- 由于MYSQL没有提供类似ORACLE中OVER()这样丰富的分析函数. 所以在MYSQL里需要实现这样的功能,我们只能用一些灵活的办法:
- 1.首先我们来创建实例数据:
- 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 |
- +-------+--------+----------+------+
- 3. 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
- ;
- 4. 结果演示
- mysql> 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
- -> ;
- +-------+--------+----------+------+
- | 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)
- MySql中取出每个分组中的前N条记录
- select a1.* from article a1
- inner join
- (select a.type,a.date from article a left join article b
- on a.type=b.type and a.date<=b.date
- group by a.type,a.date
- having count(b.date)<=2
- )b1
- on a1.type=b1.type and a1.date=b1.date
- order by a1.type,a1.date desc
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