set(t.uname)) as uname
from
 (select g.gname,u.uname
  from
   (select gid,s_uid
   from gu lateral view explode(split(uid,',')) b as s_uid) tmp 
   join g on tmp.gid=g.gid
   join u on tmp.s_uid=u.uid
 ) t
group by t.gname