MySQL经典50道练习题

YuanZhi.w YuanZhi.w 博客等级 墨衍会员 · AI 创作全网分发 墨衍智能分发 本文由作者通过墨衍一键同步至各平台 1分发平台 700+累计阅读 699平均单平台 已同步平台 优快云 微信公众号 微博 知乎 掘金 百家号 博客园 抖音 小红书 你的文章也可以这样分发 墨衍会员 ¥399起/年，写一次发全网 我也要 码龄7年
关注
原创 最新推荐文章于 2026-08-09 09:51:33 发布 699 阅读 AI 写同款· GEO 优化 ›
MySQL基础练习题及答案（表的简单查询）-02 一、查询"李"姓老师的数量 思路：在老师（teacher）表中查找老师名字并筛选出以‘李’开头的，再使用count()函数求出个数，最后重命名搜索出来的这个列名为姓"李"的老师数量 SELECT COUNT(t_name) ‘"李"姓老师的数量’ FROM teacher WHERE t_name LIKE ‘李%’ 二、查询名字中含有"风"字的学生信息 思路：搜索学生（student）表中学生信息并筛选出名字中含有“风”的学生信息 SELECT s_name FROM student WHERE s_n 阅读详情 
/*
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MySQL
 Source Server Version : 50527
 Source Host           : localhost:3306
 Source Schema         : wuxian2

 Target Server Type    : MySQL
 Target Server Version : 50527
 File Encoding         : 65001

 Date: 19/09/2022 14:57:48
*/

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course`  (
  `c_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `c_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `t_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL,
  PRIMARY KEY (`c_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Compact;

-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES ('01', '语文', '02');
INSERT INTO `course` VALUES ('02', '数学', '01');
INSERT INTO `course` VALUES ('03', '英语', '03');

SET FOREIGN_KEY_CHECKS = 1;

/*
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MySQL
 Source Server Version : 50527
 Source Host           : localhost:3306
 Source Schema         : wuxian2

 Target Server Type    : MySQL
 Target Server Version : 50527
 File Encoding         : 65001

 Date: 19/09/2022 14:58:07
*/

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher`  (
  `t_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `t_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  PRIMARY KEY (`t_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Compact;

-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES ('01', '张三');
INSERT INTO `teacher` VALUES ('02', '李四');
INSERT INTO `teacher` VALUES ('03', '王五');

SET FOREIGN_KEY_CHECKS = 1;

/*
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MySQL
 Source Server Version : 50527
 Source Host           : localhost:3306
 Source Schema         : wuxian2

 Target Server Type    : MySQL
 Target Server Version : 50527
 File Encoding         : 65001

 Date: 19/09/2022 14:58:02
*/

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student`  (
  `s_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `s_name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `s_birth` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `s_sex` varchar(10) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  PRIMARY KEY (`s_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Compact;

-- ----------------------------
-- Records of student
-- ----------------------------
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');

SET FOREIGN_KEY_CHECKS = 1;

/*
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MySQL
 Source Server Version : 50527
 Source Host           : localhost:3306
 Source Schema         : wuxian2

 Target Server Type    : MySQL
 Target Server Version : 50527
 File Encoding         : 65001

 Date: 19/09/2022 14:57:57
*/

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
-- Table structure for score
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score`  (
  `s_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `c_id` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_general_ci NOT NULL DEFAULT '',
  `s_score` int(3) NULL DEFAULT NULL,
  PRIMARY KEY (`s_id`, `c_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci ROW_FORMAT = Compact;

-- ----------------------------
-- Records of score
-- ----------------------------
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);

SET FOREIGN_KEY_CHECKS = 1;

– 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
#解1
SELECT st.,ss.
FROM student st LEFT JOIN (SELECT s1.s_id,s1.c_id, s1.s_score
FROM (SELECT s_id,c_id,s_score FROM score WHERE c_id = 01) s1,(SELECT s_id,c_id,s_score FROM score WHERE c_id = 02) s2
WHERE s1.s_id = s2.s_id
AND s1.s_score > s2.s_score
) ss
ON st.s_id = ss.s_id
WHERE ss.s_score is not NULL

#解2
SELECT *
FROM student st JOIN score sc
ON st.s_id = sc.s_id AND sc.c_id = ‘01’
LEFT JOIN score scc
ON st.s_id = scc.s_id AND scc.c_id = ‘02’
WHERE sc.s_score > scc.s_score

– 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
#解1 同上
#解2
SELECT *
FROM student st JOIN score s1
ON st.s_id = s1.s_id AND s1.c_id = ‘01’
LEFT JOIN score s2
ON st.s_id = s2.s_id and s2.c_id = ‘02’
WHERE s1.s_score < s2.s_score

– 3、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
#解1
SELECT st.,avg.
FROM student st,(SELECT s_id,AVG(s_score) avg_score FROM score GROUP BY s_id HAVING AVG(s_score) > 60) avg
WHERE st.s_id = avg.s_id
#解2
SELECT st.s_id,st.s_name,AVG(sc.s_score) avg_score
FROM student st JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id,st.s_name
HAVING AVG(sc.s_score) >= 60

– 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩(包括有成绩的和无成绩的)
#解1
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,avg.s_id avgI,avg.avg_score avgS
FROM student st JOIN (SELECT s_id,AVG(s_score) avg_score FROM score GROUP BY s_id HAVING AVG(s_score) < 60) avg
ON st.s_id = avg.s_id
UNION
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_id avgI,sc.s_score avgS FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id WHERE sc.s_id IS NULL

#解2
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_id avgI,AVG(sc.s_score) avgS
FROM student st JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id,st.s_name
HAVING AVG(sc.s_score) < 60
UNION
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_id avgI,sc.s_score avgS FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id WHERE sc.s_id IS NULL

#解3
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score
from student b left join score a
on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name
HAVING ROUND(AVG(a.s_score),2)<60
union
select a.s_id,a.s_name,0 as avg_score
from student a
where a.s_id not in (select distinct s_id from score);

– 5、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 st.s_id,st.s_name,COUNT(sc.c_id) “选课总数”,SUM(sc.s_score) “总成绩”
FROM student st LEFT JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id,st.s_name

– 6、查询"李"姓老师的数量
SELECT COUNT(*)
FROM teacher
WHERE t_name LIKE ‘李%’;

– 7、查询学过"张三"老师授课的同学的信息
SELECT *
FROM student st LEFT JOIN score sc
ON st.s_id = sc.s_id
WHERE c_id IN (SELECT co.c_id
FROM course co
WHERE co.t_id IN (SELECT t_id
FROM teacher
WHERE t_name = ‘张三’))

– 8、查询没学过"张三"老师授课的同学的信息
#解1
SELECT *
FROM student c
WHERE c.s_id NOT IN (
SELECT st.s_id
FROM student st LEFT JOIN score sc
ON st.s_id = sc.s_id
WHERE c_id IN (
SELECT co.c_id
FROM course co
WHERE co.t_id IN (
SELECT t_id
FROM teacher
WHERE t_name = ‘张三’
)
)
)

– 9、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
#解1
SELECT *
FROM (SELECT dt.*
FROM student dt LEFT JOIN score sc
ON dt.s_id = sc.s_id
WHERE sc.c_id = ‘01’
) s1
JOIN
(SELECT dt.*
FROM student dt LEFT JOIN score sc
ON dt.s_id = sc.s_id
WHERE sc.c_id = ‘02’
) s2
WHERE s1.s_id = s2.s_id
#别人写的
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’;
– 10、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1 跟别人写的一样
SELECT *
FROM student st
WHERE st.s_id IN (SELECT s_id FROM score WHERE c_id=‘01’)
AND st.s_id NOT IN (SELECT s_id FROM score WHERE c_id=‘02’)
– 11、查询没有学全所有课程的同学的信息 TODO
#解1
SELECT s1.s_id
FROM score s1 JOIN score s2 ON s1.s_id = s2.s_id AND s2.c_id = ‘02’
JOIN score s3 ON s1.s_id = s3.s_id AND s3.c_id = ‘03’
WHERE s1.c_id = 01

#别人写的
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’))

– 12、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
#解1
SELECT *
FROM student
WHERE s_id IN (
SELECT DISTINCT s_id
FROM score
WHERE c_id IN (
SELECT c_id
FROM score
WHERE s_id = ‘01’
)
)
AND s_id <> ‘01’

#别人写的
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’
)
);

– 13、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
#解1
SELECT *
FROM student st JOIN (
SELECT s_id,COUNT(c_id) munber
FROM score
GROUP BY s_id
HAVING munber = (
SELECT COUNT(c_id) munber
FROM score
WHERE s_id = ‘01’
)
) s
ON st.s_id = s.s_id
WHERE st.s_id != ‘01’

#别人写的
select a.*
from student a
where a.s_id in(
select distinct s_id
from score
where s_id!=‘01’
and c_id in(select c_id from score where s_id=‘01’)
group by s_id
having count(1)=(select count(1) from score where s_id=‘01’));

– 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
#解1 跟别人写的一样
SELECT *
FROM student
WHERE s_id NOT IN (
SELECT s_id
FROM score
WHERE c_id = (
SELECT c_id
FROM course
WHERE t_id = (
SELECT t_id
FROM teacher
WHERE t_name = ‘张三’
)
)
)

– 15、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
#解1
SELECT st.,avg_score.avg_s
FROM student st JOIN (
SELECT s_id,AVG(s_score) avg_s,COUNT()
FROM score
GROUP BY s_id
HAVING COUNT(*) >= 2
) avg_score
ON st.s_id = avg_score.s_id
WHERE avg_score.avg_s < 60

#别人写的
select a.s_id,a.s_name,ROUND(AVG(b.s_score))
from student a left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id
from score
where s_score<60
GROUP BY s_id
having count(1)>=2
)
GROUP BY a.s_id,a.s_name

– 16、检索"01"课程分数小于60，按分数降序排列的学生信息
#解1
SELECT *
FROM student st JOIN (
SELECT s_id,c_id,s_score
FROM score
WHERE c_id = ‘01’
AND s_score < 60
GROUP BY s_score DESC
) sc
ON st.s_id = sc.s_id

#别人写的
select a.*,b.c_id,b.s_score
from student a,score b
where a.s_id = b.s_id
and b.c_id=‘01’
and b.s_score < 60
ORDER BY b.s_score DESC;

– 17、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1
SELECT st.s_name,avg_s.*
FROM student st JOIN (
SELECT s_id,
(SELECT s_score FROM score s2 WHERE s1.s_id = s2.s_id AND s2.c_id = ‘01’) AS 语文,
(SELECT s_score FROM score s3 WHERE s1.s_id = s3.s_id AND s3.c_id = ‘02’) AS 数学,
(SELECT s_score FROM score s4 WHERE s1.s_id = s4.s_id AND s4.c_id = ‘03’) AS 英语,
AVG(s_score) AS 平均分
FROM score s1
GROUP BY s1.s_id
ORDER BY 平均分 DESC
) avg_s
ON st.s_id = avg_s.s_id

#别人写的
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;

– 18.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
#解1
SELECT
s.c_id “课程ID”,
c.c_name “课程名称”,
MAX(s_score) “最高分”,
MIN(s_score) “最低分”,
AVG(s_score) “平均分”,
#及格率，中等率，优良率，优秀率 略… 不知道咋算
FROM score s LEFT JOIN course c
ON s.c_id = c.c_id
GROUP BY c.c_id

#别人写的
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

– 19、按各科成绩进行排序，并显示排名(实现不完全)
– mysql没有rank函数
#解1 有问题
SELECT c_id,avgs
FROM (
SELECT c_id,s_score avgs FROM score WHERE c_id = ‘01’
UNION ALL
SELECT c_id,s_score avgs FROM score WHERE c_id = ‘02’
UNION ALL
SELECT c_id,s_score avgs FROM score WHERE c_id = ‘03’
) avg1
ORDER BY avgs DESC
#和解1一样的效果-哈哈哈
SELECT c_id,s_score
FROM score
ORDER BY s_score DESC

#别人写的-看不懂
select a.s_id,a.c_id,
@i:=@i +1 as i保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id=‘01’ GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id=‘02’ GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id=‘03’ GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s

– 20、查询学生的总成绩并进行排名
select a.s_id,
@i:=@i+1 as i,
@k:=(case when @score=a.sum_score then @k else @i end) as rank,
@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
(select @k:=0,@i:=0,@score:=0)s

– 21、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,a.t_name,a.c_id, AVG(s.s_score) avgs
FROM score s LEFT JOIN (SELECT c.c_id,c.c_name,t.t_name,t.t_id FROM course c JOIN teacher t ON c.t_id = t.t_id) a
ON s.c_id = a.c_id
GROUP BY s.c_id
ORDER BY avgs DESC

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;
– 22、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

#别人写的
select d.,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id=‘01’
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id=‘02’
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id=‘03’
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3;

– 23、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

#别人写的
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 c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as 85-100,
ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count()),2) as 百分比
from score GROUP BY c_id)b on a.c_id=b.c_id
left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as 70-85,
ROUND(100(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count()),2) as 百分比
from score GROUP BY c_id)c on a.c_id=c.c_id
left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as 60-70,
ROUND(100(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count()),2) as 百分比
from score GROUP BY c_id)d on a.c_id=d.c_id
left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as 0-60,
ROUND(100(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id

– 24、查询学生平均成绩及其名次
SELECT st.s_id,st.s_name,avgss.avgs
FROM student st JOIN (
SELECT s_id,AVG(s_score) avgs
FROM score
GROUP BY s_id
ORDER BY avgs DESC
) avgss
ON st.s_id = avgss.s_id

    select a.s_id,
            @i:=@i+1 as '不保留空缺排名',
            @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
            @avg_score:=avg_s as '平均分'
    from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;

– 25、查询各科成绩前三名的记录
SELECT s1.*
FROM score s1 LEFT JOIN score s2
ON s1.c_id = s2.c_id AND s1.s_score < s2.s_score
GROUP BY s1.s_id,s1.c_id,s1.s_score
HAVING COUNT(s2.s_id) < 3
ORDER BY s1.c_id,s1.s_score DESC

        -- 1.选出b表比a表成绩大的所有组
        -- 2.选出比当前id成绩大的 小于三个的

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)❤️
ORDER BY a.c_id,a.s_score DESC

– 26、查询每门课程被选修的学生数
SELECT c_id,COUNT(s_id)
FROM score
GROUP BY c_id

– 27、查询出只有两门课程的全部学生的学号和姓名
SELECT *
FROM student
WHERE s_id IN (
SELECT s_id
FROM score
GROUP BY s_id
HAVING COUNT(c_id) = 2
)

– 28、查询男生、女生人数
SELECT s_sex,COUNT(s_sex)
FROM student
GROUP BY s_sex

– 29、查询名字中含有"风"字的学生信息
SELECT *
FROM student
WHERE s_name LIKE ‘%风%’

– 30、查询同名同性学生名单，并统计同名人数
SELECT s1.*
FROM student s1 JOIN student s2
WHERE s1.s_id != s2.s_id
AND s1.s_name = s2.s_name
AND s1.s_sex = s2.s_sex

    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

– 31、查询1990年出生的学生名单
SELECT *
FROM student
WHERE s_birth LIKE ‘1990%’

SELECT *
FROM student
WHERE s_birth BETWEEN ‘1990’ AND ‘1991’

– 32、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 c_id,AVG(s_score)
FROM score
GROUP BY c_id
ORDER BY s_score DESC,c_id ASC

– 33、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)
FROM student s LEFT JOIN score sc
ON s.s_id = sc.s_id
GROUP BY sc.s_id
HAVING AVG(sc.s_score) > 85

– 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
SELECT st.s_name,sc.s_score
FROM student st RIGHT JOIN score sc
ON st.s_id = sc.s_id
WHERE sc.c_id = (SELECT c_id FROM course WHERE c_name = ‘数学’)
AND sc.s_score < 60

– 35、查询所有学生的课程及分数情况；
SELECT st.s_id,st.s_name,SUM(CASE co.c_name WHEN ‘语文’ THEN sc.s_score ELSE 0 END) “语文”,
SUM(CASE co.c_name WHEN ‘数学’ THEN sc.s_score ELSE NULL END) “数学”,
SUM(CASE co.c_name WHEN ‘英语’ THEN sc.s_score END) “英语”
FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id
LEFT JOIN course co ON sc.c_id = co.c_id
GROUP BY st.s_id,st.s_name

– 36、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 s_name,c_name,s_score
FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id
LEFT JOIN course co ON sc.c_id = co.c_id
WHERE s_score >= 70

– 37、查询不及格的课程
SELECT st.s_name,co.c_name,sc.s_score
FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id
LEFT JOIN course co ON sc.c_id = co.c_id
WHERE sc.s_score < 60

– 38、查询课程编号为01且课程成绩在70分以上的学生的学号和姓名；
SELECT st.s_id,st.s_name,sc.s_score
FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id
WHERE sc.c_id = ‘01’
AND sc.s_score >= 70

– 39、求每门课程的学生人数
SELECT COUNT(*)
FROM score
GROUP BY c_id

– 40、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
#解2
SELECT st.s_id,st.s_name,sc.s_score
FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id
LEFT JOIN course co ON sc.c_id = co.c_id
LEFT JOIN teacher te ON co.t_id = te.t_id
WHERE te.t_name = ‘张三’
ORDER BY sc.s_score DESC
LIMIT 0,1

#解1
SELECT st.s_id,st.s_name,MAX(tmp.s_score )
FROM student st,(
SELECT s_id ,c_id,s_score
FROM score
WHERE c_id = (
SELECT c_id
FROM course
WHERE t_id = (
SELECT t_id
FROM teacher
WHERE t_name = ‘张三’
)
)
) tmp
WHERE st.s_id = tmp.s_id

– 41、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 s1.*
FROM score s1,score s2
WHERE s1.c_id != s2.c_id
AND s1.s_score = s2.s_score

select 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

– 42、查询每门功成绩最好的前两名
– 牛逼的写法
SELECT s1.s_id,s1.c_id,s1.s_score
FROM score s1
WHERE (
SELECT COUNT(*)
FROM score s2
WHERE s1.c_id = s2.c_id
AND s2.s_score >= s1.s_score
) <= 2
ORDER BY s1.c_id

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

– 43、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

– 44、检索至少选修两门课程的学生学号
select s_id,count(*) as sel
from score
GROUP BY s_id
HAVING sel>=2

– 45、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))

– 46、查询各学生的年龄
– 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 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;

– 47、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'))

– 48、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)

– 49、查询本月过生日的学生

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

– 50、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)

– 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
#解1
SELECT st.,ss.
FROM student st LEFT JOIN (SELECT s1.s_id,s1.c_id, s1.s_score
FROM (SELECT s_id,c_id,s_score FROM score WHERE c_id = 01) s1,(SELECT s_id,c_id,s_score FROM score WHERE c_id = 02) s2
WHERE s1.s_id = s2.s_id
AND s1.s_score > s2.s_score
) ss
ON st.s_id = ss.s_id
WHERE ss.s_score is not NULL

#解2
SELECT *
FROM student st JOIN score sc
ON st.s_id = sc.s_id AND sc.c_id = ‘01’
LEFT JOIN score scc
ON st.s_id = scc.s_id AND scc.c_id = ‘02’
WHERE sc.s_score > scc.s_score

– 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
#解1 同上
#解2
SELECT *
FROM student st JOIN score s1
ON st.s_id = s1.s_id AND s1.c_id = ‘01’
LEFT JOIN score s2
ON st.s_id = s2.s_id and s2.c_id = ‘02’
WHERE s1.s_score < s2.s_score

– 3、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
#解1
SELECT st.,avg.
FROM student st,(SELECT s_id,AVG(s_score) avg_score FROM score GROUP BY s_id HAVING AVG(s_score) > 60) avg
WHERE st.s_id = avg.s_id
#解2
SELECT st.s_id,st.s_name,AVG(sc.s_score) avg_score
FROM student st JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id,st.s_name
HAVING AVG(sc.s_score) >= 60

– 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩(包括有成绩的和无成绩的)
#解1
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,avg.s_id avgI,avg.avg_score avgS
FROM student st JOIN (SELECT s_id,AVG(s_score) avg_score FROM score GROUP BY s_id HAVING AVG(s_score) < 60) avg
ON st.s_id = avg.s_id
UNION
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_id avgI,sc.s_score avgS FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id WHERE sc.s_id IS NULL

#解2
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_id avgI,AVG(sc.s_score) avgS
FROM student st JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id,st.s_name
HAVING AVG(sc.s_score) < 60
UNION
SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_id avgI,sc.s_score avgS FROM student st LEFT JOIN score sc ON st.s_id = sc.s_id WHERE sc.s_id IS NULL

#解3
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score
from student b left join score a
on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name
HAVING ROUND(AVG(a.s_score),2)<60
union
select a.s_id,a.s_name,0 as avg_score
from student a
where a.s_id not in (select distinct s_id from score);

– 5、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 st.s_id,st.s_name,COUNT(sc.c_id) “选课总数”,SUM(sc.s_score) “总成绩”
FROM student st LEFT JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id,st.s_name

– 6、查询"李"姓老师的数量
SELECT COUNT(*)
FROM teacher
WHERE t_name LIKE ‘李%’;

– 7、查询学过"张三"老师授课的同学的信息
SELECT *
FROM student st LEFT JOIN score sc
ON st.s_id = sc.s_id
WHERE c_id IN (SELECT co.c_id
FROM course co
WHERE co.t_id IN (SELECT t_id
FROM teacher
WHERE t_name = ‘张三’))

– 8、查询没学过"张三"老师授课的同学的信息
SELECT *
FROM student c
WHERE c.s_id NOT IN (
SELECT st.s_id
FROM student st LEFT JOIN score sc
ON st.s_id = sc.s_id
WHERE c_id IN (
SELECT co.c_id
FROM course co
WHERE co.t_id IN (
SELECT t_id
FROM teacher
WHERE t_name = ‘张三’
)
)
)

– 9、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 dt.*
FROM student dt LEFT JOIN score sc
ON dt.s_id = sc.s_id
WHERE sc.c_id = ‘01’ sc.c_id = ‘02’
GROUP BY dt.s_id

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’;

– 10、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

select a.* from
student a
where a.s_id in (select s_id from score where c_id=‘01’ ) and a.s_id not in(select s_id from score where c_id=‘02’)

– 11、查询没有学全所有课程的同学的信息

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’))

– 12、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’)
);

– 13、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’ and c_id in(select c_id from score where s_id=‘01’)
group by s_id
having count(1)=(select count(1) from score where s_id=‘01’));
– 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select a.s_name from student a where a.s_id not in (
select s_id from score where c_id =
(select c_id from course where t_id =(
select t_id from teacher where t_name = ‘张三’))
group by s_id);

– 15、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.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a
left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2)
GROUP BY a.s_id,a.s_name

– 16、检索"01"课程分数小于60，按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from
student a,score b
where a.s_id = b.s_id and b.c_id=‘01’ and b.s_score<60 ORDER BY b.s_score DESC;

– 17、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 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;

– 18.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

– 19、按各科成绩进行排序，并显示排名(实现不完全)
– mysql没有rank函数
select a.s_id,a.c_id,
@i:=@i +1 as i保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id=‘01’ GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id=‘02’ GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id=‘03’ GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s

– 20、查询学生的总成绩并进行排名
select a.s_id,
@i:=@i+1 as i,
@k:=(case when @score=a.sum_score then @k else @i end) as rank,
@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
(select @k:=0,@i:=0,@score:=0)s

– 21、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;

– 22、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

        select d.*,c.排名,c.s_score,c.c_id from (
            select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'    
        )c
        left join student d on c.s_id=d.s_id
        where 排名 BETWEEN 2 AND 3
        UNION
        select d.*,c.排名,c.s_score,c.c_id from (
            select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'    
        )c
        left join student d on c.s_id=d.s_id
        where 排名 BETWEEN 2 AND 3
        UNION
        select d.*,c.排名,c.s_score,c.c_id from (
            select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'    
        )c
        left join student d on c.s_id=d.s_id
        where 排名 BETWEEN 2 AND 3;

– 23、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 c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
                                        ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
                            from score GROUP BY c_id)b on a.c_id=b.c_id
            left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
                                        ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
                            from score GROUP BY c_id)c on a.c_id=c.c_id
            left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
                                        ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
                            from score GROUP BY c_id)d on a.c_id=d.c_id
            left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
                                        ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
                            from score GROUP BY c_id)e on a.c_id=e.c_id
            left join course f on a.c_id = f.c_id

– 24、查询学生平均成绩及其名次

    select a.s_id,
            @i:=@i+1 as '不保留空缺排名',
            @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
            @avg_score:=avg_s as '平均分'
    from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;

– 25、查询各科成绩前三名的记录
– 1.选出b表比a表成绩大的所有组
– 2.选出比当前id成绩大的 小于三个的
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)❤️
ORDER BY a.c_id,a.s_score DESC

– 26、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

– 27、查询出只有两门课程的全部学生的学号和姓名
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);

– 28、查询男生、女生人数
select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex

– 29、查询名字中含有"风"字的学生信息

    select * from student where s_name like '%风%';

– 30、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

– 31、查询1990年出生的学生名单

    select s_name from student where s_birth like '1990%'

– 32、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 c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC

– 33、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 a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
    left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85

– 34、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

– 35、查询所有学生的课程及分数情况；

    select a.s_id,a.s_name,
                SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
                SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
                SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
                SUM(b.s_score) as  '总分'
    from student a left join score b on a.s_id = b.s_id 
    left join course c on b.c_id = c.c_id 
    GROUP BY a.s_id,a.s_name

– 36、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 a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
left join student a on a.s_id=c.s_id where c.s_score>=70

– 37、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

–38、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

– 39、求每门课程的学生人数
select count(*) from score GROUP BY c_id;

– 40、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

    -- 查询老师id   
    select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
    -- 查询最高分（可能有相同分数）
    select MAX(s_score) from score where c_id='02'
    -- 查询信息
    select a.*,b.s_score,b.c_id,c.c_name from student a
        LEFT JOIN score b on a.s_id = b.s_id
        LEFT JOIN course c on b.c_id=c.c_id
        where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
        and b.s_score in (select MAX(s_score) from score where c_id='02')

– 41、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

– 42、查询每门功成绩最好的前两名
– 牛逼的写法
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

– 43、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

– 44、检索至少选修两门课程的学生学号
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2

– 45、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))

–46、查询各学生的年龄
– 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 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;

– 47、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'))

– 48、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)

– 49、查询本月过生日的学生

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

– 50、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)

收录于
DB

当前文章被收录于：

DB
2 篇文章 0 人学习

Mysql、Redis、ES、HBase等相关数据库学习

订阅专栏 查看详情
标签
#mysql #数据库 #java
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
MySQL学习：50道经典题目（含参考答案）(更新中） 首先根据course_id将成绩表分为两张表，筛选出" 1 "课程比" 2 "课程成绩高的学生，再与学生表关联。1. 查询" 1 "课程比" 2 "课程成绩高的学生的信息（学号、姓名、性别、出生日期）及课程分数。利用自连接，统计比当前分数高的分数有几个，排名就是几加1。使用case when语句，多个排序条件时用逗号分隔。利用课程数量，计算出所有课程数和每个学生的选课数量。先筛选与" 1 "号同学选课数量相同的同学，再剔除。10. 查询没有学全所有课程的同学的信息。选了" 1 "号同学没选的课的同学。 阅读详情

相关推荐

[数据库]50道经典SQL练习题,使用MySQL5.7解答

[数据库]50道经典SQL练习题,使用MySQL5.7解答 网上流传较广的50道SQL训练, win10系统,MySQL5.7版本解答,全部实现 尽可能的使用了相对简洁和直接的解法, 当然也肯定会存在其它解答方法欢迎一起学习的朋友进行讨论和解法优化 数据表介绍 --1.学生表 Student(SId,Sname,Sage,Ssex) --SId 学生编号,Sname 学生姓名,Sage 出生年...

关注后厂程序员 2314

SQL学习笔记之SQL查询练习题1

（网络搜集） 0x00 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....

dengxiangbao3167的博客 603

MySQL练习题及答案-11页

MySQL练习题及答案

MySQL之经典50道题

MySQL之经典50道题目一、创建数据表并插入数据 一、创建数据表并插入数据 1、学生表 Student(s_id,s_name,s_birth,s_sex) ：学生编号、姓名、年月、性别 -- 1、学生表 -- Student(s_id,s_name,s_birth,s_sex) ：学生编号、姓名、年月、性别 CREATE TABLE IF NOT EXISTS `Student` ( `s_id` VARCHAR (20), `s_name` VARCHAR (20) NOT NULL DEFA

qq_40216188的博客 1万+

50道MySQL经典练习题

这个是网上流传的50道SQL练习题，最近拿来练习，刚做完，这里把我做的答案给大家做个参考，如有错误，还请告知。 另外，做题的时候先不要看答案（答案不唯一，只要满足要求即可），要有自己的一个思考过程，这样做出来才能达到最好的训练效果。（感觉做这种题是会上瘾的……） 我使用的Mysql版本是5.7.19。SQL语句可能会因数据库系统的不同会有少许差异。 这里放上MySQL语法执行的先后顺序： 7:SELECT 查询列表 1:FROM 表 2:连接类型 JOIN 表2 3:ON 连接条件 4:WHERE 筛选条件 5:GROUP BY 分组 6:HAV

VC Spyglass实战指南：从零开始掌握CDC约束配置与调试

本文详细介绍了如何使用VC Spyglass进行CDC约束配置与调试，从基础环境搭建到典型CDC问题调试实战，帮助开发者快速掌握跨时钟域验证技术。通过TCL脚本和SDC约束的深度解析，提升CDC验证效率，避免常见设计风险。

weixin_26757925的博客 294

50道经典MySQL练习题（含解答）

本文摘要：本篇主要分享50道经典MySQL练习题（含解答）。

请你喝杯Java 1万+

AI Agent核心架构解析：从Tool、Skill到MCP协议的智能体构建实战

在人工智能领域，大语言模型（LLM）赋予了AI自主规划与决策的能力，推动了智能体（Agent）技术的演进。其核心原理在于构建一个具备感知、规划、行动与反思能力的自主系统。这一架构的技术价值在于将AI从被动应答升级为主动执行复杂任务的工作伙伴，显著提升了自动化水平与问题解决效率。应用场景广泛覆盖了市场分析、自动化办公、智能客服与垂直领域集成等。本文聚焦于实现这一能力的关键组件：基础工具（Tool）、组合技能（Skill）与标准化协议（MCP）。其中，Skill通过封装多个Tool形成可复用的高级能力，而MCP

chuannaoxuan4674的博客 318

MySQL经典50道练习题及全网最详细解析

1.学表Student(SId,Sname,Sage,Ssex)SId学编号Sname学姓名Sage出年Ssex学性别2.课程表Course(CId,Cname,TId)CId课程编号Cname课程名称TId教师编号3.教师表Teacher(TId,Tname)TId教师编号Tname教师姓名4.成绩表SC(SId,CId,score)SId学编号CId课程编号score分数。...

m0_67401417的博客 2459

JOOQ用法和实例

jOOQ，是一个ORM框架，利用其生成的Java代码和流畅的API，可以快速构建有类型约束的安全的SQL语句。jOOQ的核心优势是可以将数据库表结构映射为Java类，包含表的基本描述和所有表字段。通过jOOQ提供的API，配合生成的Java代码，可以很方便的进行数据库操作生成的Java代码字段类型是根据数据库映射成的Java类型，在进行设置和查询操作时，因为是Java代码，都会有强类型校验，所以对于数据的输入，是天然安全的，极大的减少了SQL注入的风险。

m0_67403076的博客 1376

手把手配置：用Synopsys VDK为英飞凌TC4x系列MCU做虚拟原型开发

本文详细介绍了如何使用Synopsys VDK为英飞凌TC4x系列MCU进行虚拟原型开发，涵盖开发环境搭建、关键外设虚拟化配置、AUTOSAR栈集成及虚拟调试与性能调优全流程。通过虚拟原型技术，开发者可提前6-12个月启动软件开发，显著提升汽车电子项目的开发效率。

weixin_30606461的博客 290

mysql经典50道练习题

问题及描述： –1.学生表 Student(SID,Sname,Sage,Ssex) --SID 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别 –2.课程表 Course(CID,Cname,TID) --CID --课程编号,Cname 课程名称,TID 教师编号 –3.教师表 Teacher(TID,Tname) --TID 教师编号,Tname 教师姓名 –4.成绩......

weixin_43930865的博客 2万+

pnpm 卸载教程：彻底移除 pnpm 及全局缓存的方法

卸载 pnpm = 全局包 → CLI → 环境变量 → store 缓存只要按这个顺序来，基本不会留下任何坑。如果你后面还打算重新安装 pnpm / 切换 npm、yarn / 多包管理器共存，这些细节反而更重要，环境一旦干净，后续问题会少很多。

代码简单说 Vue、JAVA、PHP、Node.js 熟练运用，接口、架构、性能全搞定。 925

50道经典SQL练习题 MYSQL8.0实现

使用MYSQL8.0完成了“50道经典SQL练习题”，答案可能有误，欢迎指正。 参考了多位网友的答案 超经典SQL练习题，做完这些你的ＳＱＬ就过关了 50道SQL练习题 50道SQL练习题及答案与详细分析 准备数据 CREATE TABLE student( id VARCHAR(10), name VARCHAR(10), age DATETIME, sex VARCHAR(10...

博客 4829

基于C#的工业机器人上位机控制程序开发（搭配松下PLC+MC协议）

基于C#的工业机器人上位机控制程序开发，核心是上位机与松下PLC的MC协议通信和机器人控制的安全逻辑，而不是单纯的界面编写。在工控现场，程序的稳定性和安全性永远是第一位的——比如解耦通信逻辑、处理跨线程问题、添加安全互锁、记录操作日志，这些都是保证产线稳定运行的关键。新手在学习时，建议先跟着步骤搭建好环境，跑通基础的PLC通信和机器人控制功能，再逐步尝试进阶优化。数据模型化、通信封装化、逻辑模块化、UI解耦化。

这里只有干货：从 Modbus 通信到 MES 对接，从 YOLO 训练到工控机部署，带你搞定工业软件开发全流程。 1239

【AI智能体】AI 编程 + 飞书CLI + Skills 重塑开发新范式实战操作详解

AI 编程 + 飞书CLI + Skills 重塑开发新范式实战操作详解

congge 3665

mysql50到例题_MySQL最经典50道练习题

表名和字段学生表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。课程表Course(c_id,c_name,t_id)：课程编号、课程名称、教师编号。教师表Teacher(t_id,t_name)：教师编号、教师姓名。成绩表Score(s_id,c_id,s_score)：学生编号、课程编号、分数。建表语句# 建表# 学生表CREATE...

weixin_39923599的博客 345

Android 系统低内存lowmemorykiller关于阈值修改 minfreeAbsKbytes优化

在系统rom定制化开发中，在对于运行内存ram 1G和2G的设备中，对于内存使用高峰期时，minfreeAbsKbytes优化很关键 调整合理的minfree，让内存合理的使用很关键,接下来分析相关功能的实现

安卓兼职framework和app工程师的博客 524

MySQL50道经典练习题，自我练习及思路讲解【上】

由于一直在互联网开发项目，在优先高并发的前提下一般表的设计都是通过冗余字段、简化表之间依赖，尽量使用简单查询来提升DB吞吐量。但实际很多传统项目、内部应用、统计报表等系统由于并发量不大，需求逻辑复杂，仍然十分依赖关系型数据库范式设计，通过数据库强大的SQL处理能力实现复杂逻辑查询。所以对于SQL的编写使用十分重要。个人汇总了网上很流行的SQL练习题，基本上涵盖的SQL查询的各个方面, 搞懂了这50题，对于SQL的基本使用算是入门了。

xiaoxiaohen2的博客 903

android 蓝牙通讯demo

android 下蓝牙通讯demo，实现了发现蓝牙，使能蓝牙，通过蓝牙发送消息。

大中型超市收银系统(可直接运行)

光步前台收银系统系统功能：商品销售、商品退货、挂单、收银员权限、会员销售、单品取消、整单取消、商品查询、脱机销售、脱机数据上传，基本资料下载，开钱箱、小票打印、收款结算，支持多币种付款方式（现金、储值卡、礼券、微信等）,支持多种促销。运行环境：.net 4.0登陆信息用户名:1001 密码:123456内置商品信息，商品编码可输入00001000 - 00010000,测试效果相关功能键F5 结算 F9挂单 F11商品查询 F8退货 Insert单品取消 Del整单取消 F12开钱箱F2 会员 Ctrl+Z退出系统 Home 锁屏 相关源码待公开

MySQL50道经典_MySQL最经典50道练习题

表名和字段学生表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。课程表Course(c_id,c_name,t_id)：课程编号、课程名称、教师编号。教师表Teacher(t_id,t_name)：教师编号、教师姓名。成绩表Score(s_id,c_id,s_score)：学生编号、课程编号、分数。建表语句# 建表语句# 学生表CREA...

weixin_29363945的博客 372

精雕软件(JDpaint) 5.5 绿色版

精雕软件(JDpaint) 5.5 绿色版精雕软件(JDpaint) 5.5 绿色版精雕软件(JDpaint) 5.5 绿色版

MySQL练习：50道经典SQL练习题(1.4w字包全面)

此查询通过连接学生表和成绩表，找出了在"01"课程中成绩高于"02"课程的学生，并显示了他们的详细信息和两门课程的分数。此查询通过分组和条件筛选，找出了同时选修"01"和"02"课程的学生，并计算了他们在这两门课程的成绩。此查询通过连接学生表和成绩表，计算了每个学生的平均成绩，并筛选出平均成绩大于等于85的学生。此查询通过自连接成绩表，找出了不同课程但成绩相同的学生的学生编号、课程编号和学生成绩。此查询找出了未选修"01"课程但选修了"02"课程的学生，并显示了他们的成绩。

Argonaut_的博客 3586

MySql经典50道SQL练习题

SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别CId 课程编号,Cname 课程名称,TId 教师编号TId 教师编号,Tname 教师姓名SId 学生编号,CId 课程编号,score 分数。

weixin_55076626的博客 2319
上一篇： Msql生成随机数
下一篇： 大数据简介与概论
点赞 0
评论 0
书签 书签 3
YuanZhi.w
YuanZhi.w
博客等级 码龄7年 501粉丝 25原创
+ 关注 专栏目录 订阅专栏
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
表情包 插入表情
表情包 代码片
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付：10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

YuanZhi.w

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付：¥1
获取中
扫码支付

您的余额不足，请更换扫码支付或充值

打赏作者

实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明：

1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

余额充值