select 查询结果，如: [学号,平均成绩：组函数avg(成绩)]from 从哪张表中查找数据，如: [涉及到成绩：成绩表score]where 查询条件，如: [b.课程号='0003' and b.成绩>80]group by 分组，如: [每个学生的平均：按学号分组](oracle,SQL server中出现在select 子句后的非分组函数，必须出现在group by子句后出现),MySQL中可以不用having 对分组结果指定条件，如: [大于60分]order by 对查询结果排序，如: [增序: 成绩 ASC / 降序: 成绩 DESC];limit 使用limt子句返回topN（对应这个问题返回的成绩前两名），如: [ limit 2 ==>从0索引开始读取2个]limit==>从0索引开始[0,N-1]
select * from table limit 2,1;
-- 含义是跳过2条取出1条数据，limit后面是从第2条开始读，读取1条信息，即读取第3条数据
select * from table limit 2 offset 1;
-- 含义是从第1条（不包括）数据开始取出2条数据，limit后面跟的是2条数据，offset后面是从第1条开始读取，即读取第2,3条
组函数: 去重 distinct() 统计总数sum() 计算个数count() 平均数avg() 最大值max() 最小数min()
多表连接: 内连接(省略默认inner) join ...on..左连接left join tableName as b on a.key ==b.key右连接right join 连接union(无重复(过滤去重))和union all(有重复[不过滤去重])
union 并集 union all(有重复) oracle(SQL server)数据库
intersect 交集 minus(except) 相减(差集)
oracle
一、数据库对象：表(table) 视图(view) 序列(sequence) 索引(index) 同义词(synonym)
1. 视图: 存储起来的 select 语句
create view emp_vw
as
select employee_id, last_name, salary
from employees
where department_id = 90;
select * from emp_vw;
可以对简单视图进行 DML 操作
update emp_vw
set last_name = 'HelloKitty'
where employee_id = 100;
select * from employees
where employee_id = 100;
1). 复杂视图
create view emp_vw2
as
select department_id, avg(salary) avg_sal
from employees
group by department_id;
select * from emp_vw2;
复杂视图不能进行 DML 操作
update emp_vw2
set avg_sal = 10000
where department_id = 100;
2. 序列：用于生成一组有规律的数值。（通常用于为主键设置值）
create sequence emp_seq1
start with 1
increment by 1
maxvalue 10000
minvalue 1
cycle
nocache;
select emp_seq1.currval from dual;
select emp_seq1.nextval from dual;
问题：裂缝，原因：
当多个表共用同一个序列时。 rollback 发生异常
create table emp1(
id number(10),
name varchar2(30)
);
insert into emp1
values(emp_seq1.nextval, '张三');
select * from emp1;
3. 索引：提高查询效率
自动创建：Oracle 会为具有唯一约束(唯一约束，主键约束)的列，自动创建索引
create table emp2(
id number(10) primary key,
name varchar2(30)
)
手动创建
create index emp_idx
on emp2(name);
create index emp_idx2
on emp2(id, name);
4. 同义词
create synonym d1 for departments;
select * from d1;
5. 表：
DDL ：数据定义语言 create table .../ drop table ... / rename ... to..../ truncate table.../alter table ...
DML : 数据操纵语言
insert into ... values ...
update ... set ... where ...
delete from ... where ...
【重要】
select ...组函数(MIN()/MAX()/SUM()/AVG()/COUNT())from ...join ... on ...左外连接：left join ... on ... 右外连接: right join ... on ...where ...group by ...(oracle,SQL server中出现在select 子句后的非分组函数，必须出现在 group by子句后)having ...用于过滤 组函数order by ...asc 升序， desc 降序limit (0,4)限制N条数据 如: topN数据
union 并集 union all(有重复) intersect 交集 minus 相减
DCL : 数据控制语言 commit : 提交 / rollback : 回滚 / 授权grant...to... /revoke
索引
何时创建索引:
一、
select employee_id, last_name, salary, department_id
from employees
where department_id in (70, 80) --> 70:1 80:34
union 并集 union all(有重复部分) intersect 交集 minus 相减
select employee_id, last_name, salary, department_id
from employees
where department_id in (80, 90) --> 90:4 80:34
问题：查询工资大于149号员工工资的员工的信息
select *
from employees
where salary > (
select salary
from employees
where employee_id = 149
)
问题：查询与141号或174号员工的manager_id和department_id相同的其他员工的employee_id, manager_id, department_id
select employee_id, manager_id, department_id
from employees
where manager_id in (
select manager_id
from employees
where employee_id in(141, 174)
) and department_id in (
select department_id
from employees
where employee_id in(141, 174)
) and employee_id not in (141, 174);
select employee_id, manager_id, department_id
from employees
where (manager_id, department_id) in (
select manager_id, department_id
from employees
where employee_id in (141, 174)
) and employee_id not in(141, 174);
from 子句中使用子查询
select max(avg(salary))
from employees
group by department_id;
select max(avg_sal)
from (
select avg(salary) avg_sal
from employees
group by department_id
) e
问题：返回比本部门平均工资高的员工的last_name, department_id, salary及平均工资
select last_name, department_id, salary, (select avg(salary) from employees where department_id = e1.department_id)
from employees e1
where salary > (
select avg(salary)
from employees e2
where e1.department_id = e2.department_id
)
select last_name, e1.department_id, salary, avg_sal
from employees e1, (
select department_id, avg(salary) avg_sal
from employees
group by department_id
) e2
where e1.department_id = e2.department_id
and e1.salary > e2.avg_sal;
case...when ... then... when ... then ... else ... end
查询：若部门为10 查看工资的 1.1 倍，部门号为 20 工资的1.2倍，其余 1.3 倍
SELECT
employee_id,
last_name,
salary,
CASE
department_id
WHEN 10 THEN
salary * 1.1
WHEN 20 THEN
salary * 1.2 ELSE salary * 1.3
END "new_salary"
FROM
employees;
SELECT
employee_id,
last_name,
salary,
decode( department_id, 10, salary * 1.1, 20, salary * 1.2, salary * 1.3 ) "new_salary"
FROM
employees;
问题：显式员工的employee_id,last_name和location。其中，若员工department_id与location_id为1800的department_id相同，则location为’Canada’,其余则为’USA’。
select employee_id, last_name, case department_id when (
select department_id
from departments
where location_id = 1800
) then 'Canada' else 'USA' end "location"
from employees;
问题：查询员工的employee_id,last_name,要求按照员工的department_name排序
select employee_id, last_name
from employees e1
order by (
select department_name
from departments d1
where e1.department_id = d1.department_id
)
SQL 优化：能使用 EXISTS 就不要使用 IN
问题：查询公司管理者的employee_id,last_name,job_id,department_id信息
select employee_id, last_name, job_id, department_id
from employees
where employee_id in (
select manager_id
from employees
)
select employee_id, last_name, job_id, department_id
from employees e1
where exists (
select 'x'
from employees e2
where e1.employee_id = e2.manager_id
)
问题：查询departments表中，不存在于employees表中的部门的department_id和department_name
select department_id, department_name
from departments d1
where not exists (
select 'x'
from employees e1
where e1.department_id = d1.department_id
)
更改 108 员工的信息: 使其工资变为所在部门中的最高工资, job 变为公司中平均工资最低的 job
update employees e1
set salary = (
select max(salary)
from employees e2
where e1.department_id = e2.department_id
), job_id = (
select job_id
from employees
group by job_id
having avg(salary) = (
select min(avg(salary))
from employees
group by job_id
)
)
where employee_id = 108;
删除 108 号员工所在部门中工资最低的那个员工.
delete from employees e1
where salary = (
select min(salary)
from employees
where department_id = (
select department_id
from employees
where employee_id = 108
)
)
select * from employees where employee_id = 108;
select * from employees where department_id = 100
order by salary;
rollback;
—完—
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