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Mysql partition by 实现及详解

2014-09-05 09:57阅读:
解决方案:
drop table if exists heyf_t10;
create table heyf_t10 (empid int ,deptid int ,salary decimal(10,2) );
insert into heyf_t10 values
(1,10,5500.00),
(2,10,4500.00),
(3,20,1900.00),
(4,20,4800.00),
(5,40,6500.00),
(6,40,14500.00),
(7,40,44500.00),
(8,50,6500.00),
(9,50,7500.00);
2. 确定需求: 根据部门来分组,显示各员工在部门里按薪水排名名次.
显示结果预期如下:
+-------+--------+----------+------+
| empid | deptid | salary | rank |
+-------+--------+----------+------+
| 1 | 10 | 5500.00 | 1 |
| 2 | 10 | 4500.00 | 2 |
| 4 | 20 | 4800.00 | 1 |
| 3 | 20 | 1900.00 | 2 |
| 7 | 40 | 44500.00 | 1 |
| 6 | 40 | 14500.00 | 2 |
| 5 | 40 | 6500.00 | 3 |
| 9 | 50 | 7500.00 | 1 |
| 8 | 50
| 6500.00 | 2 |
-----------------------------------------------
做好的sql,SQL 实现

select empid,deptid,salary,rank from (
select heyf_tmp.empid,heyf_tmp.deptid,heyf_tmp.salary,@rownum:=@rownum+1 ,
if(@pdept=heyf_tmp.deptid,@rank:=@rank+1,@rank:=1) as rank,
@pdept:=heyf_tmp.deptid
from (
select empid,deptid,salary from heyf_t10 order by deptid asc ,salary desc
) heyf_tmp ,
(select @rownum :=0 , @pdept := null ,@rank:=0) a
) result ;
结果演示
+-------+--------+----------+------+
| empid | deptid | salary | rank |
+-------+--------+----------+------+
| 1 | 10 | 5500.00 | 1 |
| 2 | 10 | 4500.00 | 2 |
| 4 | 20 | 4800.00 | 1 |
| 3 | 20 | 1900.00 | 2 |
| 7 | 40 | 44500.00 | 1 |
| 6 | 40 | 14500.00 | 2 |
| 5 | 40 | 6500.00 | 3 |
| 9 | 50 | 7500.00 | 1 |
| 8 | 50 | 6500.00 | 2 |
+-------+--------+----------+------+
9 rows in set (0.00 sec)
详解:
执行顺序:FROM > WHERE > GROUP BY > HAVING > SELECT > ORDER BY > LIMIT
特别主要需要将去重的记录通过order by 汇总到一起

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