group_concat([distinct] 字段名 [order by 排序字段 asc/desc] [separator '分隔符'])

create database mydb4;
use mydb4;
create table emp(
    emp_id int primary key auto_increment comment '编号',
    emp_name char(20) not null default '' comment '姓名',
    salary decimal(10,2) not null default 0 comment '工资',
    department char(20) not null default '' comment '部门'
insert into emp(emp_name,salary,department) values('张晶晶',5000,'财务部'),('王飞飞',5800,'财务部'),('赵刚',6200,'财务部'),('刘小贝',5700,'人事部'),('王大鹏',6700,'人事部'),('张小斐',5200,'人事部'),('刘云云',7500,'销售部'),('刘云鹏',7200,'销售部'),('刘云鹏',7800,'销售部');
-- 将所有员工的名字合并成一行 
select group_concat(emp_name) from emp;

-- 指定分隔符，分组合并 
select department,group_concat(emp_name separator ';' ) from emp group by department;

-- 指定排序方式和分隔符 
select department,group_concat(emp_name order by salary desc separator ';') from emp group by department;

use mydb4; 
-- 创建订单表
create table orders(
 oid int primary key, -- 订单id
 price double, -- 订单价格
 payType int -- 支付类型(1:微信支付 2:支付宝支付 3:银行卡支付 4：其他)
insert into orders values(1,1200,1);
insert into orders values(2,1000,2);
insert into orders values(3,200,3);
insert into orders values(4,3000,1);
insert into orders values(5,1500,2);
-- 方式1
select 
  when payType=1 then '微信支付' 
    when payType=2 then '支付宝支付' 
    when payType=3 then '银行卡支付' 
    else '其他支付方式' 
end  as payTypeStr
from orders;
-- 方式2
select 
case payType
  when 1 then '微信支付' 
    when 2 then '支付宝支付' 
    when 3 then '银行卡支付' 
    else '其他支付方式' 
end  as payTypeStr
from orders;

window_function ( expr ) OVER ( 
  PARTITION BY ... 
  ORDER BY ... 
  frame_clause 
)

row_number()|rank()|dense_rank() over ( 
  partition by ... 
  order by ... 
)

use mydb4; 
create table employee( 
   dname varchar(20), -- 部门名 
   eid varchar(20), 
   ename varchar(20), 
   hiredate date, -- 入职日期 
   salary double -- 薪资
insert into employee values('研发部','1001','刘备','2021-11-01',3000);
insert into employee values('研发部','1002','关羽','2021-11-02',5000);
insert into employee values('研发部','1003','张飞','2021-11-03',7000);
insert into employee values('研发部','1004','赵云','2021-11-04',7000);
insert into employee values('研发部','1005','马超','2021-11-05',4000);
insert into employee values('研发部','1006','黄忠','2021-11-06',4000);
insert into employee values('销售部','1007','曹操','2021-11-01',2000);
insert into employee values('销售部','1008','许褚','2021-11-02',3000);
insert into employee values('销售部','1009','典韦','2021-11-03',5000);
insert into employee values('销售部','1010','张辽','2021-11-04',6000);
insert into employee values('销售部','1011','徐晃','2021-11-05',9000);
insert into employee values('销售部','1012','曹洪','2021-11-06',6000);
-- 对每个部门的员工按照薪资排序，并给出排名
select dname,ename,salary,row_number() over(partition by dname order by salary desc) as rn from employee;

-- 对每个部门的员工按照薪资排序，并给出排名 dense-rank
select dname,ename,salary,dense_rank() over(partition by dname order by salary desc) as rn 
from employee;

--求出每个部门薪资排在前三名的员工- 分组求TOPN
select * from 
    select 
     dname,
     ename,
     salary,
     dense_rank() over(partition by dname order by salary desc)  as rn
    from employee
)t where t.rn <= 3;

-- 对所有员工进行全局排序（不分组）
-- 不加partition by表示全局排序
select 
     dname,
     ename,
     salary,
     dense_rank() over( order by salary desc)  as rn
from employee;

select  
 dname,
 ename,
 salary,
 sum(salary) over(partition by dname order by hiredate) as pv1 
from employee;
select cookieid,createtime,pv,
sum(pv) over(partition by cookieid) as pv3
from itcast_t1;  -- 如果没有order  by排序语句  默认把分组内的所有数据进行sum操作
select  
 dname,
 ename,
 salary,
 sum(salary) over(partition by dname order by hiredate  rows between unbounded preceding and current row) as c1 
from employee;
select  
 dname,
 ename,
 salary,
 sum(salary) over(partition by dname order by hiredate   rows between 3 preceding and current row) as c1 
from employee;  --c1列为上面3行（含本行）的工资总和
select  
 dname,
 ename,
 salary,
 sum(salary) over(partition by dname order by hiredate   rows between 3 preceding and 1 following) as c1 
from employee;  --c1列为上面3行（含本行）和下面1行的工资总和
select  
 dname,
 ename,
 salary,
 sum(salary) over(partition by dname order by hiredate   rows between current row and unbounded following) as c1 
from employee;

select  
 dname,
 ename,
 salary,
 cume_dist() over(order by salary) as rn1, -- 没有partition语句 所有的数据位于一组
 cume_dist() over(partition by dept order by salary) as rn2 
from employee;
rn1: 没有partition,所有数据均为1组，总行数为12，
     第一行：小于等于3000的行数为3，因此，3/12=0.25
     第二行：小于等于4000的行数为5，因此，5/12=0.4166666666666667
rn2: 按照部门分组，dname='研发部'的行数为6,
     第一行：研发部小于等于3000的行数为1，因此，1/6=0.16666666666666666
*/

select 
 dname,
 ename,
 salary,
 rank() over(partition by dname order by salary desc ) as rn,
 percent_rank() over(partition by dname order by salary desc ) as rn2
from employee;
  第一行: (1 - 1) / (6 - 1) = 0
  第二行: (1 - 1) / (6 - 1) = 0
  第三行: (3 - 1) / (6 - 1) = 0.4
*/

-- lag的用法
select 
 dname,
 ename,
 hiredate,
 salary,
 lag(hiredate,1,'2000-01-01') over(partition by dname order by hiredate) as last_1_time,
 lag(hiredate,2) over(partition by dname order by hiredate) as last_2_time 
from employee;
last_1_time: 指定了往上第1行的值，default为'2000-01-01'  
                         第一行，往上1行为null,因此取默认值 '2000-01-01'
                         第二行，往上1行值为第一行值，2021-11-01 
                         第三行，往上1行值为第二行值，2021-11-02 
last_2_time: 指定了往上第2行的值，为指定默认值
                         第一行，往上2行为null
                         第二行，往上2行为null
                         第四行，往上2行为第二行值，2021-11-01 
                         第七行，往上2行为第五行值，2021-11-02 
*/

-- lead的用法
select 
 dname,
 ename,
 hiredate,
 salary,
 lead(hiredate,1,'2000-01-01') over(partition by dname order by hiredate) as last_1_time,
 lead(hiredate,2) over(partition by dname order by hiredate) as last_2_time 
from employee;

-- 注意,  如果不指定ORDER BY，则进行排序混乱，会出现错误的结果
select
  dname,
  ename,
  hiredate,
  salary,
  first_value(salary) over(partition by dname order by hiredate) as first,
  last_value(salary) over(partition by dname order by  hiredate) as last 
from  employee;

-- 查询每个部门截止目前薪资排在第二和第三的员工信息
select 
  dname,
  ename,
  hiredate,
  salary,
  nth_value(salary,2) over(partition by dname order by hiredate) as second_score,
  nth_value(salary,3) over(partition by dname order by hiredate) as third_score
from employee

-- 根据入职日期将每个部门的员工分成3组
select 
  dname,
  ename,
  hiredate,
  salary,
ntile(3) over(partition by dname order by  hiredate) as rn 
from employee;

-- 取出每个部门的第一组员工
select
    SELECT 
        dname,
        ename,
        hiredate,
        salary,
    NTILE(3) OVER(PARTITION BY dname ORDER BY  hiredate  ) AS rn 
    FROM employee
where t.rn = 1;