sql语句 订阅
结构化查询语言(Structured Query Language)简称SQL,结构化查询语言是一种数据库查询和程序设计语言,用于存取数据以及查询、更新和管理关系数据库系统;sql 语句就是对数据库进行操作的一种语言。 展开全文
结构化查询语言(Structured Query Language)简称SQL,结构化查询语言是一种数据库查询和程序设计语言,用于存取数据以及查询、更新和管理关系数据库系统;sql 语句就是对数据库进行操作的一种语言。
信息
插    入
insert into table1(field1,
接插入
field2) values(value1,value2)
删    除
delete from table1 where 范围
选    择
select * from table1 where范围
中文名
结构化查询语言
总    数
select count(*) as totalcount
外文名
Structured Query Language
接总数
from table1
sql语句程序功能
创建数据库CREATE DATABASE database-name删除数据库drop database dbname创建新表create table tabname(col1 type1 [not null] [primary key],col2 type2 [not null],..)删除新表drop table tabname增加一个列Alter table tabname add column col type添加主键Alter table tabname add primary key(col)删除主键Alter table tabname drop primary key(col)创建索引create [unique] index idxname on tabname(col….)删除索引drop index idxname创建视图create view viewname as select statement删除视图drop view viewname
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  • 【数据库SQL系列】sql语句执行顺序,你理解了吗

    万次阅读 多人点赞 2020-01-13 15:29:00
    记得前几年,还是初级的时候,面试官问到,请你讲一下sql语句的执行顺序。当时我以为就是按照sql的关键字排列顺序来执行的。当时说完,面试官心里估计已经直接pass我了吧。今天复习的时候,突然想起这个基础知识点,...

    热门系列


     1.序言

    记得前几年,还是初级的时候,面试官问到,请你讲一下sql语句的执行顺序。当时我以为就是按照sql的关键字排列顺序来执行的。当时说完,面试官心里估计已经直接pass我了吧。今天复习的时候,突然想起这个基础知识点,所以有时间就来发表一下了。


    2.sql语句的执行顺序

    2.1 为什么要了解sql语句执行顺序

    了解一个sql语句的执行顺序,可以让我们清楚到sql执行时的操作顺序,进而有利于我们更好的优化自己的sql语句,提升程序性能。

    举个栗子:

    --sql 1
    select a.* from table_a a left join table_b b on a.id = b.a_id where b.name = 'john';
    
    --sql 2
    select * from table_a where id = (select a_id from table_b where name = 'john');

    倘若说table_a和table_b都是几十几百万数据的表。而name=‘john’这个过滤字段可以查到table_b的唯一数据。那么此时,一定是sql2的执行效率要高于sql1的。为什么?通过sql执行顺序可以知道,table_a和table_b会优先执行联表操作,两个都是大表。其查询出来的结果集虚拟表也会很大。而sql2中,table_a只用通过table_b查询出来的唯一数据更快的获取到指定结果。

    所以,如果你不知道sql执行顺序,同样的业务需求,也许,你就使用了性能不够好的sql1了。

    2.2 sql语句执行顺序说明

    (8) SELECT (9)DISTINCT<select_list>
    (1) FROM <left_table>
    (3) <join_type> JOIN <right_table>
    (2)         ON <join_condition>
    (4) WHERE <where_condition>
    (5) GROUP BY <group_by_list>
    (6) WITH {CUBE|ROLLUP}
    (7) HAVING <having_condition>
    (10) ORDER BY <order_by_list>
    (11) LIMIT <limit_number>

    (1) FROM:对FROM子句中的左表<left_table>和右表<right_table>执行笛卡儿积,产生虚拟表VT1;
    (2) ON: 对虚拟表VT1进行ON筛选,只有那些符合<join_condition>的行才被插入虚拟表VT2;
    (3) JOIN: 如果指定了OUTER JOIN(如LEFT OUTER JOIN、RIGHT OUTER JOIN),那么保留表中未匹配的行作为外部行添加到虚拟表VT2,产生虚拟表VT3。如果FROM子句包含两个以上的表,则对上一个连接生成的结果表VT3和下一个表重复执行步骤1~步骤3,直到处理完所有的表;
    (4) WHERE: 对虚拟表VT3应用WHERE过滤条件,只有符合<where_condition>的记录才会被插入虚拟表VT4;
    (5) GROUP By: 根据GROUP BY子句中的列,对VT4中的记录进行分组操作,产生VT5;如果应用了group by,那么后面的所有步骤都只能得到的vt5的列或者是聚合函数(count、sum、avg等)。原因在于最终的结果集中只为每个组包含一行。这一点请牢记。
    (6) CUBE|ROllUP: 对VT5进行CUBE或ROLLUP操作,产生表VT6;
    (7) HAVING: 对虚拟表VT6应用HAVING过滤器,只有符合<having_condition>的记录才会被插入到VT7;
    (8) SELECT: 第二次执行SELECT操作,选择指定的列,插入到虚拟表VT8中;
    (9) DISTINCT: 去除重复,产生虚拟表VT9;
    (10) ORDER BY: 将虚拟表VT9中的记录按照<order_by_list>进行排序操作,产生虚拟表VT10;
    (11) LIMIT: 取出指定街行的记录,产生虚拟表VT11,并返回给查询用户

     

    本博客皆为学习、分享、探讨为本,欢迎各位朋友评论、点赞、收藏、关注,一起加油!

     

    展开全文
  • sql语句练习50题(Mysql版)

    万次阅读 多人点赞 2017-12-19 00:11:31
    习题来源于网络,sql语句是自己写的。欢迎指正。 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, ...

    习题来源于网络,sql语句是自己写的,部分有参考。欢迎指正。


    2019.3.29更新
    写完后一年没有看过,没想到这篇文章有这么多人点击。博主工作到一半去考研了,目前已上岸某中部985,也算是比较幸运。非常感谢大家在评论里的留言,留言太多不能一一回复,希望大家见谅。这两天根据评论把文章中的某些错误或者不足的地方更新了下,,希望大家能够继续指出不足之处。


    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.成绩表
    Score(s_id,c_id,s_score) --学生编号,课程编号,分数

    测试数据

    --建表
    --学生表
    CREATE TABLE `Student`(
    	`s_id` VARCHAR(20),
    	`s_name` VARCHAR(20) NOT NULL DEFAULT '',
    	`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
    	`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
    	PRIMARY KEY(`s_id`)
    );
    --课程表
    CREATE TABLE `Course`(
    	`c_id`  VARCHAR(20),
    	`c_name` VARCHAR(20) NOT NULL DEFAULT '',
    	`t_id` VARCHAR(20) NOT NULL,
    	PRIMARY KEY(`c_id`)
    );
    --教师表
    CREATE TABLE `Teacher`(
    	`t_id` VARCHAR(20),
    	`t_name` VARCHAR(20) NOT NULL DEFAULT '',
    	PRIMARY KEY(`t_id`)
    );
    --成绩表
    CREATE TABLE `Score`(
    	`s_id` VARCHAR(20),
    	`c_id`  VARCHAR(20),
    	`s_score` INT(3),
    	PRIMARY KEY(`s_id`,`c_id`)
    );
    --插入学生表测试数据
    insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
    insert into Student values('02' , '钱电' , '1990-12-21' , '男');
    insert into Student values('03' , '孙风' , '1990-05-20' , '男');
    insert into Student values('04' , '李云' , '1990-08-06' , '男');
    insert into Student values('05' , '周梅' , '1991-12-01' , '女');
    insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
    insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
    insert into Student values('08' , '王菊' , '1990-01-20' , '女');
    --课程表测试数据
    insert into Course values('01' , '语文' , '02');
    insert into Course values('02' , '数学' , '01');
    insert into Course values('03' , '英语' , '03');
    
    --教师表测试数据
    insert into Teacher values('01' , '张三');
    insert into Teacher values('02' , '李四');
    insert into Teacher values('03' , '王五');
    
    --成绩表测试数据
    insert into Score values('01' , '01' , 80);
    insert into Score values('01' , '02' , 90);
    insert into Score values('01' , '03' , 99);
    insert into Score values('02' , '01' , 70);
    insert into Score values('02' , '02' , 60);
    insert into Score values('02' , '03' , 80);
    insert into Score values('03' , '01' , 80);
    insert into Score values('03' , '02' , 80);
    insert into Score values('03' , '03' , 80);
    insert into Score values('04' , '01' , 50);
    insert into Score values('04' , '02' , 30);
    insert into Score values('04' , '03' , 20);
    insert into Score values('05' , '01' , 76);
    insert into Score values('05' , '02' , 87);
    insert into Score values('06' , '01' , 31);
    insert into Score values('06' , '03' , 34);
    insert into Score values('07' , '02' , 89);
    insert into Score values('07' , '03' , 98);
    

    练习题和sql语句

    -- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数	
    	
    select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
    student a 
    	join score b on a.s_id=b.s_id and b.c_id='01'
    	left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score
    	
    --也可以这样写
    	select a.*,b.s_score as 01_score,c.s_score as 02_score 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' 
    			and b.s_score>c.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
    	
    select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
    	student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL 
    	 join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.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
    select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
    	student b 
    	join score a on b.s_id = a.s_id
    	GROUP BY b.s_id,b.s_name HAVING avg_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)
    		
    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 avg_score <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 a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from 
    	student a 
    	left join score b on a.s_id=b.s_id
    	GROUP BY a.s_id,a.s_name;
    			
    			
    -- 6、查询"李"姓老师的数量 
    select count(t_id) 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 a.* from 
    	student a 
    	join score b on a.s_id=b.s_id where b.c_id in(
    		select c_id from course where t_id =(
    			select t_id from teacher where t_name = '张三'));
    
    -- 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 a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(
            select a.c_id from course a join teacher b on a.t_id = b.t_id 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 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 
    --@wendiepei的写法
    select s.* from student s 
    left join Score s1 on s1.s_id=s.s_id
    group by s.s_id having count(s1.c_id)<(select count(*) from course)	
    --@k1051785839的写法
    select *
    from student
    where s_id not in(
    select s_id from score t1  
    group by s_id having count(*) =(select count(distinct c_id)  from course)) 
    -- 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 
    --@ouyang_1993的写法
    SELECT
     Student.*
    FROM
     Student
    WHERE
     s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
        #下面的语句是找到'01'同学学习的课程数
        SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
       )
     )
    AND s_id NOT IN (
     #下面的语句是找到学过‘01’同学没学过的课程,有哪些同学。并排除他们
     SELECT s_id FROM Score
     WHERE c_id IN(
       #下面的语句是找到‘01’同学没学过的课程
       SELECT DISTINCT c_id FROM Score
       WHERE c_id NOT IN (
         #下面的语句是找出‘01’同学学习的课程
         SELECT c_id FROM Score WHERE s_id = '01'
        )
      ) GROUP BY s_id
    ) #下面的条件是排除01同学
    AND s_id NOT IN ('01')
    --@k1051785839的写法
    SELECT
     t3.*
    FROM
     (
      SELECT
       s_id,
       group_concat(c_id ORDER BY c_id) group1
      FROM
       score
      WHERE
       s_id &lt;> '01'
      GROUP BY
       s_id
     ) t1
    INNER JOIN (
     SELECT
      group_concat(c_id ORDER BY c_id) group2
     FROM
      score
     WHERE
      s_id = '01'
     GROUP BY
      s_id
    ) t2 ON t1.group1 = t2.group2
    INNER JOIN student t3 ON t1.s_id = t3.s_id
    
    -- 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')));
    
    -- 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;
    --@喝完这杯还有一箱的写法
    SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文, 
    MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学, 
    MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语, 
    avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 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
    -- 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 GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
    )a,(select @k:=0,@i:=0,@score:=0)s
    --@k1051785839的写法
    (select * from (select 
    t1.c_id,
    t1.s_score,
    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
    FROM score t1 where t1.c_id='01'
    order by t1.s_score desc) t1)
    union
    (select * from (select 
    t1.c_id,
    t1.s_score,
    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
    FROM score t1 where t1.c_id='02'
    order by t1.s_score desc) t2)
    union
    (select * from (select 
    t1.c_id,
    t1.s_score,
    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank
    FROM score t1 where t1.c_id='03'
    order by t1.s_score desc) t3)
    -- 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'  
    								ORDER BY a.s_score DESC  
                )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'  
    								ORDER BY a.s_score DESC
                )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' 
    								ORDER BY a.s_score DESC
                )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 ORDER BY avg_s DESC)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)<3
    			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 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)
    
    
    展开全文
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    展开全文
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    declare @end_date datetime 
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    展开全文
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