MySQL45个常用语句练习
本文最后更新于：2 个月前
MySQL45个常用语句练习
前置知识
关于各种连接的解释
https://blog.csdn.net/qq_36501591/article/details/116234694
数据准备
create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
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('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2017-12-30' , '女');
insert into Student values('12' , '赵六' , '2017-01-01' , '女');
insert into Student values('13' , '孙七' , '2018-01-01' , '女');
create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
练习 — 45 个sql练习题
-- 1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
SELECT
s.*,
a.score AS score_01,
b.score AS score_02
FROM
student s
JOIN sc a ON s.SId = a.SId
AND a.CId = '01'
JOIN sc b ON s.SId = b.SId
AND b.CId = '02'
WHERE
a.score > b.score;
SELECT
s.*,
a.score,
b.score
FROM
student s
LEFT JOIN ( SELECT * FROM sc WHERE CId = '01' ) a ON s.sid = a.sid
INNER JOIN ( SELECT * FROM sc WHERE CId = '02' ) b ON a.SId = b.SId
WHERE
a.score > b.score
-- 1.1 查询同时存在" 01 "课程和" 02 "课程的情况
SELECT
*
FROM
( SELECT * FROM sc WHERE CId = '01' ) a
JOIN ( SELECT * FROM sc WHERE CId = '02' ) b ON a.SId = b.SId SELECT
*
FROM
sc a
JOIN sc b ON a.sid = b.sid
AND a.cid = '01'
AND b.cid = '02'
-- 1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )
SELECT
*
FROM
( SELECT * FROM sc WHERE cid = '01' ) a
LEFT JOIN ( SELECT * FROM sc WHERE cid = '02' ) b ON a.sid = b.sid
-- 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
SELECT
*
FROM
(
SELECT
*
FROM
sc
WHERE
sid NOT IN ( SELECT sid FROM sc WHERE cid = '01' )) a
INNER JOIN sc b ON a.sid = b.sid
WHERE
b.cid = '02';
-- 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
SELECT
*
FROM
student s
INNER JOIN ( SELECT sid, AVG( score ) AS avg_score FROM sc GROUP BY sid HAVING avg_score > 60 ) c ON s.sid = c.sid
-- 3.查询在 SC 表存在成绩的学生信息
SET @@GLOBAL.sql_mode = 'STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_ENGINE_SUBSTITUTION';
SELECT
b.*
FROM
sc a
LEFT JOIN student b ON a.sid = b.sid
GROUP BY
b.sid
-- 4.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(没成绩的显示为 null ) 4.1 查有成绩的学生信息
SELECT
a.sid,
a.sname,
b.cons,
b.sum_score
FROM
student a
LEFT JOIN ( SELECT sid, count( cid ) AS cons, sum( score ) AS sum_score FROM sc GROUP BY sid ) b ON a.sid = b.sid
-- 5.查询「李」姓老师的数量
SELECT
COUNT(*)
FROM
teacher
WHERE
tname LIKE "李%"
-- 6.查询学过「张三」老师授课的同学的信息
SELECT
s.*
FROM
student s
JOIN sc c ON s.sid = c.sid
WHERE
cid = ( SELECT a.cid FROM course a JOIN teacher b ON a.tid = b.tid WHERE b.tname = "张三" ) SELECT
b.*
FROM
sc a
LEFT JOIN student b ON a.sid = b.sid
WHERE
cid IN ( SELECT cid FROM course WHERE tid =( SELECT tid FROM teacher WHERE tname = '张三' ) )
GROUP BY
sid;
-- 7.查询没有学全所有课程的同学的信息
SELECT
*
FROM
student a
JOIN sc b ON a.sid = b.sid
GROUP BY
a.sid
HAVING
COUNT( b.cid ) < ( SELECT COUNT(*) FROM course )
-- 8.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
JOIN sc c ON s.sid = c.sid
WHERE
c.cid IN ( SELECT cid FROM sc WHERE sid = 1 ) SELECT DISTINCT
b.*
FROM
sc a
LEFT JOIN student b ON a.sid = b.sid
WHERE
cid IN ( SELECT cid FROM sc WHERE sid = '01' );
-- 9.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
-- 要没有学习01号同学学习课程外的其他课程， 而且学习的课程数量相同
SELECT
*
FROM
( SELECT * FROM sc WHERE cid IN ( SELECT cid FROM sc WHERE sid = '01' ) AND sid != '01' ) a
LEFT JOIN student s ON a.sid = s.sid
GROUP BY
a.sid
HAVING
count( cid ) = ( SELECT count( 1 ) FROM sc WHERE sid = '01' )
-- 10.查询没学过"张三"老师讲授的任一门课程的学生姓名
-- 查询 张三 老师教的课程cid
-- 先查询学过 张三老师课程的学生
SELECT
sid
FROM
sc a
LEFT JOIN course c ON a.cid = c.cid
JOIN teacher d ON c.tid = d.tid
WHERE
d.tname = "张三" SELECT
sname
FROM
student
WHERE
sid NOT IN (
SELECT
sid
FROM
sc a
LEFT JOIN course c ON a.cid = c.cid
JOIN teacher d ON c.tid = d.tid
WHERE
d.tname = "张三"
)
-- 11.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
a.sid,
a.sname
FROM
student a
JOIN ( SELECT * FROM sc WHERE score < 60 GROUP BY sid HAVING count( 1 ) >= 2 ) b ON a.sid = b.sid SELECT
a.sid,
a.sname,
avg( b.score ) AS avg_score
FROM
student a
LEFT JOIN sc b ON a.sid = b.sid
WHERE
b.score < 60 GROUP BY a.sid HAVING count( a.sid ) >= 2
-- 12.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 a
JOIN sc b ON a.sid = b.sid
WHERE
b.cid = '01'
AND b.score < 60
ORDER BY
b.score DESC
-- 13.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
SELECT
c.sid,
c.sname,
c.cid,
c.score,
d.avg_score
FROM
(
SELECT
a.sid,
a.sname,
b.cid,
b.score
FROM
student a
LEFT JOIN sc b ON a.sid = b.sid
) c
LEFT JOIN ( SELECT sid, avg( score ) AS avg_score FROM sc GROUP BY sid ) d ON c.sid = d.sid
ORDER BY
d.avg_score DESC SELECT
c.sid,
c.cid,
c.score,
d.avg_sco
FROM
(
SELECT
a.sid,
b.cid,
b.score
FROM
student a
LEFT JOIN sc b ON a.sid = b.sid
) c
LEFT JOIN ( SELECT sid, avg( a.score ) AS avg_sco FROM sc a GROUP BY a.sid ) d ON c.sid = d.sid
ORDER BY
avg_sco DESC;
-- 14.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 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
avg( score ) AS avg_score
FROM
sc
GROUP BY
cid SELECT
max( score ) AS avg_score
FROM
sc
GROUP BY
cid SELECT
min( score ) AS avg_score
FROM
sc
GROUP BY
cid SELECT
cid,
max( score ) AS max_score,
min( score ) AS min_score,
avg( score ) AS avg_score,
count(*) AS cons,
sum(
IF
( score >= 60, 1, 0 )) / count( 1 ) AS jige,
sum(
IF
( score >= 70 AND score < 80, 1, 0 )) / count( 1 ) AS zhongdeng,
sum(
IF
( score >= 80 AND score < 90, 1, 0 )) / count( 1 ) AS youliang,
sum(
IF
( score >= 90, 1, 0 )) / count( 1 ) AS youxiu
FROM
sc
GROUP BY
cid DESC SELECT
cid,
max( score ) AS max_sc,
min( score ) AS min_sc,
avg( score ) AS avg_sc,
count( 1 ) AS cons,
sum(
IF
( score >= 60, 1, 0 ))/ count( 1 ) AS jige,
sum(
IF
( 70 <= score AND score <= 80, 1, 0 ))/ count( 1 ) AS zd,
sum(
IF
( 80 <= score AND score <= 90, 1, 0 ))/ count( 1 ) AS yl,
sum(
IF
( 90 >= score, 1, 0 ))/ count( 1 ) AS yx
FROM
sc
GROUP BY
cid;
-- 15.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺
SELECT
sid,
cid,
score,
@rank := @rank + 1 AS rn
FROM
sc,(
SELECT
@rank := 0
) AS t
ORDER BY
score DESC;
-- 16.查询学生的总成绩，并进行排名，总分重复时保留名次空缺 16.1 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺
SELECT
s.*,
CASE
WHEN @sco = scos THEN
'' ELSE @rank := @rank + 1
END AS rn,
@sco := scos
FROM
( SELECT sid, sum( score ) AS scos FROM sc GROUP BY sid ORDER BY scos DESC ) s,
( SELECT @rank := 0, @sco := NULL ) AS t
-- 17.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]，[60-0] 及所占百分比
SELECT
cid,
max( score ) AS max_sc,
min( score ) AS min_sc,
avg( score ) AS avg_sc,
sum(
IF
( score >= 60, 1, 0 ))/ count( 1 ) AS jige,
sum(
IF
( 70 <= score AND score <= 80, 1, 0 ))/ count( 1 ) AS zd,
sum(
IF
( 80 <= score AND score <= 90, 1, 0 ))/ count( 1 ) AS yl,
sum(
IF
( 90 >= score, 1, 0 ))/ count( 1 ) AS yx
FROM
sc
GROUP BY
cid;
SELECT
a.cid,
b.cname,
max( a.score ) AS max_sc,
min( a.score ) AS min_sc,
avg( a.score ) AS avg_sc,
sum(
IF
( a.score < 60, 1, 0 )) AS bjg,
sum(
IF
( a.score >= 60 AND a.score < 70, 1, 0 )) AS jg,
sum(
IF
( 70 <= a.score AND a.score < 85, 1, 0 )) AS zd,
sum(
IF
( 85 <= a.score AND a.score <= 100, 1, 0 )) AS yl,
sum(
IF
( a.score < 60, 1, 0 )) / count( 1 ) AS bjg_p,
sum(
IF
( a.score >= 60 AND a.score < 70, 1, 0 ))/ count( 1 ) AS jg_p,
sum(
IF
( 70 <= a.score AND a.score < 85, 1, 0 ))/ count( 1 ) AS zd_p,
sum(
IF
( 85 <= a.score AND a.score <= 100, 1, 0 ))/ count( 1 ) AS yl_p
FROM
sc a
JOIN course b ON a.cid = b.cid
GROUP BY
a.cid
-- 18.查询各科成绩前三名的记录
-- 转换思路，前三名转化为若大于此成绩的数量少于3即为前三名
SELECT
*
FROM
sc a
WHERE
( SELECT count( 1 ) FROM sc b WHERE a.cid = b.cid AND b.score > a.score ) < 3
-- 19.查询每门课程被选修的学生数
SELECT
cid,
count( 1 ) AS count_
FROM
sc
GROUP BY
cid
-- 20.查询出只选修两门课程的学生学号和姓名
SELECT
b.sid,
b.sname
FROM
sc a
JOIN student b ON a.sid = b.sid
GROUP BY
a.sid
HAVING
count( 1 ) = 2 SELECT
student.SId,
student.Sname
FROM
sc,
student
WHERE
student.SId = sc.SId
GROUP BY
sc.SId
HAVING
count(*)= 2
-- 21.查询男生、女生人数
SELECT
count( 1 )
FROM
student
GROUP BY
ssex SELECT
sum(
IF
( ssex = "男", 1, 0 )) AS man,
sum(
IF
( ssex = "女", 1, 0 )) AS woman
FROM
student
-- 22.查询名字中含有「风」字的学生信息
SELECT
*
FROM
student
WHERE
sname LIKE "%风%"
-- 23.查询同名同性学生名单，并统计同名人数
SELECT
a.*,
count(*) AS sum
FROM
student a
LEFT JOIN student b ON a.sname = b.sname
WHERE
a.sid != b.sid
-- 24.查询 1990 年出生的学生名单
SELECT
*
FROM
student
WHERE
YEAR ( sage ) = '1990'
-- 25.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
cid,
avg( score ) AS avg_score
FROM
sc
GROUP BY
cid
ORDER BY
avg_score DESC,
cid
-- 26.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
b.sid,
b.sname,
avg( a.score ) AS avg_score
FROM
sc a
JOIN student b ON a.sid = b.sid
GROUP BY
a.sid
HAVING
avg_score >= 85
-- 27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数
SELECT
b.sname,
a.score
FROM
sc a
JOIN student b ON a.sid = b.sid
JOIN course c ON a.cid = c.cid
AND c.cname = "数学"
WHERE
a.score < 60
-- 28.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）
SELECT
*
FROM
student a
LEFT JOIN sc b ON a.sid = b.sid
-- 29.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
*
FROM
student a
LEFT JOIN sc b ON a.sid = b.sid
JOIN course c ON b.cid = c.cid
WHERE
b.score > 70
-- 30.查询不及格的课程
SELECT
*
FROM
sc a
JOIN course b ON a.cid = b.cid
WHERE
a.score < 60
GROUP BY
b.cid
-- 31.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
b.sid,
b.sname
FROM
sc a
JOIN student b ON a.sid = b.sid
WHERE
a.score > 80
AND a.cid = '01' SELECT
student.SId,
student.Sname
FROM
student,
sc
WHERE
sc.CId = '01'
AND student.SId = sc.SId
AND sc.score > 80
-- 32.求每门课程的学生人数
SELECT
b.cid,
b.cname,
count( 1 ) AS cons
FROM
sc a
JOIN course b ON a.cid = b.cid
GROUP BY
cid
-- 33.假设成绩不重复，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
b.*,
c.cname,
a.score
FROM
sc a
JOIN student b ON a.sid = b.sid
JOIN course c ON c.cid = a.cid
JOIN teacher d ON c.tid = d.tid
WHERE
d.tname = "张三"
ORDER BY
a.score DESC
LIMIT 0,
1
-- 34.假设成绩有重复的情况下，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
-- 35.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
*
FROM
sc a
INNER JOIN sc b ON a.sid = b.sid
AND a.score = b.score
AND a.cid != b.cid
GROUP BY
a.sid,
a.cid SELECT
*
FROM
sc a
INNER JOIN sc b ON a.sid = b.sid
WHERE
a.cid != b.cid
AND a.score = b.score
GROUP BY
a.sid,
a.cid
-- 36.查询每门功成绩最好的前两名
SELECT
*
FROM
sc a
WHERE
( SELECT count( 1 ) FROM sc b WHERE a.cid = b.cid AND b.score > a.score ) <= 1 SELECT
*
FROM
sc a
WHERE
( SELECT count( 1 ) FROM sc b WHERE a.cid = b.cid AND b.score > a.score ) <= 1;
-- 37.统计每门课程的学生选修人数（超过 5 人的课程才统计）。
SELECT cid, count( 1 ) FROM sc GROUP BY cid HAVING count( 1 ) > 5
-- 38.检索至少选修两门课程的学生学号
SELECT
sid
FROM
sc
GROUP BY
sid
HAVING
count( 1 ) >= 2
-- 39.查询选修了全部课程的学生信息
SELECT
sid
FROM
sc
GROUP BY
sid
HAVING
count( 1 ) = ( SELECT count( 1 ) FROM course )
-- 40.查询各学生的年龄，只按年份来算
SELECT
*,
YEAR (
now()) - YEAR ( sage ) AS age
FROM
student -- 41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
SELECT
*,
CASE
WHEN substr( sage, 6, 5 )< substr( now(), 6, 5 ) THEN
YEAR (
now())- YEAR ( sage )
WHEN substr( sage, 6, 5 )>= substr( now(), 6, 5 ) THEN
YEAR (
now())- YEAR ( sage ) - 1
END AS age
FROM
student;
-- 42.查询本周过生日的学生
SELECT
YEARWEEK(
CURDATE()) -- 202411
SELECT
SUBSTR( YEARWEEK( CURDATE()), 5, 2 ) -- 11
SELECT
*,
substr( YEARWEEK( student.Sage ), 5, 2 ) AS birth_week,
substr( YEARWEEK( CURDATE()), 5, 2 ) AS now_week
FROM
student
WHERE
substr( YEARWEEK( student.Sage ), 5, 2 )= substr( YEARWEEK( CURDATE()), 5, 2 );
-- 43.查询下周过生日的学生
SELECT
*,
substr( YEARWEEK( student.Sage ), 5, 2 ) AS birth_week,
substr( YEARWEEK( CURDATE()), 5, 2 ) AS now_week
FROM
student
WHERE
substr( YEARWEEK( student.Sage ), 5, 2 )= substr( YEARWEEK( CURDATE()), 5, 2 )+ 1;
-- 44.查询本月过生日的学生
SELECT
*,
MONTH ( sage ) AS birth_month,
MONTH (
now()) AS now_month
FROM
student
WHERE
MONTH ( sage )= MONTH (
now())
-- 45.查询下月过生日的学生
SELECT
*,
MONTH ( sage ) AS birth_month,
MONTH (
now()) AS now_month
FROM
student
WHERE
MONTH ( sage )= MONTH (
now()) + 1本博客所有文章除特别声明外，均采用 CC BY-SA 4.0 协议 ，转载请注明出处！
