year??month amount
1991?? 1???? 1.1
1991?? 2???? 1.2
1991?? 3???? 1.3
1991?? 4???? 1.4
1992?? 1???? 2.1
1992?? 2???? 2.2
1992?? 3???? 2.3
1992?? 4???? 2.4
查成这样一个结果
year m1??m2??m3??m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4
提示：这个与工资条非常类似，与学生的科目成绩也很相似。
准备sql语句：
drop table if exists sales;
create table sales(id int auto_increment primary key,year varchar(10), month varchar(10), amount float(2,1));
insert into sales values
(null,'1991','1',1.1),
(null,'1991','2',1.2),
(null,'1991','3',1.3),
(null,'1991','4',1.4),
(null,'1992','1',2.1),
(null,'1992','2',2.2),
(null,'1992','3',2.3),
(null,'1992','4',2.4);
答案一、
select sales.year ,
(select t.amount from sales t where t.month='1' and t.year= sales.year) '1',
(select t.amount from sales t where t.month='1' and t.year= sales.year) '2',
(select t.amount from sales t where t.month='1' and t.year= sales.year) '3',
(select t.amount from sales t where t.month='1' and t.year= sales.year) as '4'
from sales ?group by year;
表：id,title,postuser,postdate,parentid
准备sql语句：
drop table if exists articles;
create table articles(id int auto_increment primary key,title varchar(50), postuser varchar(10), postdate datetime,parentid int references articles(id));
insert into articles values
(null,'第一条','张三','1998-10-10 12:32:32',null),
(null,'第二条','张三','1998-10-10 12:34:32',null),
(null,'第一条回复1','李四','1998-10-10 12:35:32',1),
(null,'第二条回复1','李四','1998-10-10 12:36:32',2),
(null,'第一条回复2','王五','1998-10-10 12:37:32',1),
(null,'第一条回复3','李四','1998-10-10 12:38:32',1),
(null,'第二条回复2','李四','1998-10-10 12:39:32',2),
(null,'第一条回复4','王五','1998-10-10 12:39:40',1);
答案：
select a.title,a.postuser,
(select max(postdate) from articles where parentid=a.id) reply
from articles a where a.parentid is null;
注释：子查询可以用在选择列中，也可用于where的比较条件中，还可以用于from从句中。
2.学生表 如下:
id号?? 学号?? 姓名 课程编号 课程名称 分数
1??????? 2005001? 张三? 0001????? 数学??? 69
2??????? 2005002? 李四? 0001????? 数学??? 89
3??????? 2005001? 张三? 0001????? 数学??? 69
A: delete?from?tablename where id号 not in(select min(id号) from tablename group by 学号,姓名,课程编号,课程名称,分数)
实验：
create table student2(id int auto_increment primary key,code varchar(20),name varchar(20));
insert into student2 values(null,'2005001','张三'),(null,'2005002','李四'),(null,'2005001','张三');
//如下语句，mysql报告错误，可能删除依赖后面统计语句，而删除又导致统计语句结果不一致。
delete from student2 where id not in(select min(id) from student2 group by name);
//但是，如下语句没有问题：
select * ?from student2 where id not in(select min(id) from student2 group by name);
//于是，我想先把分组的结果做成虚表，然后从虚表中选出结果，最后再将结果作为删除的条件数据。
delete from student2 where id not in(select mid from (select min(id) mid
from student2 group by name) as t);
或者：
delete from student2 where id not in(select min(id) from (select * from s
tudent2) as t group by t.name);
表结构如下：
flight{flightID,StartCityID ,endCityID,StartTime}
city{cityID, CityName)
实验环境：
create table city(cityID int auto_increment primary key,cityName varchar(20));
create table flight (flightID int auto_increment primary key,
StartCityID int references city(cityID),
endCityID ?int references city(cityID),
StartTime timestamp);
//航班本来应该没有日期部分才好，但是下面的题目当中涉及到了日期
insert into city values(null,'北京'),(null,'上海'),(null,'广州');
insert into flight values
(null,1,2,'9:37:23'),(null,1,3,'9:37:23'),(null,1,2,'10:37:23'),(null,2,3,'10:37:23');
1、查询起飞城市是北京的所有航班，按到达城市的名字排序
参与运算的列是我起码能够显示出来的那些列，但最终我不一定把它们显示出来。各个表组合出来的中间结果字段中必须包含所有运算的字段。
??select ?* from flight f,city c
where f.endcityid = c.cityid and startcityid =
(select c1.cityid from city c1 where c1.cityname = "北京")
order by c.cityname asc;
mysql> select flight.flightid,'北京' startcity, e.cityname from flight,city e wh
ere flight.endcityid=e.cityid and flight.startcityid=(select cityid from city wh
ere cityname='北京');
mysql> select flight.flightid,s.cityname,e.cityname from flight,city s,city e wh
ere flight.startcityid=s.cityid and s.cityname='北京' and flight.endCityId=e.cit
yID order by e.cityName desc;
2、查询北京到上海的所有航班纪录（起飞城市，到达城市，起飞时间，航班号）
select c1.CityName,c2.CityName,f.StartTime,f.flightID
from city c1,city c2,flight f
where f.StartCityID=c1.cityID
and f.endCityID=c2.cityID
and c1.cityName='北京'
and c2.cityName='上海'
3、查询具体某一天（2005-5-8）的北京到上海的的航班次数
select count(*) from
(select c1.CityName,c2.CityName,f.StartTime,f.flightID
from city c1,city c2,flight f
where f.StartCityID=c1.cityID
and f.endCityID=c2.cityID
and c1.cityName='北京'
and c2.cityName='上海'
and 查帮助获得的某个日期处理函数(startTime) like '2005-5-8%'
mysql中提取日期部分进行比较的示例代码如下：
select * from flight where date_format(starttime,'%Y-%m-%d')='1998-01-02'
Drop table if not exists employees;
create table employees(id int primary key auto_increment,name varchar(50)
,salary int,managerid int references employees(id));
insert into employees values (null,' lhm',10000,null), (null,' zxx',15000,1
),(null,'flx',9000,1),(null,'tg',10000,2),(null,'wzg',10000,3);
Wzg大于flx,lhm大于zxx
解题思路：
?????根据sql语句的查询特点，是逐行进行运算，不可能两行同时参与运算。
涉及了员工薪水和经理薪水，所有，一行记录要同时包含两个薪水，所有想到要把这个表自关联组合一下。
?????首先要组合出一个包含有各个员工及该员工的经理信息的长记录，譬如，左半部分是员工，右半部分是经理。而迪卡尔积会组合出很多垃圾信息，先去除这些垃圾信息。
select e.* from employees e,employees m where e.managerid=m.id and e.sala
ry>m.salary;
数据库中有3个表 teacher 表，student表，tea_stu关系表。
teacher 表 teaID name age
student 表 stuID name age
teacher_student表 teaID stuID
要求用一条sql查询出这样的结果
1.显示的字段要有老师name,?age 每个老师所带的学生人数
2 只列出老师age为40以下，学生age为12以上的记录
预备知识：
??????1.sql语句是对每一条记录依次处理，条件为真则执行动作（select,insert,delete,update）
???????2.只要是迪卡尔积，就会产生“垃圾”信息，所以，只要迪卡尔积了，我们首先就要想到清除“垃圾”信息
实验准备：
drop table if exists tea_stu;
drop table if exists teacher;
drop table if exists student;
??????create table teacher(teaID int primary key,name varchar(50),age int);
??????create table student(stuID int primary key,name varchar(50),age int);
??????create table tea_stu(teaID int references teacher(teaID),stuID int references student(stuID));
insert into teacher values(1,'zxx',45), (2,'lhm',25) , (3,'wzg',26) , (4,'tg',27);
insert into student values(1,'wy',11), (2,'dh',25) , (3,'ysq',26) , (4,'mxc',27);
insert into tea_stu values(1,1), (1,2), (1,3);
insert into tea_stu values(2,2), (2,3), (2,4);
?insert into tea_stu values(3,3), (3,4), (3,1);
insert into tea_stu values(4,4), (4,1), (4,2) , (4,3);
结果：2à3,3à2,4à3
解题思路：（真实面试答题时，也要写出每个分析步骤，如果纸张不够，就找别人要）
1要会统计分组信息，统计信息放在中间表中：
select teaid,count(*) from tea_stu group by teaid;
2接着其实应该是筛除掉小于12岁的学生，然后再进行统计，中间表必须与student关联才能得到12岁以下学生和把该学生记录从中间表中剔除，代码是：
select tea_stu.teaid,count(*) total from student,tea_stu
where student.stuid=tea_stu.stuid and student.age>12 group by tea_stu.teaid
3.接着把上面的结果做成虚表与teacher进行关联，并筛除大于45的老师
select teacher.teaid,teacher.name,total from teacher ,(select tea_stu.tea
id,count(*) total from student,tea_stu where student.stuid=tea_stu.stuid and stu
dent.age>12 group by tea_stu.teaid) as tea_stu2?where teacher.teaid=tea_stu2.tea
id and teacher.age<45;
select authorid,count(*) total from articles
group by authorid
having total=
(select max(total2) from (select count(*) total2 from articles group by authorid) as t);
select t.authorid,max(t.total) from
（select authorid,count(*) total from articles ）as t
这条语句不行，因为max只有一列，不能与其他列混淆。
select authorid,count(*) total from articles
group by authorid having?total=max(total)也不行。
alter table drop column score;
alter table add colunm score int;
可能会很快，但是需要试验，试验不能拿真实的环境来操刀，并且要注意，
这样的操作时无法回滚的，在我的印象中，只有inert update delete等DML语句才能回滚，
对于create table,drop table ,alter table等DDL语句是不能回滚。
解决方案一，update user set score=0;
解决方案二，假设上面的代码要执行好长时间，超出我们的容忍范围，那我就alter table user drop column score;alter table user add column score int。
下面代码实现每年的那个凌晨时刻进行清零。
Runnable runnable =
new Runnable(){
public void run(){
clearDb();
schedule(this,new Date(new Date().getYear()+1,0,0));
}
};
schedule(runnable,
new Date(new Date().getYear()+1,0,1));
select count(*) as num,tb.id
from
?tb,
?(select role?from tb where id=xxx) as t1
where
?tb.role?= t1.role?and tb.id != t1.id
group by tb.id
having
num = select count(role)?from tb where id=xxx;
Table?EMPLOYEES Structure:
EMPLOYEE_ID ?????NUMBER ???????Primary Key,
FIRST_NAME ??????VARCHAR2(25),
LAST_NAME ??????VARCHAR2(25),
Salary number(8,2),
HiredDate DATE,
Departmentid number(2)
Table?Departments?Structure:
Departmentid number(2)????????Primary Key,
DepartmentName ?VARCHAR2(25).
?(2）基于上述EMPLOYEES表写出查询：写出雇用日期在今年的，或者工资在[1000,2000]之间的，或者员工姓名（last_name）以’Obama’打头的所有员工，列出这些员工的全部个人信息。（4分）
select * from employees
where Year(hiredDate) = Year(date())
or (salary between 1000 and 200)
or left(last_name,3)='abc';
(3) 基于上述EMPLOYEES表写出查询：查出部门平均工资大于1800元的部门的所有员工，列出这些员工的全部个人信息。（4分）
mysql> select id,name,salary,deptid did from employee1 where (select avg(salary)
?from employee1 where deptid = did) > 1800;
(4)?基于上述EMPLOYEES表写出查询：查出个人工资高于其所在部门平均工资的员工，列出这些员工的全部个人信息及该员工工资高出部门平均工资百分比。（5分）
select employee1.*,(employee1.salary-t.avgSalary)*100/employee1.salary
from employee1,
(select deptid,avg(salary) avgSalary from employee1 group by deptid) as t
where employee1.deptid = t.deptid and employee1.salary>t.avgSalary;