mysql笔试题大餐---1、组合查询方式及having
mysql笔试题大餐---1、组合查询方式及having
一、总结
一句话总结：
实践：我之前的mysql真的学的太浅了，这种情况下，依据实践（做题）才是唯一能把它学好的方式
学的暂时够了，以实践而学
项目中可以考虑多使用原生查询：简单，方便，查询效率高，记忆负担小，有用（考都是考这个）
1、查询考了100分的学生的信息（三种查询方式）？
#嵌套查询：小括号括住嵌套部分：select * from student s where s.s_id in (select ss_s_id from student_score where ss_score=100)
#普通多表查询：from后接多个表，where后接表相连条件：select * from student s,student_score ss where s.s_id=ss.ss_s_id and ss.ss_score=100
#链接查询：比如inner join ... on ...：select * from student s inner join student_score ss on s.s_id=ss.ss_s_id where ss.ss_score=100
2、group by如何使用？
1、注意group by的字段要被选出去：select 类别, sum(数量) as 数量之和 from A group by 类别
2、函数比如count，sum是作用在group by的那个字段上的，也就是group by选好的基础上的
select 类别, sum(数量) as 数量之和 from A group by 类别
3、查询每个学生分别考了几门课程（group by的实例）？
注意group by的字段为ss_s_id的字段，count作用的也是ss_s_id字段
链接查询：select student.s_name,count(ss_s_id) from student inner join student_score on student.s_id=student_score.ss_s_id group by ss_s_id;
普通多表查询：select s.s_name,count(ss.ss_s_id) from student s,student_score ss where s.s_id=ss.ss_s_id group by ss.ss_s_id
#链接查询 ###注意group by的字段为ss_s_id的字段，count作用的也是id字段 select student.s_name,count(ss_s_id) from student inner join student_score on student.s_id=student_score.ss_s_id group by ss_s_id; #普通多表查询 select s.s_name,count(ss.ss_s_id) from student s,student_score ss where s.s_id=ss.ss_s_id group by ss.ss_s_id
4、Group By 和 Order By一起怎么使用？
select 类别, sum(数量) AS 数量之和 from A group by 类别 order by sum(数量) desc
注意你想看到什么字段的排序结果，你就用什么字段排序：包括函数作用的group by的字段，比如order by count(ss_s_id) asc
查询每个学生分别考了几门课程，并且按课程数量从低到高排序：select student.s_name,count(ss_s_id) from student inner join student_score on student.s_id=student_score.ss_s_id group by ss_s_id order by count(ss_s_id) asc;
5、字段如何起别名？
空格：原始字段 别名：count(sc_s_id) num
mysql中的别名都是以空格来连接
select student.s_name,count(sc_s_id) num from student inner join student_score on student.s_id=student_score.sc_s_id group by sc_s_id order by num asc;
6、如何给数据库起别名？
空格：原始数据库名 数据库别名：from student s
mysql中的别名都是以空格来连接
select s.s_name,count(sc_s_id) num from student s inner join student_score sc on s.s_id=sc.sc_s_id group by sc_s_id order by num asc;
7、where后面的多个条件如何连接？
用脑子：直接and：where s.s_id=ss.ss_s_id and ss.ss_score=100
select * from student s,student_score ss where s.s_id=ss.ss_s_id and ss.ss_score=100
8、普通多表查询 和 链接查询 的区别？
普通多表查询：where s.s_id=ss.ss_s_id；直接from两表
链接查询：on s.s_id=ss.ss_s_id：inner join另外一个表
#普通多表查询 select * from student s,student_score ss where s.s_id=ss.ss_s_id and ss.ss_score=100 #链接查询 select * from student s inner join student_score ss on s.s_id=ss.ss_s_id where ss.ss_score=100
9、嵌套查询怎么写？
直接把嵌套的内容放到小括号里面即可
查询考了100分的学生的信息：select * from student s where s.s_id in (select ss_s_id from student_score where ss_score=100)
#查询考了100分的学生的信息 #嵌套查询 select * from student s where s.s_id in (select ss_s_id from student_score where ss_score=100)
10、多表链接查询？
多个inner join连在一起即可
select * from student s inner join student_score ss on s.s_id=ss.ss_s_id inner join course c on c.c_id=ss.ss_c_id order by s.s_name;
11、mysql中的符号说明（空格和逗号）？
并列用 逗号：select s.s_name,count(sc_s_id) num from
别名用 空格：select s.s_name,count(sc_s_id) num from
12、查询“01:牛顿力学”课程比“2:狂人日记”课程成绩高的所有学生的学号； (多条记录之间比较大小)？
1、多条记录之间比较大小，用的是嵌套查询：将嵌套查询的部分起别名：(select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1) a
2、嵌套查询出来的内容可以直接起别名后用点的方式查看：select a.ss_s_id from (select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1) a,
3、最后的比较条学生分数，又比较学生id：where a.ss_score>b.ss_score and a.ss_s_id=b.ss_s_id
4、嵌套查询部分如何起别名：from (select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1) a,
select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1 select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=2 select a.ss_s_id from (select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1) a,(select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=2) b where a.ss_score>b.ss_score and a.ss_s_id=b.ss_s_id
13、mysql中的having是什么意思？
和where一样选条件：和where一样都是选条件的，但是 因为having是从前筛选的字段再筛选，而where是从数据表中的字段直接进行的筛选的。
出错条件：select goods_price,goods_name from sw_goods having goods_price > 100：上面的having可以用的前提是我已经筛选出了goods_price字段，在这种情况下和where的效果是等效的，但是如果我没有select goods_price 就会报错！！
误区：不要错误的认为having和group by 必须配合使用。
14、having和where的区别是什么?
having是从前筛选的字段再筛选：上面的having可以用的前提是我已经筛选出了goods_price字段，在这种情况下和where的效果是等效的，但是如果我没有select goods_price 就会报错！！
where是从数据表中的字段直接进行的筛选的：
having是从前筛选的字段再筛选：select goods_price,goods_name from sw_goods having goods_price > 100：上面的having可以用的前提是我已经筛选出了goods_price字段，在这种情况下和where的效果是等效的，但是如果我没有select goods_price 就会报错！！ where是从数据表中的字段直接进行的筛选的：
15、where和having都可以使用的场景？
字段被筛选出来了：比如 select goods_price之后，where和having都可以对goods_price操作
select goods_price,goods_name from sw_goods where goods_price > 100 select goods_price,goods_name from sw_goods having goods_price > 100 解释：上面的having可以用的前提是我已经筛选出了goods_price字段，在这种情况下和where的效果是等效的，但是如果我没有select goods_price 就会报错！！因为having是从前筛选的字段再筛选，而where是从数据表中的字段直接进行的筛选的。
16、为什么having总是和group by搭配使用？
1、因为group by一定会保证having的字段被选
2、group by只能和having搭配：因为group by出来的东西数据库没有，比如avg(ss.ss_score)，所以不能用where
3、我猜having速度比where快：因为having是从前筛选的字段再筛选，而where是从数据表中的字段直接进行的筛选的。
#2、查询平均成绩大于60分的同学的学号和平均成绩； select ss.ss_s_id,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60;
17、如何复制 navicat 查询出来的结果？
选工具栏中 导出当前结果 即可：可以有多种导出方式
18、group by + having的使用场景是什么？
group by之后，要对依赖group by的字段（比如avg(ss.ss_score)）进行选择判断：比如 查询平均成绩大于60分的同学的学号和平均成绩；
select ss.ss_s_id,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60;
19、对学生分组查询的时候，能够选择select *么？
可以，不推荐：select ss.*,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60;
20、查询平均成绩大于60分的同学的学号和平均成绩？
平均成绩怎么弄（每个人学科数量不一样）：group by + having +avg：group by ss.ss_s_id having avg(ss.ss_score) > 60;
原理：group by将每个人的信息分组，avg将group by出来的每个人的分数平均，having来选这个平均分
#2、查询平均成绩大于60分的同学的学号和平均成绩； ##平均成绩怎么弄（每个人学科数量不一样）：group by + having +avg：group by ss.ss_s_id having avg(ss.ss_score) > 60; ##原理：group by将每个人的信息分组，avg将group by出来的每个人的分数平均，having来选这个平均分 #having select ss.ss_s_id,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60; select ss.*,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60;
二、对应数据库结构及sql
1、查询
select * from student; #查询每个学生的课程成绩-链接查询 select * from student inner join student_score on student.s_id=student_score.ss_s_id; #查询每个学生分别考了几门课程-链接查询 ###注意group by的字段为count的字段 select student.s_name,count(ss_s_id) from student inner join student_score on student.s_id=student_score.ss_s_id group by ss_s_id; #普通多表查询 select s.s_name,count(ss.ss_s_id) from student s,student_score ss where s.s_id=ss.ss_s_id group by ss.ss_s_id #查询每个学生分别考了几门课程，并且按课程数量从低到高排序 ###注意你想要什么字段，就用什么字段排序 select student.s_name,count(ss_s_id) from student inner join student_score on student.s_id=student_score.ss_s_id group by ss_s_id order by count(ss_s_id) asc; #字段别名 select student.s_name,count(ss_s_id) num from student inner join student_score on student.s_id=student_score.ss_s_id group by ss_s_id order by num asc; #数据库别名 select s.s_name,count(ss_s_id) num from student s inner join student_score ss on s.s_id=ss.ss_s_id group by ss_s_id order by num asc; #查询考了100分的学生的信息 #嵌套查询 select * from student s where s.s_id in (select ss_s_id from student_score where ss_score=100) #普通多表查询 select * from student s,student_score ss where s.s_id=ss.ss_s_id and ss.ss_score=100 #链接查询 select * from student s inner join student_score ss on s.s_id=ss.ss_s_id where ss.ss_score=100 #将学生的成绩 按照学生姓名排到一块 select * from student s inner join student_score ss on s.s_id=ss.ss_s_id inner join course c on c.c_id=ss.ss_c_id inner join teacher t on c.c_t_id=t.t_id order by s.s_id,c.c_id asc; #1、查询“01:牛顿力学”课程比“2:狂人日记”课程成绩高的所有学生的学号； (多条记录之间比较大小) ##多条记录之间比较大小，用的是嵌套查询：将嵌套查询的部分起别名 ##嵌套查询出来的内容可以直接用点的方式查看：select a.ss_s_id from (select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1) a, ##最后的比较条学生分数，又比较学生id：where a.ss_score>b.ss_score and a.ss_s_id=b.ss_s_id select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1 select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=2 select a.ss_s_id from (select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=1) a,(select ss.ss_s_id,ss.ss_score from student_score ss where ss.ss_c_id=2) b where a.ss_score>b.ss_score and a.ss_s_id=b.ss_s_id #2、查询平均成绩大于60分的同学的学号和平均成绩； ##平均成绩怎么弄（每个人学科数量不一样）：group by + having +avg：group by ss.ss_s_id having avg(ss.ss_score) > 60; ##原理：group by将每个人的信息分组，avg将group by出来的每个人的分数平均，having来选这个平均分 #having select ss.ss_s_id,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60; select ss.*,avg(ss.ss_score) from student_score ss group by ss.ss_s_id having avg(ss.ss_score) > 60;
2、数据库结构
/* Navicat Premium Data Transfer Source Server : localhost_3306 Source Server Type : MySQL Source Server Version : 50553 Source Host : localhost:3306 Source Schema : test_2019_06_10 Target Server Type : MySQL Target Server Version : 50553 File Encoding : 65001 Date: 10/06/2019 13:55:45 */ SET NAMES utf8mb4; SET FOREIGN_KEY_CHECKS = 0; -- ---------------------------- -- Table structure for course -- ---------------------------- DROP TABLE IF EXISTS `course`; CREATE TABLE `course` ( `c_id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT COMMENT '课程表id', `c_name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci DEFAULT NULL, `c_t_id` int(11) DEFAULT NULL COMMENT '课程对应的老师的id', PRIMARY KEY (`c_id`) USING BTREE ) ENGINE = MyISAM AUTO_INCREMENT = 5 CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic; -- ---------------------------- -- Records of course -- ---------------------------- INSERT INTO `course` VALUES (1, '牛顿力学', 1); INSERT INTO `course` VALUES (2, '狂人日记', 2); INSERT INTO `course` VALUES (3, '鲁迅作品集', 2); INSERT INTO `course` VALUES (4, '史记', 3); -- ---------------------------- -- Table structure for student -- ---------------------------- DROP TABLE IF EXISTS `student`; CREATE TABLE `student` ( `s_id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT COMMENT '学生表主键', `s_name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci DEFAULT NULL, `s_age` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci DEFAULT NULL, `s_sex` tinyint(2) DEFAULT NULL, PRIMARY KEY (`s_id`) USING BTREE ) ENGINE = MyISAM AUTO_INCREMENT = 4 CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic; -- ---------------------------- -- Records of student -- ---------------------------- INSERT INTO `student` VALUES (1, '张三', '12', 0); INSERT INTO `student` VALUES (2, '王丽丽', '16', 1); INSERT INTO `student` VALUES (3, '李四', '58', 0); -- ---------------------------- -- Table structure for student_score -- ---------------------------- DROP TABLE IF EXISTS `student_score`; CREATE TABLE `student_score` ( `ss_id` int(11) NOT NULL AUTO_INCREMENT, `ss_s_id` int(11) DEFAULT NULL, `ss_c_id` int(11) DEFAULT NULL, `ss_score` int(11) DEFAULT NULL, PRIMARY KEY (`ss_id`) USING BTREE ) ENGINE = MyISAM AUTO_INCREMENT = 7 CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Fixed; -- ---------------------------- -- Records of student_score -- ---------------------------- INSERT INTO `student_score` VALUES (1, 1, 1, 75); INSERT INTO `student_score` VALUES (2, 2, 1, 98); INSERT INTO `student_score` VALUES (3, 3, 3, 54); INSERT INTO `student_score` VALUES (4, 1, 4, 100); INSERT INTO `student_score` VALUES (5, 2, 2, 43); INSERT INTO `student_score` VALUES (6, 2, 3, 100); -- ---------------------------- -- Table structure for teacher -- ---------------------------- DROP TABLE IF EXISTS `teacher`; CREATE TABLE `teacher` ( `t_id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT, `t_name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci DEFAULT NULL, PRIMARY KEY (`t_id`) USING BTREE ) ENGINE = MyISAM AUTO_INCREMENT = 4 CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic; -- ---------------------------- -- Records of teacher -- ---------------------------- INSERT INTO `teacher` VALUES (1, '牛顿'); INSERT INTO `teacher` VALUES (2, '鲁迅'); INSERT INTO `teacher` VALUES (3, '司马迁'); SET FOREIGN_KEY_CHECKS = 1;
三、某硕笔试题mysql数据库部分（较为全面）
转自或参考：某硕笔试题mysql数据库部分（较为全面）
https://www.cnblogs.com/zjfjava/p/6012443.html
Student(S#,Sname,Sage,Ssex) 学生表 Course(C#,Cname,T#) 课程表 SC(S#,C#,score) 成绩表 Teacher(T#,Tname) 教师表 问题： 1、查询“001”课程比“002”课程成绩高的所有学生的学号； select a.S# from (select s#,score from SC where C#='001') a,(select s#,score from SC where C#='002') b where a.score>b.score and a.s#=b.s#; 2、查询平均成绩大于60分的同学的学号和平均成绩； select S#,avg(score) from sc group by S# having avg(score) >60; 3、查询所有同学的学号、姓名、选课数、总成绩； select Student.S#,Student.Sname,count(SC.C#),sum(score) from Student left Outer join SC on Student.S#=SC.S# group by Student.S#,Sname 4、查询姓“李”的老师的个数； select count(distinct(Tname)) from Teacher where Tname like '李%'; 5、查询没学过“叶平”老师课的同学的学号、姓名； select Student.S#,Student.Sname from Student where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平'); 6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名； select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002'); 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名； select S#,Sname from Student where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平')); 8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名； Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2 from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score; 9、查询所有课程成绩小于60分的同学的学号、姓名； select S#,Sname from Student where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60); 10、查询没有学全所有课的同学的学号、姓名； select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course); 11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名； select S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='1001'; 12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名； select distinct SC.S#,Sname from Student,SC where Student.S#=SC.S# and C# in (select C# from SC where S#='001'); 13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩； update SC set score=(select avg(SC_2.score) from SC SC_2 where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平'); 14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名； select S# from SC where C# in (select C# from SC where S#='1002') group by S# having count(*)=(select count(*) from SC where S#='1002'); 15、删除学习“叶平”老师课的SC表记录； Delect SC from course ,Teacher where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平'; 16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、 号课的平均成绩； Insert SC select S#,'002',(Select avg(score) from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002'); 17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分 SELECT S# as 学生ID ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语 ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩 FROM SC AS t GROUP BY S# ORDER BY avg(t.score) 18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分 SELECT L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分 FROM SC L ,SC AS R WHERE L.C# = R.C# and L.score = (SELECT MAX(IL.score) FROM SC AS IL,Student AS IM WHERE L.C# = IL.C# and IM.S#=IL.S# GROUP BY IL.C#) AND R.Score = (SELECT MIN(IR.score) FROM SC AS IR WHERE R.C# = IR.C# GROUP BY IR.C# ); 19、按各科平均成绩从低到高和及格率的百分数从高到低顺序 SELECT t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩 ,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数 FROM SC T,Course where t.C#=course.C# GROUP BY t.C# ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理（001），马克思（002），OO&UML （003），数据库（004） SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分 ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分 ,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数 ,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分 ,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数 ,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分 ,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数 FROM SC 21、查询不同老师所教不同课程平均分从高到低显示 SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ＩＤ,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩 FROM SC AS T,Course AS C ,Teacher AS Z where T.C#=C.C# and C.T#=Z.T# GROUP BY C.C# ORDER BY AVG(Score) DESC 22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004） [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩 SELECT DISTINCT top 3 SC.S# As 学生学号, Student.Sname AS 学生姓名 , T1.score AS 企业管理, T2.score AS 马克思, T3.score AS UML, T4.score AS 数据库, ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分 FROM Student,SC LEFT JOIN SC AS T1 ON SC.S# = T1.S# AND T1.C# = '001' LEFT JOIN SC AS T2 ON SC.S# = T2.S# AND T2.C# = '002' LEFT JOIN SC AS T3 ON SC.S# = T3.S# AND T3.C# = '003' LEFT JOIN SC AS T4 ON SC.S# = T4.S# AND T4.C# = '004' WHERE student.S#=SC.S# and ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) NOT IN (SELECT DISTINCT TOP 15 WITH TIES ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) FROM sc LEFT JOIN sc AS T1 ON sc.S# = T1.S# AND T1.C# = 'k1' LEFT JOIN sc AS T2 ON sc.S# = T2.S# AND T2.C# = 'k2' LEFT JOIN sc AS T3 ON sc.S# = T3.S# AND T3.C# = 'k3' LEFT JOIN sc AS T4 ON sc.S# = T4.S# AND T4.C# = 'k4' ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60] SELECT SC.C# as 课程ID, Cname as 课程名称 ,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85] ,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70] ,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60] ,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -] FROM SC,Course where SC.C#=Course.C# GROUP BY SC.C#,Cname; 24、查询学生平均成绩及其名次 SELECT 1+(SELECT COUNT( distinct 平均成绩) FROM (SELECT S#,AVG(score) AS 平均成绩 FROM SC GROUP BY S# ) AS T1 WHERE 平均成绩 > T2.平均成绩) as 名次, S# as 学生学号,平均成绩 FROM (SELECT S#,AVG(score) 平均成绩 FROM SC GROUP BY S# ) AS T2 ORDER BY 平均成绩 desc; 25、查询各科成绩前三名的记录:(不考虑成绩并列情况) SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数 FROM SC t1 WHERE score IN (SELECT TOP 3 score FROM SC WHERE t1.C#= C# ORDER BY score DESC ) ORDER BY t1.C#; 26、查询每门课程被选修的学生数 select c#,count(S#) from sc group by C#; 27、查询出只选修了一门课程的全部学生的学号和姓名 select SC.S#,Student.Sname,count(C#) AS 选课数 from SC ,Student where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1; 28、查询男生、女生人数 Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男'; Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女'； 29、查询姓“张”的学生名单 SELECT Sname FROM Student WHERE Sname like '张%'; 30、查询同名同性学生名单，并统计同名人数 select Sname,count(*) from Student group by Sname having count(*)>1;; 31、1981年出生的学生名单(注：Student表中Sage列的类型是datetime) select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age from student where CONVERT(char(11),DATEPART(year,Sage))='1981'; 32、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列 Select C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ; 33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩 select Sname,SC.S# ,avg(score) from Student,SC where Student.S#=SC.S# group by SC.S#,Sname having avg(score)>85; 34、查询课程名称为“数据库”，且分数低于60的学生姓名和分数 Select Sname,isnull(score,0) from Student,SC,Course where SC.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60; 35、查询所有学生的选课情况； SELECT SC.S#,SC.C#,Sname,Cname FROM SC,Student,Course where SC.S#=Student.S# and SC.C#=Course.C# ; 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 distinct student.S#,student.Sname,SC.C#,SC.score FROM student,Sc WHERE SC.score>=70 AND SC.S#=student.S#; 37、查询不及格的课程，并按课程号从大到小排列 select c# from sc where scor e <60 order by C# ; 38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名； select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003'; 39、求选了课程的学生人数 select count(*) from sc; 40、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩 select Student.Sname,score from Student,SC,Course C,Teacher where Student.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# ); 41、查询各个课程及相应的选修人数 select count(*) from sc group by C#; 42、查询不同课程成绩相同的学生的学号、课程号、学生成绩 select distinct A.S#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ; 43、查询每门功成绩最好的前两名 SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数 FROM SC t1 WHERE score IN (SELECT TOP 2 score FROM SC WHERE t1.C#= C# ORDER BY score DESC ) ORDER BY t1.C#; 44、统计每门课程的学生选修人数（超过10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，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# as 课程号,count(*) as 人数 from sc group by C# order by count(*) desc,c# 45、检索至少选修两门课程的学生学号 select S# from sc group by s# having count(*) > = 2 46、查询全部学生都选修的课程的课程号和课程名 select C#,Cname from Course where C# in (select c# from sc group by c#) 47、查询没学过“叶平”老师讲授的任一门课程的学生姓名 select Sname
from Student
where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and Tname='叶平'); 48、查询两门以上不及格课程的同学的学号及其平均成绩 select S#,avg(isnull(score,0))
from SC
where S# in (select S# from SC where score <60 group by S# having count(*)>2)
group by S#; 49、检索“004”课程分数小于60，按分数降序排列的同学学号 select S#
from SC
where C#='004'and score <60
order by score desc; 50、删除“002”同学的“001”课程的成绩 delete from Sc where S#='001'and C#='001';
