疑难点语句练习

数据库创建脚本

  1. create database scott;
  3. use scott;
  5. # 创建员工表
  6. create table emp(
  7.                     EMPNO INT NOT NULL comment '员工编号',
  8.                     ENAME VARCHAR(10) comment '员工名字',
  9.                     JOB VARCHAR(9) comment '员工职位',
  10.                     MGR int comment '员工对应领导编号',
  11.                     HIREDATE DATE comment '入职时间',
  12.                     SAL decimal(7,2) comment '基本工资',
  13.                     COMM decimal(7,2) comment '奖金',
  14.                     DEPTNO int comment '员工所在部门编号'
  15. );
  17. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7369','SMITH','CLERK','7902',date('19801217'),'800','0','20');
  18. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7944','ALLEN','SALESMAN','7698',date('19811220'),'1600','300','30');
  19. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7521','WARD','SALESMAN','7698',date('19821222'),'1250','500','30');
  20. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7566','JONES','MANAGER','7839',date('19820402'),'2975','0','20');
  21. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7654','MARTIN','SALESMAN','7698',date('19820928'),'1250','1400','30');
  22. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7698','BLAKE','MANAGER','7839',date('19810501'),'2850','0','30');
  23. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7782','CLARK','MANAGER','7839',date('19810609'),'2450','0','10');
  24. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7839','KING','PRESIDENT','',date('19811117'),'5000','0','10');
  25. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7844','TURNER','SALESMAN','7698',date('19811208'),'1500','0','30');
  26. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7900','JAMES','CLERK','7698',date('19811203'),'950','0','30');
  27. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7902','FORD','ANALYST','7566',date('19811203'),'3000','0','30');
  28. INSERT INTO emp (EMPNO,ENAME,JOB,MGR,HIREDATE,SAL,COMM,DEPTNO) VALUES ('7934','MILLER','CLERK','7782',date('19820123'),'1300','0','10');
  30. # 创建部门表
  31. create table dept(
  32.                      DEPTNO int comment '部门编号',
  33.                      DNAME VARCHAR(14) comment '部门名称',
  34.                      LOC VARCHAR(13) comment '部门位置'
  36. );
  38. INSERT INTO dept (DEPTNO,DNAME,LOC) VALUES ('30','SALES','CHICAGO');
  39. INSERT INTO dept (DEPTNO,DNAME,LOC) VALUES ('20','RESEARCH','DAllAS');
  40. INSERT INTO dept (DEPTNO,DNAME,LOC) VALUES ('10','ACCOUNTING','NEW YORK');
  41. INSERT INTO dept (DEPTNO,DNAME,LOC) VALUES ('40','OPERATIONS','BOSTON');
  43. # 创建工资等级表salgrade
  44. create table salgrade(
  45.                          GRADE int comment '工资等级',
  46.                          LOSAL int comment '此等级最低工资',
  47.                          HISAL int comment '此等级最高工资'
  48. );
  50. INSERT INTO salgrade (GRADE,LOSAL,HISAL) VALUES ('1','700','1200');
  51. INSERT INTO salgrade (GRADE,LOSAL,HISAL) VALUES ('2','1201','1400');
  52. INSERT INTO salgrade (GRADE,LOSAL,HISAL) VALUES ('3','1401','2000');
  53. INSERT INTO salgrade (GRADE,LOSAL,HISAL) VALUES ('4','2001','3000');
  54. INSERT INTO salgrade (GRADE,LOSAL,HISAL) VALUES ('5','3001','9999');
  56. # 创建工资表
  57. create table bonus(
  58.                       ENAME VARCHAR(10) comment '雇员姓名',
  59.                       JOB VARCHAR(9) comment '雇员职位',
  60.                       SAL int comment '雇员薪资',
  61.                       COMM int comment '雇员奖金'
  62. );

UNION和UNION ALL

  • 所有查询中的列数和列的顺序必须相同。
  • 数据类型必须兼容。

    UNION

    不允许重复

UNION ALL

允许重复

exists的使用

exists（sub－query）,当exists中的子查询语句能查到对应结果的时候，
意味着条件满足
相当于双层for循环 image.png

现在要查询部门编号为10和20的员工·

普通写法

  1. select * from emp where deptno = 10 or deptno = 20;

exists实现

  • 错误展示
    1. select * from emp e where exists(
    2.                                 select d.deptno from dept d
    3.                                 where (d.DEPTNO=10 or d.DEPTNO=20));

    结果查出来deptno为30的数据 EXISTS：用于检查子查询是否至少会返回一行数据，该子查询实际上并不返回任何数据，而是返回值True或False。
    EXISTS 指定一个子查询，检测 行 的存在。

EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
7369 SMITH CLERK 7902 1980-12-17 800.00 0.00 20
7944 ALLEN SALESMAN 7698 1981-12-20 1600.00 300.00 30
7521 WARD SALESMAN 7698 1982-12-22 1250.00 500.00 30
7566 JONES MANAGER 7839 1982-04-02 2975.00 0.00 20
7654 MARTIN SALESMAN 7698 1982-09-28 1250.00 1400.00 30
7698 BLAKE MANAGER 7839 1981-05-01 2850.00 0.00 30
7782 CLARK MANAGER 7839 1981-06-09 2450.00 0.00 10
7839 KING PRESIDENT 1981-11-17 5000.00 0.00 10
7844 TURNER SALESMAN 7698 1981-12-08 1500.00 0.00 30
7900 JAMES CLERK 7698 1981-12-03 950.00 0.00 30
7902 FORD ANALYST 7566 1981-12-03 3000.00 0.00 30
7934 MILLER CLERK 7782 1982-01-23 1300.00 0.00 10
  • 正确写法
    1. select * from emp e where exists(
    2.                                 select d.deptno from dept d
    3.                                 where (d.DEPTNO=10 or d.DEPTNO=20) and e.deptno = d.deptno);
    | EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO | | —- | —- | —- | —- | —- | —- | —- | —- | | 7369 | SMITH | CLERK | 7902 | 1980-12-17 | 800.00 | 0.00 | 20 | | 7566 | JONES | MANAGER | 7839 | 1982-04-02 | 2975.00 | 0.00 | 20 | | 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | 0.00 | 10 | | 7839 | KING | PRESIDENT | | 1981-11-17 | 5000.00 | 0.00 | 10 | | 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | 0.00 | 10 |

decode,case when函数使用

decode：只有oracle才有，用来做相等判断 DECODE(条件，值1，返回值1，值2，返回值2…..) case when： CASE <条件列名> WHEN <参数> THEN <结果> ELSE ‘’ END

给不同部门的人员涨薪，10部门涨10%，20部门涨20%，30部门涨30%

解法一

  1. select ename,sal,deptno,decode(deptno,10,sal*1.1,20,sal*1.2,30,sal*1.3) from emp;
  2. # mysql不支持decode

解法二

  1. select *, 
  2.        case DEPTNO 
  3.            when 10 
  4.                then SAL*1.1 
  5.            when 20 
  6.                then SAL*1.2 
  7.            when 30 
  8.                then SAL*1.3 
  9.            end  
  10. from emp;

行转列


需求
将表显示转换为
姓名 语文 数学 英语
张三 78 88 98 李四 89 76 90 王五 89 56 89

sql脚本

  1. create table STUDENT_SCORE
  2. (
  3.     name    VARCHAR(20),
  4.     subject VARCHAR(20),
  5.     score   VARCHAR(20)
  6. );
  7. insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '语文', 78.0);
  8. insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '数学', 88.0);
  9. insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '英语', 98.0);
  10. insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '语文', 89.0);
  11. insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '数学', 76.0);
  12. insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '英语', 90.0);
  13. insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '语文', 99.0);
  14. insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '数学', 66.0);
  15. insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '英语', 91.0);
name subject score
张三 语文 78.0
张三 数学 88.0
张三 英语 98.0
李四 语文 89.0
李四 数学 76.0
李四 英语 90.0
王五 语文 99.0
王五 数学 66.0
王五 英语 91.0

解法

  1. select ss.name,
  2.        max(case ss.subject
  3.                when '语文' then
  4.                    ss.score
  5.            end) 语文,
  6.        max(case ss.subject
  7.                when '数学' then
  8.                    ss.score
  9.            end) 数学,
  10.        max(case ss.subject
  11.                when '英语' then
  12.                    ss.score
  13.            end) 英语
  14. from student_score ss
  15. group by ss.name;
name 语文 数学 英语
张三 78.0 88.0 98.0
李四 89.0 76.0 90.0
王五 99.0 66.0 91.0

列转行

解法

  1. SELECT
  2.     NAME,
  3.     '语文' AS subject ,
  4.     MAX("语文") AS score
  5. FROM student1 GROUP BY NAME
  6. UNION
  7. SELECT
  8.     NAME,
  9.     '数学' AS subject ,
  10.     MAX("数学") AS score
  11. FROM student1 GROUP BY NAME
  12. UNION
  13. SELECT
  14.     NAME,
  15.     '英语' AS subject ,
  16.     MAX("英语") AS score
  17. FROM student1 GROUP BY NAME

limit的使用

limit 5 取前5条记录 limit 0, 5 取前5条记录 limit 10, 15 取11-25条记录 LIMIT 2 OFFSET 3 取4-5条记录

多表连接查询，条件在on和where后面的区别

  1. 当条件放在on后面的时候，后面的条件作为连接条件，如果不满足，则显示左的一条数据，右表的查询字段都用null填充；
  2. 当条件放在where后面的时候，后面的条件作为查询条件，是在连表以后，再过滤的条件，只要不满足就过滤掉，所以查询出来的结果为空。

    sql语句书写和执行顺序

    ```sql — 书写顺序 select distinct from join on where group by having order by limit;

— 执行顺序 from on join where group by having select distinct order by limit;

  2. ---
  4. <a name="MC5gQ"></a>
  5. # 关联查询
  6. > 数据库脚本延续上面的脚本
  8. <a name="qfWpI"></a>
  9. ## 连接类型
  11. 1. 等值连接：两个表中包含相同的列名
  12. 1. 非等值连接：两个表中没有相同的列名，但是某一个列在另一张表的列的范围之中
  13. 1. 外连接：
  14.    1. 左外连接：将左表的数据全部显示
  15.    1. 右外连接：将右表的数据全部显示
  16. 4. 自连接：将一张表当成不同的表来看待，自己关联自己
  17. <a name="qIqzQ"></a>
  18. ## 语法类型
  19. <a name="o9M1O"></a>
  20. ### 92语法
  21. 多张表的连接条件会方法where子句中，同时where需要对表进行条件过滤
  22. <a name="fMuw4"></a>
  23. #### 等值连接
  24. ```sql
  25. select * from emp e,dept d where e.deptno = d.deptno;--等值连接

�左外连接

  1. select * from emp e,dept d where e.deptno = d.deptno(+);--mysql没有此写法

�右外连接

  1. select * from emp e,dept d where e.deptno(+) = d.deptno;--mysql没有此写法

自连接

  1. select e.ename,m.ename from emp e,emp m where e.mgr = m.empno;

笛卡尔积

  1. select * from emp e,dept d; --当关联多表，但是不写关联条件

99语法

CROSS JOIN

等同于92语法的笛卡尔积

  1. select * from emp cross join dept;

NATURAL JOIN

等同于92语法的等值连接，但是不需要写连接条件，mysql会自动找到相同的列名。如果找不到则做笛卡尔积

  1. select * from emp e natural join dept d ;

LEFT OUTER JOIN / LEFT JOIN

查出左表的所有数据，右表不存在则null填充

  1. select * from emp e left outer join dept d on e.deptno = d.deptno;

RIGHT OUTER JOIN / RIGHT JOIN

查出右表的所有数据，左表不存在则null填充

  1. select * from emp e right outer join dept  d on e.deptno = d.deptno;

INNER JOIN

两张表的连接查询，只会查询出有匹配记录的数据

  1. select * from emp e inner join dept d on e.deptno = d.deptno;
  2. select * from emp e join dept d on e.deptno = d.deptno;

USING子句

除了使用on作为连接条件外，还可以使用using关键字，此时depton不属于任何

  1. select deptno from emp e join dept d using(deptno);

语句练习

题目：检索雇员名字、所在单位、薪水等级

92写法

  1. select e.ENAME, d.DNAME, s.GRADE 
  2. from emp e, dept d, salgrade s 
  3. where e.DEPTNO = d.DEPTNO 
  4.   and e.SAL between s.LOSAL and s.HISAL;

99写法

  1. select e.ename, d.loc, sg.grade
  2. from emp e
  3.          join dept d
  4.               on e.deptno = d.deptno
  5.          join salgrade sg
  6.               on e.sal between sg.losal and sg.hisal;

子查询

嵌套在其它sql语句中执行的完整sql语句

标量子查询

返回单一的值 外部查询使用=、>、<、>=、<=和<>符号进行比较判断

  1. SELECT * FROM article WHERE uid = (SELECT uid FROM user WHERE status=1 ORDER BY uid DESC LIMIT 1)
  2. SELECT * FROM t1 WHERE column1 = (SELECT MAX(column2) FROM t2)
  3. SELECT * FROM article AS t WHERE 2 = (SELECT COUNT(*) FROM article WHERE article.uid = t.uid)

列子查询

返回的结果集是N行1列 外部查询使用IN、ANY、SOME和ALL符号进行比较判断

  1. SELECT * FROM article WHERE uid IN(SELECT uid FROM user WHERE status=1)
  2. SELECT s1 FROM table1 WHERE s1 > ANY (SELECT s2 FROM table2)
  3. SELECT s1 FROM table1 WHERE s1 > ALL (SELECT s2 FROM table2)

行子查询

返回的结果集是1行N列

  1. SELECT * FROM table1 WHERE (1,2) = (SELECT column1, column2 FROM table2)
  2. 注：(1,2) 等同于 row(1,2)
  3. SELECT * FROM article WHERE (title,content,uid) = (SELECT title,content,uid FROM blog WHERE bid=2)

表子查询

返回的结果集是N行N列

  1. SELECT * FROM article WHERE (title,content,uid) IN (SELECT title,content,uid FROM blog)

语法练习

有哪些人的薪水是在整个雇员的平均薪水之上的

  1. # 1 雇员的平均薪水
  2. select avg(SAL) from emp;
  3. # 哪些人的薪水 在 1 之上
  4. select * from emp where SAL > (select avg(SAL) from emp);

�我们要查在雇员中有哪些人是经理人

  1. # 我们要查在雇员中有哪些人是经理人
  2. # 1 先查出经理编号
  3. select distinct MGR from emp;
  4. # 查在雇员中有哪些人的编号in 1
  5. select * from emp where EMPNO in (select distinct MGR from emp);

每个部门平均薪水的等级

  1. # 每个部门平均薪水的等级
  2. # 1 部门平均薪水
  3. select avg(SAL), DEPTNO from emp group by DEPTNO;
  4. # 每个部门 1 的等级
  5. --92语法
  6. select s.GRADE, d.DEPTNO from salgrade s ,(select avg(SAL) as avgSal, DEPTNO from emp group by DEPTNO) d
  7. where d.avgSal between s.LOSAL and s.HISAL;
  9. -- 99语法
  10. select s.GRADE, d.DEPTNO from salgrade s join 
  11.     (select avg(SAL) as avgSal, DEPTNO from emp group by DEPTNO) d 
  12.         on d.avgSal between s.LOSAL and s.HISAL;

每个部门平均的薪水等级

  1. # 1 求部门的平均薪水
  2. select avg(SAL) from emp group by DEPTNO;
  4. # 2 求部门平均薪水（实际意思就是说，求三个部门的各自平均薪水，再做一次平均）
  5. select avg(e.avgSal) from (select avg(SAL) as avgSal from emp group by DEPTNO) e;
  7. # 3 求 2 的等级
  8. select s.GRADE from  salgrade s join (select avg(e.avgSal) avaSall from (select avg(SAL) as avgSal from emp group by DEPTNO) e) ee
  9. on ee.avaSall between s.LOSAL and s.HISAL;

求平均薪水最高的部门的部门编号

# 1 部门的平均薪水
select avg(SAL) avgSal, e.DEPTNO from emp e group by e.DEPTNO;

# 2 求 1 最高的薪水
select ee.DEPTNO,ee.avgSal from
         (select avg(SAL) avgSal, e.DEPTNO from emp e group by e.DEPTNO) ee
order by ee.avgSal desc limit 1;

# 3 求 2 最高的部门编号
select d.DEPTNO from dept d join
    (select ee.DEPTNO,ee.avgSal from
                 (select avg(SAL) avgSal, e.DEPTNO from emp e group by e.DEPTNO) ee
    order by ee.avgSal desc limit 1) eee
on d.DEPTNO = eee.DEPTNO;

�sql语句

表结构

–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

�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);

�测试题

查询”01”课程比”02”课程成绩高的学生的信息及课程分数

select a.* ,b.s_score as 01_score,c.s_score as 02_score from
    student a
        join score b on a.s_id=b.s_id and b.c_id='01'
        left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id is null where b.s_score>c.s_score;

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

select s.s_id, s_name, a.avgScore from Student s 
    join (select sc.s_id, avg(sc.s_score) avgScore from Score sc group by sc.s_id) a 
        on s.s_id = a.s_id and a.avgScore >= 60;

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

-- 解法一
select s.s_id, s.s_name, a.countCid, a.sumScore from Student s join (select count(s2.c_id) countCid, sum(s2.s_score) sumScore, s2.s_id  from Score s2 group by s2.s_id) a
on s.s_id = a.s_id group by s.s_id;

-- 解法二
select s.s_id, s.s_name,count(s2.c_id), sum(s2.s_score) from Student s join Score s2 on s.s_id = s2.s_id group by s.s_id, s.s_name;

查询”李”姓老师的数量

select count(*) from Teacher where t_name like '李%';

查询学过”张三”老师授课的同学的信息

select a.* from
    student a
        join score b on a.s_id=b.s_id where b.c_id in(
    select c_id from course where t_id =(
        select t_id from teacher where t_name = '张三'));

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 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';

查询学过编号为”01”但是没有学过编号为”02”的课程的同学的信息

--解法1
select s.*,sc1.s_score,sc2.s_score from student s
        left join (select * from score where c_id = '01') sc1 on s.s_id = sc1.s_id
        left join (select * from score where c_id = '02') sc2 on s.s_id = sc2.s_id
where sc1.c_id = '01' and sc2.c_id is null;

--解法2
select s.* from Student s
    where s.s_id in (select sc1.s_id from Score sc1 where sc1.c_id='01') and 
          s.s_id not in (select sc2.s_id from Score sc2 where sc2.c_id='02');

查询没有学全所有课程的同学的信息

-- 解法1
select s.* from Student s left join
    (select count(c_id) countCid, s_id from Score group by s_id) a
        on s.s_id = a.s_id where a.countCid < (select count(*) from Course);
-- 解法2
select s.* from 
    student s where s.s_id in(
        select s_id from score where s_id not in(
            select a.s_id from score a 
                join score b on a.s_id = b.s_id and b.c_id='02'
                join score c on a.s_id = c.s_id and c.c_id='03'
            where a.c_id='01'))

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 * from student where s_id in(
    select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
);

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 a.* from student a where a.s_id in(
    select distinct s_id from score where s_id!='01' group by s_id
    having count(a.s_id)=(select count(1) from score where s_id='01'));

查询没学过”张三”老师讲授的任一门课程的学生姓名

select * from Student s where s.s_id not in (
    select sc.s_id from Score sc where sc.c_id = (
        select c.c_id from Course c join Teacher t on c.t_id = t.t_id where t.t_name = '张三') group by s_id);

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 * from Student s where s.s_id in 
        (select sc.s_id from Score sc where sc.s_score < 60 group by sc.s_id having count(sc.c_id) >= 2);

检索”01”课程分数小于60，按分数降序排列的学生信息

select * from Student s where s.s_id in(select sc.s_id from Score sc where sc.s_score < 60 group by sc.s_id);

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

--解法1
select sc1.s_score, sc1.s_id, sc1.c_id ,a.avgScore from Score sc1
    left join (select sc.s_id, avg(sc.s_score) avgScore from Score sc group by sc.s_id) a
        on a.s_id = sc1.s_id
order by a.avgScore desc;

--解法2
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
       (select s_score from score where s_id=a.s_id and c_id='02') as 数学,
       (select s_score from score where s_id=a.s_id and c_id='03') as 英语,
       round(avg(s_score),2) as 平均分 from score a  GROUP BY a.s_id ORDER BY 平均分 DESC;

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 a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
       ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
       ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
       ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
       ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name

查询不同老师所教不同课程平均分从高到低显示

select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
          left join score b on a.c_id=b.c_id
          left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;

统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比

select distinct f.c_name,
                a.c_id,
                b.`85-100`,
                b.百分比,
                c.`70-85`,
                c.百分比,
                d.`60-70`,
                d.百分比,
                e.`0-60`,
                e.百分比
from score a
    left join (select sc1.c_id,
                      sum(case when sc1.s_score >= 85 and sc1.s_score <= 100 then 1 else 0 end) '85-100',
                      round(sum(case when sc1.s_score >= 85 and sc1.s_score <= 100 then 1 else 0 end) / count(*) * 100,2) '百分比'
               from Score sc1 group by sc1.c_id) b on a.c_id = b.c_id
    left join (select sc1.c_id,
                      sum(case when sc1.s_score >= 70 and sc1.s_score <= 85 then 1 else 0 end) '70-85',
                      round(sum(case when sc1.s_score >= 70 and sc1.s_score <= 85 then 1 else 0 end) / count(*) * 100,2) '百分比'
               from Score sc1 group by sc1.c_id) c on a.c_id = c.c_id
    left join (select sc1.c_id,
                      sum(case when sc1.s_score >= 60 and sc1.s_score <= 70 then 1 else 0 end) '60-70',
                      round(sum(case when sc1.s_score >= 60 and sc1.s_score <= 70 then 1 else 0 end) / count(*) * 100,2) '百分比'
               from Score sc1 group by sc1.c_id) d on a.c_id = d.c_id
    left join (select sc1.c_id,
                      sum(case when sc1.s_score >= 0 and sc1.s_score <= 60 then 1 else 0 end) '0-60',
                      round(sum(case when sc1.s_score >= 0 and sc1.s_score <= 60 then 1 else 0 end) / count(*) * 100,2) '百分比'
               from Score sc1 group by sc1.c_id) e on a.c_id = e.c_id

         left join course f on a.c_id = f.c_id;

查询各科成绩前三名的记录*

select a.s_id,a.c_id,a.s_score from score a
                                        left join score b on a.c_id = b.c_id and a.s_score<b.s_score
group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
ORDER BY a.c_id,a.s_score DESC

查询每门课程被选修的学生数

select c_id,count(s_id) from score a GROUP BY c_id

查询出只有两门课程的全部学生的学号和姓名

--解法1
select sc.s_id, count(sc.c_id) a from Score sc group by sc.s_id having a = 2;

--解法2
select s_id,s_name from student where s_id in(
                select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

查询男生、女生人数

select s_sex,COUNT(s_sex) as 人数  from student GROUP BY s_sex;

查询同名同性学生名单，并统计同名人数

select a.s_name, a.s_sex, count(*) from student a JOIN student b 
    on a.s_id != b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
GROUP BY a.s_name, a.s_sex

查询1990年出生的学生名单

select s_name from student where s_birth like '1990%';

查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

select sc.c_id, avg(sc.s_score) avgScore from Score sc group by  sc.c_id order by avgScore, sc.c_id asc;

查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩

select s.s_id, s.s_name, avg(sc.s_score) avgScore from Student s left join Score sc on s.s_id  = sc.s_id group by s.s_id, s.s_name having avgScore > 85;

查询课程名称为”数学”，且分数低于60的学生姓名和分数

select a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=(
    select c_id from course where c_name ='数学') and b.s_score<60;

查询所有学生的课程及分数情况

select s.s_name, s.s_name,
       sum(case c.c_name when '语文' then sc.s_score else 0 end) 语文,
       sum(case c.c_name when '数学' then sc.s_score else 0 end) 数学,
       sum(case c.c_name when '英语' then sc.s_score else 0 end) 英语,
       SUM(sc.s_score) as  '总分'
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
group by s.s_name, s.s_name;

查询任何一门课程成绩在70分以上的姓名、课程名称和分数

SELECT
    st2.s_id,
    st2.s_name,
    c2.c_name,
    sc2.s_score
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 st2.s_id IN(
    SELECT
        st.s_id
    FROM student st
             LEFT JOIN score sc ON sc.s_id=st.s_id
    GROUP BY st.s_id HAVING MIN(sc.s_score)>=70)
ORDER BY s_id;

查询不及格的课程

select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id where a.s_score<60;

查询课程编号为01且课程成绩在80分以上的学生的学号和姓名

select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
where a.c_id = '01' and a.s_score>80

求每门课程的学生人数

select sc.s_id, count(sc.s_id) from Score sc group by sc.s_id;

查询选修”张三”老师所授课程的学生中，成绩最高的学生信息及其成绩

select s.*, sc.s_score 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 c.t_id = t.t_id
where t.t_name='张三' order by sc.s_score desc limit 1;

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

select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score

查询每门功课成绩最好的前两名*

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

统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列

select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC

检索至少选修两门课程的学生学号

select sc.s_id, count(sc.c_id) from Score sc group by sc.s_id having count(sc.c_id) >= 2;

查询选修了全部课程的学生信息

select * from student where s_id in(        
            select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))

查询各学生的年龄

select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -
                (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
from student;

查询本周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
    select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
    select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

查询下周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)

查询本月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

查询下月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

查询分组内最新的一条数据

select * from (select RECEIVE_OBJECT_TYPE 接口类型, RECEIVE_OBJECT_VALUE 接口值, MSG_CONTENT 接口入参, RETURN_PARAM 接口出参,
               STATE 执行状态, ERR_INFO 错误信息,EXECUTE_END_DATE 执行结束时间,OP_ID 操作员
               from ord_inter_msg_571_202206
               where OP_ID = 20001609
               and DONE_CODE = 57120220603094151158297
               group by 接口值,执行结束时间
               order by 执行结束时间 desc) temp group by  temp.接口值;

               select RECEIVE_OBJECT_TYPE 接口类型, RECEIVE_OBJECT_VALUE 接口值, MSG_CONTENT 接口入参, RETURN_PARAM 接口出参,
               STATE 执行状态, ERR_INFO 错误信息,EXECUTE_END_DATE 执行结束时间,OP_ID 操作员 from (select *
                                                                                 from ord_inter_msg_571_202206
                                                                                 where OP_ID = 20001609
                                                                                 and DONE_CODE = 57120220603094151158297
                                                                                 having 1
                                                                                 order by EXECUTE_END_DATE desc) as temp
 group by 接口值

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