Java面试——SQL语句题

一、行转列问题

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现有表格A，按照以下格式排列；

姓名	收入类型	收入金额
Tom	年奖金	5w
Tom	月工资	10k
Jack	年奖金	8w
Jack	月工资	12k

先需要将表格转化为：

姓名	月工资	年奖金
Tom	10k	50k
Jack	12k	80k

方法一：使用静态SQL

select '姓名',
sum(case '收入类型' when '年奖金' then '收入金额' else 0 end) 年奖金,
sum(case '收入类型' when '月工资' then '收入金额' else 0 end) 月工资
from A
group by '姓名'

方法二：使用 pivot：MySQL不支持

select * from
(
    select 姓名,收入类型,收入金额 from A
) test
pivot(sum(收入金额) for 收入类型 in ('月工资','年终奖')) pvt

二、准备工作：来自博客园（https://www.cnblogs.com/coder-wf/p/11128033.html）自己当面试先写一遍，不行再参考答案。

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【1】表名和字段

–1.学生表 
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别 
–2.课程表 
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号 
–3.教师表 
Teacher(t_id,t_name) –教师编号,教师姓名 
–4.成绩表 
Score(s_id,c_id,s_score) –学生编号,课程编号,分数

【2】测试数据

--建表
--学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT '',
`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

三、练习题

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【1】查询"01"课程比"02"课程成绩高的学生的信息及课程分数：当对一张表中的一列数据比较时，应当将一张表拆分为两张表；

SELECT st.*,sc.`s_score` AS '语文' ,sc2.`s_score` AS '数学' 
FROM student st 
LEFT JOIN score sc ON st.s_id=sc.`s_id` AND sc.`c_id`='01'
LEFT JOIN score sc2 ON st.s_id=sc2.`s_id` AND sc2.`c_id`='02'  
WHERE sc.`s_score` > sc2.`s_score`;

【2】查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩：分组在 having 之前，有函数表达式时，条件判断需要使用 having，同时主要成绩需要截取为两位；

SELECT s.`s_id`,s.`s_name`,ROUND(AVG(sc.`s_score`),2) AS '平均成绩' FROM student s
LEFT JOIN score sc ON s.`s_id` = sc.`s_id`
GROUP BY sc.`s_id`
HAVING AVG(sc.`s_score`) >= 60;

【3】查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩；

SELECT s.`s_id`,s.`s_name`,COUNT(sc.`c_id`) AS '选课总数',SUM(CASE WHEN sc.`s_score` IS NULL THEN 0 ELSE sc.`s_score` END) AS '总成绩' FROM student s 
LEFT JOIN score sc ON s.`s_id` = sc.`s_id`
GROUP BY sc.`s_id`

【4】查询学过 "张三" 老师授课的同学的信息；

SELECT s.* FROM student s 
LEFT JOIN score sc ON s.`s_id` = sc.`s_id`
LEFT JOIN course c ON sc.`c_id` = c.`c_id`
LEFT JOIN teacher t ON t.`t_id` = c.`t_id`
WHERE t.`t_name` = "张三"

【5】查询没学过"张三"老师授课的同学的信息；

 SELECT st.* FROM student st WHERE st.s_id NOT IN(
  SELECT sc.s_id FROM score sc WHERE sc.c_id IN (SELECT c.c_id FROM course c LEFT JOIN teacher t ON t.t_id=c.t_id WHERE  t.t_name="张三")
  )

【6】查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息

SELECT s.* FROM student s
INNER JOIN score sc ON s.`s_id` = sc.`s_id` 
INNER JOIN score sc1 ON s.`s_id` = sc1.`s_id`
WHERE sc.`c_id`='01' AND sc1.`c_id`='02'

--方式二
SELECT a.* 
FROM
    student a,
    score b,
    score c
WHERE
    a.s_id = b.s_id
    AND a.s_id = c.s_id
    AND b.c_id = '01'
    AND c.c_id = '02';

【7】查询至少有一门课与学号为"01"的同学所学相同的同学的信息

SELECT DISTINCT s.* FROM student s 
LEFT JOIN score c ON s.`s_id` = c.`s_id`
WHERE c.`c_id` IN (
	SELECT sc.`c_id` FROM student s
	LEFT JOIN score sc ON s.`s_id` = sc.`s_id`
	WHERE s.`s_id`='01'
);

【8】查询和"01"号的同学学习的课程完全相同的其他同学的信息

SELECT DISTINCT s.* FROM student s 
LEFT JOIN score c ON s.`s_id` = c.`s_id`
GROUP BY s.`s_id`
HAVING COUNT(c.`c_id`) = (
	SELECT COUNT(sc.`c_id`) FROM student s
	LEFT JOIN score sc ON s.`s_id` = sc.`s_id`
	WHERE s.`s_id`='01'
);

【9】查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT s.`s_name` FROM student s
WHERE s.`s_id` NOT IN(
	SELECT sc.`s_id` FROM score sc
	LEFT JOIN course c ON sc.`c_id` = c.`c_id`
	LEFT JOIN teacher t ON t.`t_id` = c.`t_id`
	WHERE t.`t_name`="张三"
)

【10】查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

SELECT s.`s_id`,s.`s_name`,AVG(sc.`s_score`) FROM student s 
INNER JOIN score sc ON s.`s_id` = sc.`s_id`
WHERE s.`s_id` IN (
	SELECT sc.`s_id` FROM score sc 
	WHERE sc.`s_score`<60
	GROUP BY sc.`s_id`
	HAVING COUNT(1)>=2
)
GROUP BY s.`s_id`

【11】按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩：这里要注意 where 和 on 的区别：on 条件是在生成临时表时使用的条件，它不管on中的条件是否为真，都会返回左（右）边表中的记录。(返回左（右）表全部记录)。此时可能会出现与右表不匹配的记录即为空的记录。即使on后边的条件不为真也会返回左（右）表中的记录。where 条件是在临时表生成好后，再对临时表进行过滤的条件。

SELECT s.`s_id`,s.`s_name`,sc.`s_score` AS "语文" ,sc1.`s_score` AS "数学",sc2.`s_score` AS "英语",AVG(sc3.`s_score`) "平均分" FROM student s 
LEFT JOIN score sc ON s.`s_id` = sc.`s_id` AND sc.`c_id` = "01"
LEFT JOIN score sc1 ON s.`s_id` = sc1.`s_id` AND sc1.`c_id` = "02"
LEFT JOIN score sc2 ON s.`s_id` = sc2.`s_id` AND sc2.`c_id` = "03"
LEFT JOIN score sc3 ON s.`s_id` = sc3.`s_id`  
GROUP BY s.`s_id`
ORDER BY AVG(sc3.`s_score`) DESC;

【12】查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程 Name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率（及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90）

SELECT c.`c_id`,c.`c_name`,MAX(s.`s_score`) "最高分",MIN(s.`s_score`) "最低分",AVG(s.`s_score`) "平均分", 
((SELECT COUNT(1) FROM score sc WHERE sc.`c_id` = c.`c_id` AND sc.`s_score` >= 60)/(SELECT COUNT(1) FROM score sc WHERE sc.c_id = c.c_id)) "及格率",
((SELECT COUNT(1) FROM score sc WHERE sc.`c_id` = c.`c_id` AND 80 >= sc.`s_score` AND sc.`s_score` >= 70)/(SELECT COUNT(1) FROM score sc WHERE sc.c_id = c.c_id)) "中等率",
((SELECT COUNT(1) FROM score sc WHERE sc.`c_id` = c.`c_id` AND 90 >= sc.`s_score` AND sc.`s_score` >= 80)/(SELECT COUNT(1) FROM score sc WHERE sc.c_id = c.c_id)) "优良率",
((SELECT COUNT(1) FROM score sc WHERE sc.`c_id` = c.`c_id` AND sc.`s_score` >= 90)/(SELECT COUNT(1) FROM score sc WHERE sc.c_id = c.c_id)) "优秀率"
FROM course c
LEFT JOIN score s ON c.`c_id` = s.`c_id`
GROUP BY c.`c_id`; 

【13】查询所有课程的成绩第2名到第3名的学生信息及该课程成绩：Union：对两个结果集进行并集操作，不包括重复行，同时进行默认规则的排序；Union All：对两个结果集进行并集操作，包括重复行，不进行排序；注意 limit下标是从0开始的。

(SELECT s.*,c.`c_name`,sc.`s_score` "成绩" FROM student s
LEFT JOIN score sc ON s.`s_id` = sc.`s_id` AND sc.`c_id`="01"
LEFT JOIN course c ON sc.`c_id` = c.`c_id`
ORDER BY sc.`s_score` DESC
LIMIT 1,2)
UNION ALL
(SELECT s.*,c.`c_name`,sc.`s_score` "成绩" FROM student s
LEFT JOIN score sc ON s.`s_id` = sc.`s_id` AND sc.`c_id`="02"
LEFT JOIN course c ON sc.`c_id` = c.`c_id` 
ORDER BY sc.`s_score` DESC
LIMIT 1,2)
UNION ALL
(SELECT s.*,c.`c_name`,sc.`s_score` "成绩" FROM student s
LEFT JOIN score sc ON s.`s_id` = sc.`s_id` AND sc.`c_id`="03"
LEFT JOIN course c ON sc.`c_id` = c.`c_id` 
ORDER BY sc.`s_score` DESC
LIMIT 1,2)

【14】查询学生平均成绩及其名次：重点是名次的获取，通过变量 @i 进行递增获取。

SET @i=0;
SELECT test.*,@i:=@i+1 "名次" FROM(
SELECT s.`s_name`,ROUND(AVG(sc.`s_score`),2) "平均成绩" FROM score sc
LEFT JOIN student s ON s.`s_id` = sc.`s_id`
GROUP BY sc.`s_id`
ORDER BY AVG(sc.`s_score`) DESC) test;

【15】查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩：思路就是先查询一条数据，然后与表中的数据比较相同的成绩，且科目号不相同的数据行，如果大于1则返回当前行即可。逐行比较；

SELECT st.s_id,st.s_name,sc.c_id,sc.s_score FROM student st 
LEFT JOIN score sc ON sc.s_id=st.s_id
LEFT JOIN course c ON c.c_id=sc.c_id
WHERE (
SELECT COUNT(1) FROM student st2 
LEFT JOIN score sc2 ON sc2.s_id=st2.s_id
LEFT JOIN course c2 ON c2.c_id=sc2.c_id
WHERE sc.s_score=sc2.s_score AND c.c_id!=c2.c_id 
)>=1

【16】 查询每门功成绩最好的前两名

SELECT a.* FROM (SELECT st.s_id,st.s_name,c.c_name,sc.s_score FROM student st
LEFT JOIN score sc ON sc.s_id=st.s_id
INNER JOIN course c ON c.c_id=sc.c_id AND c.c_id="01"
ORDER BY sc.s_score DESC LIMIT 0,2) a
UNION ALL
SELECT b.* FROM (SELECT st.s_id,st.s_name,c.c_name,sc.s_score FROM student st
LEFT JOIN score sc ON sc.s_id=st.s_id
INNER JOIN course c ON c.c_id=sc.c_id AND c.c_id="02"
ORDER BY sc.s_score DESC LIMIT 0,2) b
UNION ALL
SELECT c.* FROM (SELECT st.s_id,st.s_name,c.c_name,sc.s_score FROM student st
LEFT JOIN score sc ON sc.s_id=st.s_id
INNER JOIN course c ON c.c_id=sc.c_id AND c.c_id="03"
ORDER BY sc.s_score DESC LIMIT 0,2) c

方式二
SELECT a.s_id,a.c_id,a.s_score FROM score a
WHERE (SELECT COUNT(1) FROM score b WHERE b.c_id=a.c_id AND b.s_score>=a.s_score)<=2 ORDER BY a.c_id

【17】查询本周过生日的学生：此处可能有问题,week函数取的为当前年的第几周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期几(%w)， 再判断本周是否会持续到下一个月进行判断,太麻烦。

SELECT st.* FROM student st 
WHERE WEEK(NOW())=WEEK(DATE_FORMAT(st.s_birth,'%Y%m%d'))

【18】查询下周过生日的学生

SELECT st.* FROM student st 
WHERE WEEK(NOW())+1=WEEK(DATE_FORMAT(st.s_birth,'%Y%m%d'))

【19】查询本月过生日的学生

SELECT st.* FROM student st 
WHERE MONTH(NOW())=MONTH(DATE_FORMAT(st.s_birth,'%Y%m%d'))

【20】查询下月过生日的学生： 注意，如果当前月为12月时，用month(now())+1为13而不是1，可用 timestampadd() 函数或 mod 取模

SELECT st.* FROM student st 
WHERE MONTH(TIMESTAMPADD(MONTH,1,NOW()))=MONTH(DATE_FORMAT(st.s_birth,'%Y%m%d'))

方法二：
SELECT st.* FROM student st WHERE (MONTH(NOW()) + 1) MOD 12 = MONTH(DATE_FORMAT(st.s_birth,'%Y%m%d'))

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