MySQL经典50道练习题

/*
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 Source Server Type    : MySQL
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 Target Server Version : 50527
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 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 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 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)

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