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py知识(每日更新) 8.2

时间:2019-08-09 19:18:30      阅读:73      评论:0      收藏:0      [点我收藏+]

标签:nio   order by   员工   dep   现在   计算   连表   lex   筛选   

连表查询

# 笛卡尔积
# 先计算两张表的笛卡尔积,再根据用户给出的条件进行筛选
select * from employee,department where dep_id = department.id

# 内连接  inner join ... on 连接条件
    select * from 表1 inner join 表2 on 条件
    select * from employee inner join department on dep_id = department.id
        employee --> dep_id: 200,201,202,204
        department --> id : 200,201,202,203
    # 内连接 :只显示两张表中互相匹配的项,其他不匹配的不显示
# 外连接z
    # 左外连接 left join .. on
        select * from 表1 left join 表2 on 条件
        select * from employee left join department on dep_id = department.id
        select * from department left join employee  on dep_id = department.id
        # 不管左表中是不是匹配上都会显示所有内容
    # 右外连接 right join .. on
        select * from 表1 right join 表2 on 条件
        select * from employee right join department on dep_id = department.id
        # 不管右表中是不是匹配上都会显示所有内容
    # 全外连接
        select * from department left join employee  on dep_id = department.id
        union
        select * from department right join employee on dep_id = department.id

# 连表查询
# 原来是两张表,现在拼成一张表,所有的操作都像是操作一张表一样了
# 年龄大于25岁的员工以及员工所在的部门
select * from employee inner join department d on d.id = dep_id;

# 以内连接的方式查询employee和department表,并且以age字段的升序方式显示
select * from employee e inner join department d on d.id = e.dep_id order by age;

子查询

# 存在年龄大于25岁员工的部门
select * from employee where age>25;
select distinct dep_id from employee where age>25;
select * from department where id in (select distinct dep_id from employee where age>25);

# a.查询平均年龄在25岁以上的部门名
    # 求部门的平均年龄
    select avg(age) from employee group by dep_id
    # 求部门的平均年龄>25岁的 部门id
    select dep_id from employee group by dep_id having avg(age)>25
    # 根据部门id求部门名
    select name from department where id in (select dep_id from employee group by dep_id having avg(age)>25);

# b.查询平均年龄在25岁以上的部门名,平均年龄的值
    # 方法1:
    # 先查部门的平均年龄>25岁的部门id,平均年龄
    select dep_id,avg(age) from employee group by dep_id having avg(age) > 25
    # 查出结果在之后再连表
    select name,avg_age from department as d right join (select dep_id,avg(age) as avg_age from employee group by dep_id having avg(age) > 25) as t on d.id = t.dep_id

    # 方法2
    # 先查各部门的平均年龄
    select dep_id,avg(age) from employee group by dep_id;
    # 查出结果在之后再连表
    select name,avg_age from
        department as d inner join (select dep_id,avg(age) as avg_age from employee group by dep_id) as t
        on d.id =t.dep_id
        where avg_age>25

    # 根据a,b总结出来的规律:
    # 如果最终需要的结果只出现在一张表中,可以用子查询解决问题
    # 如果最终需要的结果出现在2表中,那么最后用的一定是连表查询

# 查看"技术"部员工姓名
    # 技术部门在department表中的id
    select id from department where name = '技术';
    # 查的是姓名 employee
    select name from employee where dep_id = (select id from department where name = '技术');

# 查看不足1人的部门名
    # 子查询
    # 先查出所有人的部门id
        select distinct dep_id from employee;
    # 看不在部门id这个集合中的部门有哪些
        select name from department where id not in (select distinct dep_id from employee);

    # 连表
    select d.name from department d left join employee e on d.id = dep_id where e.id is null;

# 查询大于所有人平均年龄的员工名与年龄
    # 先查询所有人的平均年龄
    select avg(age) from employee;
    select name,age from employee where age > (select avg(age) from employee);

# 查询大于部门内平均年龄的员工名、年龄
    # 先求各部门的平均年龄
    select dep_id,avg(age) from employee group by dep_id;
    # 拼接
    select e.name,e.age from
        employee e
    inner join
        (select dep_id,avg(age) avg_age from employee group by dep_id) t
    on e.dep_id = t.dep_id
    where e.age > t.avg_age;

select * from 表 where 字段 > 一个值

如果我们查询出来的结果是一个值,那么就可以使用 > < =

select * from 表 where 字段 in (1个值,1个值)

如果我们查询出来的结果是一列中的多个值

    # dep_id
    # 201
    # 202
    # 204
    in ;not in

如果我们查出来的是一张表,也不能作为条件,如果必须保留这两个字段,就不能用作条件,只能连表

# id name
# 1  alex

select 语句到底做了什么?

select name from emp;
select name as n,(select age from employee where name = n) from emp;

子查询处理可以放在条件中,还可以放在连表中,还可以放在select字段(要求查询的结果必须是一个单行单列的值)中.

推荐连表 : 连表的效率比子查询的效率要高

py知识(每日更新) 8.2

标签:nio   order by   员工   dep   现在   计算   连表   lex   筛选   

原文地址:https://www.cnblogs.com/lyoko1996/p/11328913.html

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