##mysql数据库事务###

数据库事务(简称：事务)是由一系列对系统中数据进行访问与更新的操作所组成的一个程序执行逻辑单元。

1. 主键 超键 候选键 外键

主 键：数据库表中对储存数据对象予以唯一和完整标识的数据列或属性的组合。一个数据列只能有一个主键，且主键的取值不能缺失，即不能为空值(Null)。

超 键：在关系中能唯一标识元组的属性集称为关系模式的超键。一个属性可以为作为一个超键，多个属性组合在一起也可以作为一个超键。超键包含候选键和主键。

候选键：是最小超键，即没有冗余元素的超键。

外 键：在一个表中存在的另一个表的主键称此表的外键。

2.数据库事务的四个特性及含义

数据库事务transanction正确执行的四个基本要素。

ACID,原子性(Atomicity)、一致性(Correspondence)、隔离性(Isolation)、持久性(Durability)。

原子性:整个事务中的所有操作，要么全部完成，要么全部不完成，不可能停滞在中间某个环节。事务在执行过程中发生错误，会被回滚(Rollback)到事务开始前的状态，就像这个事务从来没有执行过一样。

一致性:在事务开始之前和事务结束以后，数据库的完整性约束没有被破坏。

隔离性:隔离状态执行事务，使它们好像是系统在给定时间内执行的唯一操作。如果有两个事务，运行在相同的时间内，执行 相同的功能，事务的隔离性将确保每一事务在系统中认为只有该事务在使用系统。这种属性有时称为串行化，为了防止事务操作间的混淆，必须串行化或序列化请 求，使得在同一时间仅有一个请求用于同一数据。

持久性:在事务完成以后，该事务所对数据库所作的更改便持久的保存在数据库之中，并不会被回滚。

视图的作用，视图可以更改么？

1.视图的创建

第一类：create view v as select * from table;

第二类、基于不同数据库

这种情况只比上面的sql语句多一个数据库的名字，如下：

create view 数据库1.v as (select * from 数据库1.table1) union all (select * from 数据库2.table2);

或

create view 数据库2.v as (select * from 数据库1.table1) union all (select * from 数据库2.table2);

如果执行第一个sql将在数据库1下建立视图，反之亦然；

视图的作用？

视图是虚拟的表，与包含数据的表不一样，视图只包含使用时动态检索数据的查询；不包含任何列或数据。使用视图可以简化复杂的sql操作，隐藏具体的细节，保护数据；视图创建后，可以使用与表相同的方式利用它们。视图不能被索引，也不能有关联的触发器或默认值，如果视图本身内有order by 则对视图再次order by将被覆盖。

视图可以更改么？

对于某些视图比如未使用联结子查询分组聚集函数Distinct 、Union等，是可以对其更新的，对视图的更新将对基表进行更新；但是视图主要用于简化检索，保护数据，并不用于更新，而且大部分视图都不可以更新

4、drop,delete与truncate的区别

一、不同点

1.truncate table 和 delete只删除记录不删除表的结构，drop语句将删除表的结构依赖的约束(constrain)，触发器(trigger),索引(index);drop语句将删除表的结构被依赖的约束(constrain)，触发器(trigger)，索引(index);依赖于该表的存储过程/函数将保留，但是变为invalid状态

2.truncate之后的自增字段从头开始计数了，而delete的仍保留原来的最在数值。

二、总结：

1.在速度上，一般来说，drop> truncate > delete。

2.在使用drop和truncate时一定要注意，虽然可以恢复，但为了减少麻烦，还是要慎重。

3.如果想删除部分数据用delete，注意带上where子句，回滚段要足够大；

如果想删除表，当然用drop；

如果想保留表而将所有数据删除，如果和事务无关，用truncate即可；

如果和事务有关，或者想触发trigger，还是用delete

如果是整理表内部的碎片，可以用truncate跟上reuse stroage，再重新导入/插入数据

5.索引的工作原理及其种类

数据库索引，是数据库管理系统中一个排序的数据结构，以协助快速查询、更新数据库表中数据。索引的实现通常使用B树及其变种B+树

MySQL索引的建立对于MySQL的高效运行是很重要的，索引可以大大提高MySQL的检索速度。索引也是一张表，该表保存了主键与索引字段，并指向实体表的记录。上面都在说使用索引的好处，但过多的使用索引将会造成滥用。

因此索引也会有它的缺点：虽然索引大大提高了查询速度，同时却会降低更新表的速度，如对表进行INSERT、UPDATE和DELETE。因为更新表时，MySQL不仅要保存数据，还要保存一下索引文件。建立索引会占用磁盘空间的索引文件。

ALTER table tableName ADD INDEX indexName(columnName)

##表的创建、更新、修改

28、 创建表

Create table actor

Actor_id smallint(5) notnull Primary key,

First_name varchar(45) not null,

Last_name varchar(45) not null,

Last_update timestamp not null

Default (datetime('now','locatetime'))

33、

insert INTO actor #tablename

ValueS (1,PENELOPE,GUINESS,2006-02-15 12:34:33),

(2,NICK,WAHLBERG,2006-02-15 12:34:33)

34、

Insert ignore into

ValueS (1,PENELOPE,GUINESS,2006-02-15 12:34:33)

35、

Create table actor_name

As

Select first_name,last_name from actor

37、 创建索引

Create unique index unique_idx_firstname on actor(first_name)

Create index index_lastname on actor(last_name)

38、 临时试图

Create view actor_name_view

As

Select first_name as first_name_v,last_name as last_name_v

From actor

39、 增加一列

Alter table actor add column

Create_date datetime not null

Default '0000-00000 00:00:00'

40、 触发器

Create triggle audit_log

After insert into employee_test

Begin insert into audit

Values (NEW.id,NEW.NAME)

END

41、 删除重复记录，保留小的id

Select emp_no,title,from_date,to_date,min(id) as id

From title_test t

Group by emp_no,title,from_date,to_date

Delete title_test t

Where id not in (select min(id) from title_test group by emp_no)

42 更新

Update title_set set from_date='2001-01-01'and to_date = null

Where to_date = '9999-01-01'

43、

Repalce into titles_test

as

Select 5,10005,title ,from_date,to_date

From title_test where id = 5

44、 更改表明

Alter table titles_test rename to titles_2017

45、 创建外检约束

Alter table audit add foreign key(emp_no)

Reference emplyees_test(id)

46、 选出和视图一样的数据

Select em.* from emplyees as em,emp_v as ev

Where em.emp_no = ev.emp_no

Select * from emp_v

47、 获奖员工当前工资增加

Update

Salaries s

Set salary = salary *1.1

Where s.emp_no in (select emp_no from emp_bonus) and s.to_date='9999-01-01'

48

只修改列的数据类型的方法:

通常可以写成 alter table 表名 modify column 列名 新的列的类型

例如:student表中列sname的类型是char(20),现在要修改为varchar(20),SQL语句如下

alter table student modify column sname varchar(20);

同时修改列名和列的数据类型的方法:

通常可以写成 alter table 表名 change column 旧列名 新列名 新的列类型

例如:student表中列sname的类型是char(20),现在要修改为stuname varchar(20),SQL语句如下

alter table student change column sname stuname varchar(20);

###数据库查询

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 t.* , 01_score , 02_score

From student t

Inner join

(Select s1.s_id,s1.score as 01_score ,s2.score as 02_score

From score s1 ,score s2

Where s1.c_id =’01’ and s2.c_id =’02’ and s1.score > s2.score and s1.s_id =s2.s_id ) t1

On t.s_id =t1.s_id

##没选02课程的人

Select

S.*,s1.score as 01_score , s2.score as 02_score

From

(Student s

Right join

score s1 on s.s_id = s1.s_id and s1.c_id=’01’ )

Left join score s2

On s1.s_id = s2.s_id and s2.c_id=’02’ and s1.score > s2.score)

Union

###没有选择02课程成绩的同学#####

Select

S.*,s1.score as 01_score , 0 as 02_score

From

Student s

join

score s1 on s.s_id = s1.s_id and s1.c_id=’01’

join score s2

On s1.s_id = s2.s_id and s2_id not in (select s_id from score where c_id = ‘02’))

如果要的满足"01"课程比"02"课程成绩高条件的学生 所有功课

Select st.* ,’语文’,’数学’,’英语

’from student st

Join

(

Select sc.s_id ,

Sum(case sc.c_id = ‘01’ then sc.score else 0 end) as ‘语文’,

Sum(case sc.c_id = ‘02’ then temp.02_scoreelse 0 end )as ‘数学’, ##选择temp 02成绩作为数学成绩

Sum(case sc.c_id = ‘03’ then sc.score else 0 end) as ‘英语’,

From score sc

Join

###满足条件的学生id#####################

(Select

S.*,s1.score as 01_score , s2.score as 02_score

From

(Student s

Right join

score s1 on s.s_id = s1.s_id and s1.c_id=’01’ )

Left join score s2

On s1.s_id = s2.s_id and s2.c_id=’02’ and s1.score > s2.score)

Union

###没有选择02课程成绩的同学#####

Select

S.*,s1.score as 01_score , 0 as 02_score

From

(Student s

join

score s1 on s.s_id = s1.s_id and s1.c_id=’01’ )

join score s2

On s1.s_id = s2.s_id and s2_id not in (select s_id from score where c_id = ‘02’))

) temp

On sc.s_id = temp.s_id

Group by sc.s_id)

-- 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' )

join score c on a.s_id=c.s_id and c.c_id='02' where b.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 s_id ,avg(score) as svg_score

From score

Group by s_id

Having avg(score) >=60

Select s1.name,s2.s_id,round(avg(score),2) as svg_score

from student s1

Left join score s2

On s1.s_id = s2.s_id

Group by s1.s_id

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 s.name ,s.s_id, round(avg(score),2) as avg_score

From student s join score s1 on s.s_id = s1.s_id

Group by s.s_id having avg(score) < 60

Union

Select s.name ,s.s_id, 0 as avg_score

From student s

Where s_id not in (select dinstinct s_id from score)

#mysql 版本优化了，可以再having里面使用别名

Select s.name ,s.s_id, round(avg(score),2) as avg_score

From student s

Left join score s1 on s.s_id = s1.s_id

Group by s.s_id

having avg_score < 60 or avg_score is null

--19、按各科成绩进行排序，并显示排名

1. 分数相同rank相同，rank考虑是有多少人比你分数高

#先进行子查询，每一条外查询都跑一轮子查询

Select s1.c_id ,

(Select count(1)+1 from score s2 where s1.c_id =s2.c_id and s1.score < s2.score ) as rank

From score s1

Order by s1.c_id ,rank

2.分数相同rank相同，rank考虑是有多少分数比你分数高

Select s1.c_id ,

(Select count(dinstinct s2.score)+1 from score s2 where s1.c_id =s2.c_id and s1.score < s2.score ) as rank

From score s1

Order by s1.c_id ,rank

--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.name,s_id,

Date_format(now(),’%Y’)- Date_format(s.birth,’%Y’) -

(Case when Date_format(now(),’%m%d’) < Date_format(s.birth,’%m%d’) then 1 else 0 end) as age

From student s

Leetcode

##########行程################

select t.Request_at as Day,

round((sum(case when t.Status != 'completed' then 1 else 0 end))/count(*),2)

as 'Cancellation Rate' #名字有空格使用引号

from

Trips t

inner join Users u1

on t.Client_Id= u1.Users_Id and u1.Banned ='No'

inner join Users u2

on t.Driver_Id= u2.Users_Id and u2.Banned ='No'

where

convert(t.Request_at,date) between '2013-10-01' and '2013-10-03'

group by t.Request_at

order by Day

Month(t.Request_at) = 10 and Year(t.Request_at) = 2013 and day(t.Request_at) <=3 and day(t.Request_at) >= 1

DATE_FORMAT(t.Request_at,'%Y-%m-%d') between '2013-10-01' and '2013-10-03'

from_unixtime(time,'%Y-%m-%d %H:%i:%s') #unixtime 改成固定格式

Select UNIX_TIMESTAMP(’2006-11-04 12:23:00′);

Datatime = date+time

IFNULL(expr1,expr2)：如果第一个参数不为空，则返回第一个参数，否则返回第二个参数。

ISNULL(expr)：判断是否是空，是空则返回1，否则返回0。

IF(expr1,expr2,expr3)：如果第一个表达式的值为TRUE(不为0或null)，则返回第二个参数的值，否则返回第三个参数的值。

-- 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 s1.s_id ,s1.name,count(1) as num, sum(score) as sum_score

From student s1 left join score s2

On s1.s_id =s2.s_id

Group by s1.s_id

-- 6、查询学过"张三"老师授课的同学的信息

Select from s.*

Student s join score s1

On s.s_id = s1.s_id

Where s1.c_id in

(

Select Course.C_id

from

Course join Teacher

On Course.t_id = Teacher.t_id

Where Teacher.t_name = ‘张三’)

-- 7、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 S.*

From student s, score s1,score s2

Where s.s_id =s1.s_id and s1.s_id = s2.s_id

And s1.c_id = ‘01’and s2.s_id = ‘02’

Select S.*

From student s

Join score s1 on s.s_id =s1.s_id And s1.c_id = ‘01’

Join score s2 on s1.s_id = s2.s_id and s2.s_id = ‘02’

-- 8、查询学过编号为"01"并且"02"的同学所选的课程的同学的信息

Select s.*

From student s join

(Select * from (

Select distinct s_id

From score t Where t.c_id in

(Select c_id From score s

Where s.s_id = ‘01’) and t.s_id != ‘01’) t1

Where t1.c_id in

(Select c_id

From score s

Where s.s_id = ‘02’) and t1.s_id != ‘02’))t2

On s.s_id = t2.s_id

Select s.*

From student s join

(select distinct s_id From score s where s_id != ‘01’and s_id != ‘02’and (c_id in (Select c_id From score s

Where s.s_id = ‘01’) or c_id in (Select c_id From score s

Where s.s_id = ‘01’ ))

) t

On s.s_id =t.s_id

-- 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 s.*

From student s ,score s1 Where s.s_id = s1.s_id

And s1.c_id = ‘01’ and s1.s_id not in (

Select s_id from score s2 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 ,(select s_id from score s1

Group by s1.s_id

Having count(s1.c_id) < (select count(distinct s2.c_id ) from score s2)

) t

Where s.s_id = t.s_id

-- 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 distinct s.* # 这样对，但是慢

from student s ,

score s1

Where s.s_id = s1.s_id and

S1.s_id != ‘01’ and s1.c_id in

(Select c_id from score where s_id = ‘01’)[写错了，一个同学出现了很多条记录]

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’)

-- 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 s.* from student s

Join

(Select s_id from score s1 on where s1.c_id in (Select c_id from score where s_id = ‘01’)

and s_id != ‘01’

group by s.s_id

having

count(c_id) = (select count(s2.c_id) from score s2 where s2.s_id = ‘01’)

)

Tmp

On s.s_id = tmp.s_id

Leetcode 删除重复邮箱，保留id小的

没有id直接distinct 就好

Select e1.id ,e1.email

From emails e1,emails e2

Where e1.email = e2.email and e1.id < e2.id

Delete e1.id ,e1.email

from

emails e1,emails e2

Where e1.email = e2.email and e1.id > e2.id

-- 20、查询学生的总成绩并进行排名

Select

@k := (case when @score = a.sum_score then @k else @k+1 end )as rank,

@score : = a.sum_score as score

From

(Select sum(score) as sum_score from score group by s_id order by sum_score Desc) a , (select @k:=0,@score:=0)s

Leetcode

select a.score as Score, (select count(distinct b.score)+1 from Scores b

where a.score < b.score ) as Rank

from Scores a

order by Rank desc

Group_concat函数

mysql中的函数，字符串拼接的话，可以用concat()，但是此函数是针对一条记录中，可以将不同的字段拼接，并不适用多条记录的某一字段。查了一下，mysql中group_concat函数就可以获得到这样的结果。

1. group_concat只有与group by语句同时使用才能产生效果。

2. 需要将拼接的结果去重的话，可与DISTINCT结合使用即可。

SELECT

DISTINCT o.id_,o.order_sn,o.create_time,o.wait_out_storage_total,o.back,group_concat(og.goods_name) AS goods_names ,o.store_title FROM wms_orders o LEFT JOIN wms_orders_goods og ON o.id_=og.order_id WHERE o.wait_out_storage_total>0

GROUP BY o.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

s1.s_id,s1.name,round(avg(s2.score),2) as avg_score

Student s1 join score s2

On s1.s_id = s2.s_id

where s2.score < 60

Gruop by s1.s_id,s1.name

Having count(1) >=2

15.查询比30部门最高薪资的人薪资更高的所有员工信息

SELECT *

FROM emp

WHERE sal >

(SELECT MAX(sal) FROM emp WHERE deptno = '30')

查询比30部门所有人薪资都高员工信息

SELECT *

FROM emp

WHERE sal >

ALL (SELECT sal FROM emp WHERE deptno = '30')

两句话一样的查询，函数不一样，函数放的位置不一样

- 16、检索"01"课程分数小于60，按01分数降序排列的学生信息

SELECT s.* ,s1.score

from student s ,score s1

Where s.s_id = s1.s_id and s1.c_id = ‘01’and s1.score < 60

Order by s1.score

- 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 s_id ,sum(score) as sum_score ,round(avg(acore),2) as avg_score

From score s

Group by s_id) #有成绩的有选课的同学

Union

(Select s_id ,0 as sum_score ,0 as svg_score

From student s1 where s1.s_id not in (select distinct s_id from score)

每门课的成绩和平均成绩

Select

s.s_id,s.name,

sum(case when s1.c_id = ‘01’then s1.score else 0 end ) as ‘语文’，

sum(case when s1.c_id = ‘02’then s1.score else 0 end ) as ‘数学’，

sum(case when s1.c_id = ‘03’then s1.score else 0 end ) as ‘英语’，

round(avg(acore),2) as avg_score

From student s left join score s1

Group by s1.s_id

Order by avg_score 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

S.C_id,c.name

Max(score) as max_score,

Min(score) as min_score,

Round(avg(score),2)as avg_score,

Round(Sum(case when score >= 60 then 1 else 0 end )/count(1),2) as jige_rate,

Round(Sum(case when score >=70 and score < 80then 1 else 0 end )/count(1),2) as mid_rate,

Round(Sum(case when score >= 80 and score < 90 then 1 else 0 end )/count(1),2) as lianghao_rate,

Round(Sum(case when score >= 90 then 1 else 0 end )/count(1),2) as youxiu_rate,

From score s left join course c

Group by s.c_id,c.name

-- 19、按各科成绩进行排序，并显示排名(实现不完全)

Select

@i := @i+1 as index,

@k := (case when @score = a.score then @k else @i end ) as rank

@score : = a.score as 01_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

................

网易mysql

1、好评率是会员对平台评价的重要指标。现在需要统计2018年1月1日到2018年1月31日，用户'小明'提交的母婴类目"花王"品牌的好评率(好评率=“好评”评价量/总评价量):

用户评价详情表：a

字段：id(评价id，主键)，create_time(评价创建时间，格式'2017-01-01')， user_name(用户名称)，goods_id(商品id，外键) ，

sub_time(评价提交时间，格式'2017-01-01 23:10:32')，sat_name(好评率类型，包含：“好评”、“中评”、“差评”)

商品详情表：

b 字段：good_id(商品id，主键)，bu_name(商品类目), brand_name(品牌名称)

Select

A. user_name ,b.bu_name ,brand_name,

Round(sum(Case when a.sat_name = ‘好评’then 1 esle 0 end) / count(sat_name) ,2) as ‘好评率’

From a left join b on a.goods_id = b.good_id

Where b.bu_name = ‘母婴’and brand_name = ‘花王’

And a.user_name = ‘小明’and a.sub_time between '2018-01-01' and

'2018-01-31'

2、考拉运营"小明"负责多个品牌的销售业绩，请完成：

(1)请统计小明负责的各个品牌，在2017年销售最高的3天，及对应的销售额。

销售表 a:

字段：logday(日期，主键组)，SKU_ID(商品SKU，主键组)，sale_amt(销售额)

商品基础信息表 b:

字段：SKU_ID(商品SKU，主键),bu_name(商品类目)，brand_name(品牌名称)，user_name(运营负责人名称)

(2)请统计小明负责的各个品牌，在2017年连续3天增长超过50%的日期，及对应的销售额。

###通过查询类容创建表###################################

1

Create table if not exists temp

as

(Select a.* ,b.*

From a left join b on a.SKU_ID = b.SKU_ID

Where year(a.logday) = ‘2017’and b.user_name = ‘小明’)

Select a.*

From temp a

Where

(select count(b.sale_amt) from temp b where a.sale_amt < b.sale_amt and a.bu_name = b.bu_name and a.brand_name = b.brand_name ) < 3

Order by a.bu_name, a.brand_name,a.sale_amt

2.

Select

Distinct a.logday ,a.sale_amt ,a.bu_name ,a.brand_name

From temp a,temp b ,temp c

Where

a. bu_name = b.bu_name and a.brand_name = b.brand_name and

b. bu_name = b.bu_name and c.brand_name = b.brand_name

and

(a. sale_amt * 1.5 < b.sale_amt and b.sale_amt * 1.5 < c.sale_amt

Date_add(a.logday,interval 1 day) = b.logday and

Date_add(b.logday,interval 1 day) = c.logday )

Or

(b. sale_amt * 1.5 < a.sale_amt and a.sale_amt * 1.5 < c.sale_amt

And Date_add(b.logday,interval 1 day) = a.logday and

Date_add(a.logday,interval 1 day) = c.logday )

)

or

(b.sale_amt * 1.5 < c.sale_amt and c.sale_amt * 1.5 < a.sale_amt

Date_add(b.logday,interval 1 day) = c.logday and

Date_add(c.logday,interval 1 day) = a.logday )

)

Order by a.bu_name, a.brand_name,a.sale_amt

查询student表中重名的学生，结果包含id和name，按name,id升序

Select s1.name ,s1.id

From student s1 ,student s2

Where s1.name = s2.name and s1.id != s2.id

Order by s1.name ,s1.id

select id,name

from student

where name in (select name from student group by name having(count(*) > 1)

) order by name,id;

查询student表中重名的学生，保留id小的

Select s1.name ,s1.id

From student s1 ,student s2

Where s1.name = s2.name and s1.id < s2.id

Order by s1.name ,s1.id

Select s.name ,min(s.id) as id

From student s

Group by s.name

Order by s.name ,s.id

Delete

s1.name ,s1.id

From student s1 ,student s2

Where s1.name = s2.name and s1.id > s2.id

Order by s1.name ,s1.id

总成绩最高的学生，结果列出学生id、姓名和总成绩

Select s.s_id ,s.name,sum(score) as sum_score

From student s join score s1 on s.s_id = s1.s_id

Group by s.s_id

Having sum(s1.score) >= (select max(sum(score))from score s2 group by s_id )

Select s.s_id ,s.name,sum(score) as sum_score

From student s join score s1 on s.s_id = s1.s_id

Group by s.s_id,s.name

order by sum_score desc limit 1

在student_course表查询课程1成绩第2高的学生，如果第2高的不止一个则列出所有的学生

Select a.* from

(Select s.* ,s1.score

From (student s join score s1 on s.s_id = s1.s_id )

join course c on s1.c_id = c.c_Id and c_c_id = ‘01’

###多次的join

##或者 之前的作为表temp，score s1 where temp.s_id = s1.s_Id

)a

Where

(select count(distinct score) from

(Select s.* From student s join course c on s.c_id = c.c_id and c_c_id = ‘01’)b

Where a.score < b.score ) =1

新思路： 对成绩分组

Select st.*,sc.score

From student st join score sc on st.s_id = sc.s_id

Where sc.score =

(Select s1.score

From (score s join course c on s.c_id = c.c_id and c.c_id = ‘01’)

join student s1 on s.s_id = s1.s_Id

Group by s1.score

Order by s1.score desc

Limit 1,1)