常用30种SQL查询语句优化方法。
1、应尽量避免在 where 子句中使用!=或<>操作符，否则将引擎放弃使用索引而进行全表扫描。
2、对查询进行优化，应尽量避免全表扫描，首先应考虑在 where 及 order by 涉及的列上建立索引。
3、应尽量避免在 where 子句中对字段进行 null 值判断，否则将导致引擎放弃使用索引而进行全表扫描。如：
select id from t where num is null
可以在num上设置默认值0，确保表中num列没有null值，然后这样查询：
select id from t where num=0
4、尽量避免在 where 子句中使用 or 来连接条件，否则将导致引擎放弃使用索引而进行全表扫描，如：
select id from t where num=10 or num=20
可以这样查询：
select id from t where num=10
union all
select id from t where num=20
5、下面的查询也将导致全表扫描：(不能前置百分号)
select id from t where name like ‘%c%’
下面走索引
select id from t where name like ‘c%’
若要提高效率，可以考虑全文检索。
6、in 和 not in 也要慎用，否则会导致全表扫描，如：
select id from t where num in(1,2,3)
对于连续的数值，能用 between 就不要用 in 了：
select id from t where num between 1 and 3
7、如果在 where 子句中使用参数，也会导致全表扫描。因为SQL只有在运行时才会解析局部变量，但优化程序不能将访问计划的选择推迟到运行时；它必须在编译时进行选择。然 而，如果在编译时建立访问计划，变量的值还是未知的，因而无法作为索引选择的输入项。如下面语句将进行全表扫描：
select id from t where num=@num
可以改为强制查询使用索引：
select id from t with(index(索引名)) where num=@num
8、应尽量避免在 where 子句中对字段进行表达式操作，这将导致引擎放弃使用索引而进行全表扫描。如：
select id from t where num/2=100
应改为:
select id from t where num=1002
9、应尽量避免在where子句中对字段进行函数操作，这将导致引擎放弃使用索引而进行全表扫描。如：
select id from t where substring(name,1,3)=’abc’ –name以abc开头的id
select id from t where datediff(day,createdate,’2005-11-30′)=0 –’2005-11-30′生成的id
应改为:
select id from t where name like ‘abc%’
select id from t where createdate>=’2005-11-30′ and createdate<’2005-12-1′
10、不要在 where 子句中的“=”左边进行函数、算术运算或其他表达式运算，否则系统将可能无法正确使用索引。
11、在使用索引字段作为条件时，如果该索引是复合索引，那么必须使用到该索引中的第一个字段作为条件时才能保证系统使用该索引，否则该索引将不会被使 用，并且应尽可能的让字段顺序与索引顺序相一致。
12、不要写一些没有意义的查询，如需要生成一个空表结构：
select col1,col2 into #t from t where 1=0
这类代码不会返回任何结果集，但是会消耗系统资源的，应改成这样：
create table #t(…)
13、很多时候用 exists 代替 in 是一个好的选择：
select num from a where num in(select num from b)
用下面的语句替换：
select num from a where exists(select 1 from b where num=a.num)
14、并不是所有索引对查询都有效，SQL是根据表中数据来进行查询优化的，当索引列有大量数据重复时，SQL查询可能不会去利用索引，如一表中有字段 sex，male、female几乎各一半，那么即使在sex上建了索引也对查询效率起不了作用。
15、索引并不是越多越好，索引固然可以提高相应的 select 的效率，但同时也降低了 insert 及 update 的效率，因为 insert 或 update 时有可能会重建索引，所以怎样建索引需要慎重考虑，视具体情况而定。一个表的索引数较好不要超过6个，若太多则应考虑一些不常使用到的列上建的索引是否有 必要。
16.应尽可能的避免更新 clustered 索引数据列，因为 clustered 索引数据列的顺序就是表记录的物理存储顺序，一旦该列值改变将导致整个表记录的顺序的调整，会耗费相当大的资源。若应用系统需要频繁更新 clustered 索引数据列，那么需要考虑是否应将该索引建为 clustered 索引。
17、尽量使用数字型字段，若只含数值信息的字段尽量不要设计为字符型，这会降低查询和连接的性能，并会增加存储开销。这是因为引擎在处理查询和连接时会 逐个比较字符串中每一个字符，而对于数字型而言只需要比较一次就够了。
18、尽可能的使用 varchar/nvarchar 代替 char/nchar ，因为首先变长字段存储空间小，可以节省存储空间，其次对于查询来说，在一个相对较小的字段内搜索效率显然要高些。
19、任何地方都不要使用 select * from t ，用具体的字段列表代替“”，不要返回用不到的任何字段。
20、尽量使用表变量来代替临时表。如果表变量包含大量数据，请注意索引非常有限（只有主键索引）。
21、避免频繁创建和删除临时表，以减少系统表资源的消耗。
22、临时表并不是不可使用，适当地使用它们可以使某些例程更有效，例如，当需要重复引用大型表或常用表中的某个数据集时。但是，对于一次性事件，较好使 用导出表。
23、在新建临时表时，如果一次性插入数据量很大，那么可以使用 select into 代替 create table，避免造成大量 log ，以提高速度；如果数据量不大，为了缓和系统表的资源，应先create table，然后insert。
24、如果使用到了临时表，在存储过程的最后务必将所有的临时表显式删除，先 truncate table ，然后 drop table ，这样可以避免系统表的较长时间锁定。
25、尽量避免使用游标，因为游标的效率较差，如果游标操作的数据超过1万行，那么就应该考虑改写。
26、使用基于游标的方法或临时表方法之前，应先寻找基于集的解决方案来解决问题，基于集的方法通常更有效。
27、与临时表一样，游标并不是不可使用。对小型数据集使用 FAST_FORWARD 游标通常要优于其他逐行处理方法，尤其是在必须引用几个表才能获得所需的数据时。在结果集中包括“合计”的例程通常要比使用游标执行的速度快。如果开发时 间允许，基于游标的方法和基于集的方法都可以尝试一下，看哪一种方法的效果更好。
28、在所有的存储过程和触发器的开始处设置 SET NOCOUNT ON ，在结束时设置 SET NOCOUNT OFF 。无需在执行存储过程和触发器的每个语句后向客户端发送 DONEINPROC 消息。
29、尽量避免向客户端返回大数据量，若数据量过大，应该考虑相应需求是否合理。
30、尽量避免大事务操作，提高系统并发能力。
SQL 语句的多表查询方式
例如：按照 department_id 查询 employees(员工表)和 departments(部门表)的信息。
方式一(通用型):SELECT … FROM … WHERE
SELECT e.last_name,e.department_id,d.department_name FROM employees e,departments d
where e.department_id = d.department_id
方式二：SELECT … FROM … NATURAL JOIN …
有局限性：会自动连接两个表中相同的列 ( 可能有多个 :department_id 和
manager_id)
SELECT last_name,department_id,department_name
FROM employees
NATURAL JOIN departments
方式三：SELECT … JOIN … USING …
有局限性：好于方式二，但若多表的连接列列名不同，此法不合适
SELECT last_name,department_id,department_name
FROM employees
JOIN departments
USING(department_id)
方式四：SELECT … FROM … JOIN … ON …
常用方式，较方式一，更易实现外联接(左、右、满)
SELECT last_name,e.department_id,department_name
FROM employees e
JOIN departments d
ON e.department_id = d.department_id
–内连接
1）
–等值连接
–不等值连接
2）
–非自连接
–自连接
–外连接
–左外连接、右外连接、满外连接
1
创建和管理表(DDL)
CRAETE TABLE /TRUNCATE TABLE /ALTER TABLE /REANME…TO/DROP TABLE …操作完以后，自动 commit;所以，rollback 对其操作，没有效果
1.创建表
1）直接创建
create table emp1(
name varchar2(20),
salary number(8,2)default 1000,
id number(4),
hire_date date
);
2）通过子查询的方式创建
create table emp2
as
select last_name name,employee_id id,hire_date from employees;
或者
create table emp2
as
select last_name name,employee_id id,hire_date from employees
where department_id = 80;/where 1=2;
2.修改表
1)增加新的列
alter table emp1
add(birthday date)
2)修改现有的列
alter table emp1
modify(name varchar2(25) default ‘abc’)
3)重命名现有的列
alter table emp1
rename column salary to sal;
4)删除现有的列
alter table emp1
2
】
drop column birthday;
3.清空表中的数据(与 delete from table_name 区分开)
truncate table emp2;
4.重命名表
rename emp2 to emp3;
5.删除表
drop table emp3;
数据处理 DML
1）增
1.1 增添一条记录
insert into 表名
values(,)
1.2 从其它表中拷贝数据
insert into [表名]
select … from [另一个表]
where …
2）改
update [表名]
set …
where …
3）删
delete from [表名]
where …
4）查(最常用的数据库操作)
select …
from …
where ….
group by …
having …
order by ….
约 束
对创建的表的列属性、字段进行的限制。诸如 :not null/unique/primary key/foreign key/check
3
1.如何定义约束—在创建表的同时，添加对应属性的约束
1.1 表级约束 & 列级约束
create table emp1(
employee_id number(8),
salary number(8),
–列级约束
hire_date date not null,
dept_id number(8),
email varchar2(8) constraint emp1_email_uk unique,
name varchar2(8) constaint emp1_name_uu not null,
first_name varchar2(8),
–表级约束
constraint emp1_emp_id_pk primary key(employee_id), constraint emp1_fir_name_uk unique(first_name),
constraint emp1_dept_id_fk foreign key(dept_id) references departments(department_id) ON DELETE CASCADE
)
1.2 只有 not null 只能使用列级约束。其他的约束两种方式皆可
2.添加和删除表的约束–在创建表以后，只能添加和删除，不能修改
2.1 添加
alter table emp1
add constaint emp1_sal_ck check(salary > 0)
2.1.1 对于 not null 来讲，不用 add，需要使用 modify:
alter table emp1
modify (salary not null)
2.2 删除
alter table emp1
drop constaint emp1_sal_ck
2.3 使某一个约束失效:此约束还存在于表中，只是不起作用
alter table emp1
disable constraint emp1_email_uk;
2.4 使某一个约束激活：激活以后，此约束具有约束力
alter table emp1
enable constraint emp1_email_uk;
题目练习
问题及描述：
–1.学生表
Student(SID,Sname,Sage,Ssex) --SID 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
–2.课程表
Course(CID,Cname,TID) --CID --课程编号,Cname 课程名称,TID 教师编号
–3.教师表
Teacher(TID,Tname) --TID 教师编号,Tname 教师姓名
–4.成绩表
SC(SID,CID,score) --SID 学生编号,CID 课程编号,score 分数
/
–创建测试数据
create table Student(SID varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10));
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’);
create table Course(CID varchar(10),Cname nvarchar(10),TID varchar(10));
insert into Course values(‘01’ , ‘语文’ , ‘02’);
insert into Course values(‘02’ , ‘数学’ , ‘01’);
insert into Course values(‘03’ , ‘英语’ , ‘03’);
create table Teacher(TID varchar(10),Tname nvarchar(10));
insert into Teacher values(‘01’ , ‘张三’);
insert into Teacher values(‘02’ , ‘李四’);
insert into Teacher values(‘03’ , ‘王五’);
create table SC(SID varchar(10),CID varchar(10),score decimal(18,1));
decimal(18,1)表示小数和整数加起来是18位，其中小数是1位。
insert into SC values(‘01’ , ‘01’ , 80);
insert into SC values(‘01’ , ‘02’ , 90);
insert into SC values(‘01’ , ‘03’ , 99);
insert into SC values(‘02’ , ‘01’ , 70);
insert into SC values(‘02’ , ‘02’ , 60);
insert into SC values(‘02’ , ‘03’ , 80);
insert into SC values(‘03’ , ‘01’ , 80);
insert into SC values(‘03’ , ‘02’ , 80);
insert into SC values(‘03’ , ‘03’ , 80);
insert into SC values(‘04’ , ‘01’ , 50);
insert into SC values(‘04’ , ‘02’ , 30);
insert into SC values(‘04’ , ‘03’ , 20);
insert into SC values(‘05’ , ‘01’ , 76);
insert into SC values(‘05’ , ‘02’ , 87);
insert into SC values(‘06’ , ‘01’ , 31);
insert into SC values(‘06’ , ‘03’ , 34);
insert into SC values(‘07’ , ‘02’ , 89);
insert into SC values(‘07’ , ‘03’ , 98);
–1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
–1.1、查询同时存在"01"课程和"02"课程的情况
select a. , b.score 课程01的分数,c.score 课程02的分数 from Student a , SC b , SC c
where a.SID = b.SID and a.SID = c.SID and b.CID = ‘01’ and c.CID = ‘02’ and b.score > c.score
–1.2、查询同时存在"01"课程和"02"课程的情况和存在"01"课程但可能不存在"02"课程的情况(不存在时显示为null)(以下存在相同内容时不再解释)
select a.* , b.score 课程01的分数,c.score 课程02的分数 from Student a
left join SC b on a.SID = b.SID and b.CID = ‘01’
left join SC c on a.SID = c.SID and c.CID = ‘02’
where b.score > isnull(c.score,0)
是ifnull
–2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
–2.1、查询同时存在"01"课程和"02"课程的情况
select a.* , b.score 课程01的分数 ,c.score 课程02的分数 from Student a , SC b , SC c
where a.SID = b.SID and a.SID = c.SID and b.CID = ‘01’ and c.CID = ‘02’ and b.score < c.score
–2.2、查询同时存在"01"课程和"02"课程的情况和不存在"01"课程但存在"02"课程的情况
select a.* , b.score 课程01的分数 ,c.score 课程02的分数 from Student a
left join SC b on a.SID = b.SID and b.CID = ‘01’
left join SC c on a.SID = c.SID and c.CID = ‘02’
where isnull(b.score,0) < c.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 a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60
order by a.SID
其中Cast的函数用于转换类型。
Cast(字段名 as 转换的类型 )。
decimal(5,2)中的“2”表示小数部分的位数，如果插入的值未指定小数部分或者小数部分不足两位则会自动补到2位小数，若插入的值小数部分超过了2为则会发生截断，截取前2位小数。
“5”指的是整数部分加小数部分的总长度，也即插入的数字整数部分不能超过“5-2”位，否则不能成功插入，会报超出范围的错误。
–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
–4.1、查询在sc表存在成绩的学生信息的SQL语句。
select a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60
order by a.SID
–4.2、查询在sc表中不存在成绩的学生信息的SQL语句。
select a.SID , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0) avg_score
from Student a left join sc b
on a.SID = b.SID
group by a.SID , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60
order by a.SID
–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
–5.1、查询所有有成绩的SQL。
select a.SID 学生编号 , a.Sname 学生姓名 , count(b.CID) 选课总数, sum(score) 所有课程的总成绩
from Student a , SC b
where a.SID = b.SID
group by a.SID,a.Sname
order by a.SID
–5.2、查询所有(包括有成绩和无成绩)的SQL。
select a.SID 学生编号 , a.Sname 学生姓名 , count(b.CID) 选课总数, sum(score) 所有课程的总成绩
from Student a left join SC b
on a.SID = b.SID
group by a.SID,a.Sname
order by a.SID
–6、查询"李"姓老师的数量
–方法1
select count(Tname) 李姓老师的数量 from Teacher where Tname like ‘李%’
–方法2
select count(Tname) 李姓老师的数量 from Teacher where left(Tname,1) = ‘李’
–7、查询学过"张三"老师授课的同学的信息
select distinct Student.* from Student , SC , Course , Teacher
where Student.SID = SC.SID and SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = ‘张三’
order by Student.SID
–8、查询没学过"张三"老师授课的同学的信息
select m.* from Student m where SID not in (select distinct SC.SID from SC , Course , Teacher where SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = ‘张三’) order by m.SID
–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
–方法1
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘01’ and exists (Select 1 from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘02’) order by Student.SID
–方法2
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘02’ and exists (Select 1 from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘01’) order by Student.SID
–方法3
select m.* from Student m where SID in
(
select SID from
(
select distinct SID from SC where CID = ‘01’
union all
select distinct SID from SC where CID = ‘02’
) t group by SID having count(1) = 2
)
order by m.SID
–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
–方法1
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘01’ and not exists (Select 1 from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘02’) order by Student.SID
–方法2
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘01’ and Student.SID not in (Select SC_2.SID from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘02’) order by Student.SID
–11、查询没有学全所有课程的同学的信息
–11.1、
select Student.*
from Student , SC
where Student.SID = SC.SID
group by Student.SID , Student.Sname , Student.Sage , Student.Ssex having count(CID) < (select count(CID) from Course)
–11.2
select Student.*
from Student left join SC
on Student.SID = SC.SID
group by Student.SID , Student.Sname , Student.Sage , Student.Ssex having count(CID) < (select count(CID) 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 distinct Student.* from Student , SC where Student.SID = SC.SID and SC.CID in (select CID from SC where SID = ‘01’) and Student.SID <> ‘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 Student.* from Student where SID in
(select distinct SC.SID from SC where SID <> ‘01’ and SC.CID in (select distinct CID from SC where SID = ‘01’)
group by SC.SID having count(1) = (select count(1) from SC where SID=‘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 student.* from student where student.SID not in
(select distinct sc.SID from sc , course , teacher where sc.CID = course.CID and course.TID = teacher.TID and teacher.tname = ‘张三’)
order by student.SID
–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 student.SID , student.sname , cast(avg(score) as decimal(18,2)) avg_score from student , sc
where student.SID = SC.SID and student.SID in (select SID from SC where score < 60 group by SID having count(1) >= 2)
group by student.SID , student.sname
–16、检索"01"课程分数小于60，按分数降序排列的学生信息
select student.* , sc.CID , sc.score from student , sc
where student.SID = SC.SID and sc.score < 60 and sc.CID = ‘01’
order by sc.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
–17.1 SQL 2000 静态
select a.SID 学生编号 , a.Sname 学生姓名 ,
max(case c.Cname when ‘语文’ then b.score else null end) 语文 ,
max(case c.Cname when ‘数学’ then b.score else null end) 数学 ,
max(case c.Cname when ‘英语’ then b.score else null end) 英语 ,
cast(avg(b.score) as decimal(18,2)) 平均分
from Student a
left join SC b on a.SID = b.SID
left join Course c on b.CID = c.CID
group by a.SID , a.Sname
order by 平均分 desc
–17.2 SQL 2000 动态
declare @sql nvarchar(4000)
set @sql = ‘select a.SID ’ + ‘学生编号’ + ’ , a.Sname ’ + ‘学生姓名’
select @sql = @sql + ‘,max(case c.Cname when ‘’’+Cname+’’’ then b.score else null end) ‘+Cname+’ ’
from (select distinct Cname from Course) as t
set @sql = @sql + ’ , cast(avg(b.score) as decimal(18,2)) ’ + ‘平均分’ + ’ from Student a left join SC b on a.SID = b.SID left join Course c on b.CID = c.CID
group by a.SID , a.Sname order by ’ + ‘平均分’ + ’ desc’
exec(@sql)
–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
–方法1
select m.CID 课程编号 , m.Cname 课程名称 ,
max(n.score) 最高分 ,
min(n.score) 最低分 ,
cast(avg(n.score) as decimal(18,2)) 平均分 ,
cast((select count(1) from SC where CID = m.CID and score >= 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 及格率 ,
cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 中等率 ,
cast((select count(1) from SC where CID = m.CID and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优良率 ,
cast((select count(1) from SC where CID = m.CID and score >= 90)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优秀率
from Course m , SC n
where m.CID = n.CID
group by m.CID , m.Cname
order by m.CID
–方法2
select m.CID 课程编号 , m.Cname 课程名称 ,
(select max(score) from SC where CID = m.CID) 最高分 ,
(select min(score) from SC where CID = m.CID) 最低分 ,
(select cast(avg(score) as decimal(18,2)) from SC where CID = m.CID) 平均分 ,
cast((select count(1) from SC where CID = m.CID and score >= 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 及格率,
cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 中等率 ,
cast((select count(1) from SC where CID = m.CID and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优良率 ,
cast((select count(1) from SC where CID = m.CID and score >= 90)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优秀率
from Course m
order by m.CID
–19、按各科成绩进行排序，并显示排名
–19.1 sql 2000用子查询完成
–Score重复时保留名次空缺
select t.* , px = (select count(1) from SC where CID = t.CID and score > t.score) + 1 from sc t order by t.cid , px
–Score重复时合并名次
select t.* , px = (select count(distinct score) from SC where CID = t.CID and score >= t.score) from sc t order by t.cid , px
–19.2 sql 2005用rank,DENSE_RANK完成
–Score重复时保留名次空缺(rank完成)
select t.* , px = rank() over(partition by cid order by score desc) from sc t order by t.CID , px
–Score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t order by t.CID , px
–20、查询学生的总成绩并进行排名
–20.1 查询学生的总成绩
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
order by 总成绩 desc
–20.2 查询学生的总成绩并进行排名，sql 2000用子查询完成，分总分重复时保留名次空缺和不保留名次空缺两种。
select t1.* , px = (select count(1) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t1
order by px
select t1.* , px = (select count(distinct 总成绩) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t2 where 总成绩 >= t1.总成绩) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t1
order by px
–20.3 查询学生的总成绩并进行排名，sql 2005用rank,DENSE_RANK完成，分总分重复时保留名次空缺和不保留名次空缺两种。
select t.* , px = rank() over(order by 总成绩 desc) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t
order by px
select t.* , px = DENSE_RANK() over(order by 总成绩 desc) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(sum(score),0) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t
order by px
–21、查询不同老师所教不同课程平均分从高到低显示
select m.TID , m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score
from Teacher m , Course n , SC o
where m.TID = n.TID and n.CID = o.CID
group by m.TID , m.Tname
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
–22.1 sql 2000用子查询完成
–Score重复时保留名次空缺
select * from (select t.* , px = (select count(1) from SC where CID = t.CID and score > t.score) + 1 from sc t) m where px between 2 and 3 order by m.cid , m.px
–Score重复时合并名次
select * from (select t.* , px = (select count(distinct score) from SC where CID = t.CID and score >= t.score) from sc t) m where px between 2 and 3 order by m.cid , m.px
–22.2 sql 2005用rank,DENSE_RANK完成
–Score重复时保留名次空缺(rank完成)
select * from (select t.* , px = rank() over(partition by cid order by score desc) from sc t) m where px between 2 and 3 order by m.CID , m.px
–Score重复时合并名次(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t) m where px between 2 and 3 order by m.CID , m.px
–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
–23.1 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
–横向显示
select Course.CID 课程编号 , Cname as 课程名称 ,
sum(case when score >= 85 then 1 else 0 end) 85-100 ,
sum(case when score >= 70 and score < 85 then 1 else 0 end) 70-85 ,
sum(case when score >= 60 and score < 70 then 1 else 0 end) 60-70 ,
sum(case when score < 60 then 1 else 0 end) 0-60
from sc , Course
where SC.CID = Course.CID
group by Course.CID , Course.Cname
order by Course.CID
–纵向显示1(显示存在的分数段)
select m.CID 课程编号 , m.Cname 课程名称 , 分数段 = (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end) ,
count(1) 数量
from Course m , sc n
where m.CID = n.CID
group by m.CID , m.Cname , (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end)
order by m.CID , m.Cname , 分数段
–纵向显示2(显示存在的分数段，不存在的分数段用0显示)
select m.CID 课程编号 , m.Cname 课程名称 , 分数段 = (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end) ,
count(1) 数量
from Course m , sc n
where m.CID = n.CID
group by all m.CID , m.Cname , (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end)
order by m.CID , m.Cname , 分数段
–23.2 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 , <60 及所占百分比
–横向显示
select m.CID 课程编号, m.Cname 课程名称,
(select count(1) from SC where CID = m.CID and score < 60) 0-60 ,
cast((select count(1) from SC where CID = m.CID and score < 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比 ,
(select count(1) from SC where CID = m.CID and score >= 60 and score < 70) 60-70 ,
cast((select count(1) from SC where CID = m.CID and score >= 60 and score < 70)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比 ,
(select count(1) from SC where CID = m.CID and score >= 70 and score < 85) 70-85 ,
cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 85)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比 ,
(select count(1) from SC where CID = m.CID and score >= 85) 85-100 ,
cast((select count(1) from SC where CID = m.CID and score >= 85)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比
from Course m
order by m.CID
–纵向显示1(显示存在的分数段)
select m.CID 课程编号 , m.Cname 课程名称 , 分数段 = (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end) ,
count(1) 数量 ,
cast(count(1) * 100.0 / (select count(1) from sc where CID = m.CID) as decimal(18,2)) 百分比
from Course m , sc n
where m.CID = n.CID
group by m.CID , m.Cname , (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end)
order by m.CID , m.Cname , 分数段
–纵向显示2(显示存在的分数段，不存在的分数段用0显示)
select m.CID 课程编号 , m.Cname 课程名称 , 分数段 = (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end) ,
count(1) 数量 ,
cast(count(1) * 100.0 / (select count(1) from sc where CID = m.CID) as decimal(18,2)) 百分比
from Course m , sc n
where m.CID = n.CID
group by all m.CID , m.Cname , (
case when n.score >= 85 then ‘85-100’
when n.score >= 70 and n.score < 85 then ‘70-85’
when n.score >= 60 and n.score < 70 then ‘60-70’
else ‘0-60’
end)
order by m.CID , m.Cname , 分数段
–24、查询学生平均成绩及其名次
–24.1 查询学生的平均成绩并进行排名，sql 2000用子查询完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。
select t1.* , px = (select count(1) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(cast(avg(score) as decimal(18,2)),0) 平均成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t2 where 平均成绩 > t1.平均成绩) + 1 from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(cast(avg(score) as decimal(18,2)),0) 平均成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t1
order by px
select t1.* , px = (select count(distinct 平均成绩) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(cast(avg(score) as decimal(18,2)),0) 平均成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t2 where 平均成绩 >= t1.平均成绩) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(cast(avg(score) as decimal(18,2)),0) 平均成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t1
order by px
–24.2 查询学生的平均成绩并进行排名，sql 2005用rank,DENSE_RANK完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。
select t.* , px = rank() over(order by 平均成绩 desc) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(cast(avg(score) as decimal(18,2)),0) 平均成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t
order by px
select t.* , px = DENSE_RANK() over(order by 平均成绩 desc) from
(
select m.SID 学生编号 ,
m.Sname 学生姓名 ,
isnull(cast(avg(score) as decimal(18,2)),0) 平均成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
) t
order by px
–25、查询各科成绩前三名的记录
–25.1 分数重复时保留名次空缺
select m.* , n.CID , n.score from Student m, SC n where m.SID = n.SID and n.score in
(select top 3 score from sc where CID = n.CID order by score desc) order by n.CID , n.score desc
–25.2 分数重复时不保留名次空缺，合并名次
–sql 2000用子查询实现
select * from (select t.* , px = (select count(distinct score) from SC where CID = t.CID and score >= t.score) from sc t) m where px between 1 and 3 order by m.Cid , m.px
–sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by Cid order by score desc) from sc t) m where px between 1 and 3 order by m.CID , m.px
–26、查询每门课程被选修的学生数
select Cid , count(SID) 学生数 from sc group by CID
–27、查询出只有两门课程的全部学生的学号和姓名
select Student.SID , Student.Sname
from Student , SC
where Student.SID = SC.SID
group by Student.SID , Student.Sname
having count(SC.CID) = 2
order by Student.SID
–28、查询男生、女生人数
select count(Ssex) as 男生人数 from Student where Ssex = N’男’
select count(Ssex) as 女生人数 from Student where Ssex = N’女’
select sum(case when Ssex = N’男’ then 1 else 0 end) 男生人数 ,sum(case when Ssex = N’女’ then 1 else 0 end) 女生人数 from student
select case when Ssex = N’男’ then N’男生人数’ else N’女生人数’ end 男女情况 , count(1) 人数 from student group by case when Ssex = N’男’ then N’男生人数’ else N’女生人数’ end
–29、查询名字中含有"风"字的学生信息
select * from student where sname like N’%风%’
select * from student where charindex(N’风’ , sname) > 0
–30、查询同名同性学生名单，并统计同名人数
select Sname 学生姓名 , count() 人数 from Student group by Sname having count() > 1
–31、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)
select * from Student where year(sage) = 1990
select * from Student where datediff(yy,sage,‘1990-01-01’) = 0
select * from Student where datepart(yy,sage) = 1990
select * from Student where convert(varchar(4),sage,120) = ‘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 m.CID , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n
where m.CID = n.CID
group by m.CID , m.Cname
order by avg_score desc, m.CID 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.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85
order by a.SID
–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 sname , score
from Student , SC , Course
where SC.SID = Student.SID and SC.CID = Course.CID and Course.Cname = N’数学’ and score < 60
–35、查询所有学生的课程及分数情况；
select Student.* , Course.Cname , SC.CID , SC.score
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID
order by Student.SID , SC.CID
–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 Student.* , Course.Cname , SC.CID , SC.score
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID and SC.score >= 70
order by Student.SID , SC.CID
–37、查询不及格的课程
select Student.* , Course.Cname , SC.CID , SC.score
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID and SC.score < 60
order by Student.SID , SC.CID
–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 Student.* , Course.Cname , SC.CID , SC.score
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID and SC.CID = ‘01’ and SC.score >= 80
order by Student.SID , SC.CID
–39、求每门课程的学生人数
select Course.CID , Course.Cname , count() 学生人数
from Course , SC
where Course.CID = SC.CID
group by Course.CID , Course.Cname
order by Course.CID , Course.Cname
–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
–40.1 当最高分只有一个时
select top 1 Student. , Course.Cname , SC.CID , SC.score
from Student, SC , Course , Teacher
where Student.SID = SC.SID and SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = N’张三’
order by SC.score desc
–40.2 当最高分出现多个时
select Student.* , Course.Cname , SC.CID , SC.score
from Student, SC , Course , Teacher
where Student.SID = SC.SID and SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = N’张三’ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = N’张三’)
–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
–方法1
select m.* from SC m ,(select CID , score from SC group by CID , score having count(1) > 1) n
where m.CID= n.CID and m.score = n.score order by m.CID , m.score , m.SID
–方法2
select m.* from SC m where exists (select 1 from (select CID , score from SC group by CID , score having count(1) > 1) n
where m.CID= n.CID and m.score = n.score) order by m.CID , m.score , m.SID
–42、查询每门功成绩最好的前两名
select t.* from sc t where score in (select top 2 score from sc where CID = T.CID order by score desc) order by t.CID , t.score desc
–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 Course.CID , Course.Cname , count() 学生人数
from Course , SC
where Course.CID = SC.CID
group by Course.CID , Course.Cname
having count() >= 5
order by 学生人数 desc , Course.CID
–44、检索至少选修两门课程的学生学号
select student.SID , student.Sname
from student , SC
where student.SID = SC.SID
group by student.SID , student.Sname
having count(1) >= 2
order by student.SID
–45、查询选修了全部课程的学生信息
–方法1 根据数量来完成
select student.* from student where SID in
(select SID from sc group by SID having count(1) = (select count(1) from course))
–方法2 使用双重否定来完成
select t.* from student t where t.SID not in
(
select distinct m.SID from
(
select SID , CID from student , course
) m where not exists (select 1 from sc n where n.SID = m.SID and n.CID = m.CID)
)
–方法3 使用双重否定来完成
select t.* from student t where not exists(select 1 from
(
select distinct m.SID from
(
select SID , CID from student , course
) m where not exists (select 1 from sc n where n.SID = m.SID and n.CID = m.CID)
) k where k.SID = t.SID
)
–46、查询各学生的年龄
–46.1 只按照年份来算
select * , datediff(yy , sage , getdate()) 年龄 from student
–46.2 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 * , case when right(convert(varchar(10),getdate(),120),5) < right(convert(varchar(10),sage,120),5) then datediff(yy , sage , getdate()) - 1 else datediff(yy , sage , getdate()) end 年龄 from student
–47、查询本周过生日的学生
select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = 0
–48、查询下周过生日的学生
select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1
–49、查询本月过生日的学生
select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = 0
–50、查询下月过生日的学生
select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1
drop table Student,Course,Teacher,SC
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