MySQL记录

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代码

题目

获取每个部门中员工薪水最高的员工信息

构造表如下

mysql> select * from dept_emp;
+--------+---------+------------+------------+
| emp_no | dept_no | from_date  | to_date    |
+--------+---------+------------+------------+
|  10001 | d001    | 1986-06-26 | 9999-01-01 |
|  10002 | d001    | 1996-08-03 | 9999-01-01 |
|  10003 | d004    | 1995-12-03 | 9999-01-01 |
|  10004 | d004    | 1986-12-01 | 9999-01-01 |
|  10005 | d003    | 1989-09-12 | 9999-01-01 |
|  10006 | d002    | 1990-08-05 | 9999-01-01 |
|  10007 | d005    | 1989-02-10 | 9999-01-01 |
|  10009 | d006    | 1985-02-18 | 9999-01-01 |
|  10010 | d006    | 2000-06-26 | 9999-01-01 |
+--------+---------+------------+------------+
mysql> select * from salaries;
+--------+--------+------------+------------+
| emp_no | salary | from_date  | to_date    |
+--------+--------+------------+------------+
|  10001 |  88958 | 2002-06-22 | 9999-01-01 |
|  10002 |  72527 | 2001-08-02 | 9999-01-01 |
|  10003 |  43311 | 2001-12-01 | 9999-01-01 |
|  10004 |  74057 | 2001-11-27 | 9999-01-01 |
|  10005 |  94692 | 2001-09-09 | 9999-01-01 |
|  10006 |  43311 | 2001-08-02 | 9999-01-01 |
|  10007 |  88070 | 2002-02-07 | 9999-01-01 |
|  10009 |  95409 | 2002-02-14 | 9999-01-01 |
|  10010 |  94409 | 2001-11-23 | 9999-01-01 |
+--------+--------+------------+------------+

需要连接两表：

mysql> select d.dept_no,d.emp_no,s.salary
    -> from dept_emp d, salaries s
    -> where d.emp_no=s.emp_no;
+---------+--------+--------+
| dept_no | emp_no | salary |
+---------+--------+--------+
| d001    |  10001 |  88958 |
| d001    |  10002 |  72527 |
| d004    |  10003 |  43311 |
| d004    |  10004 |  74057 |
| d003    |  10005 |  94692 |
| d002    |  10006 |  43311 |
| d005    |  10007 |  88070 |
| d006    |  10009 |  95409 |
| d006    |  10010 |  94409 |
+---------+--------+--------+


mysql> select d.dept_no,s.emp_no,s.salary
    -> from dept_emp d inner join salaries s
    -> on d.emp_no=s.emp_no
    -> group by d.dept_no
    -> ;
+---------+--------+--------+
| dept_no | emp_no | salary |
+---------+--------+--------+
| d001    |  10001 |  88958 |
| d004    |  10003 |  43311 |
| d003    |  10005 |  94692 |
| d002    |  10006 |  43311 |
| d005    |  10007 |  88070 |
| d006    |  10009 |  95409 |
+---------+--------+--------+


mysql> select d.dept_no,d.emp_no,max(s.salary) salary
    ->      from dept_emp d, salaries s
    ->      where d.emp_no=s.emp_no
    ->      group by d.dept_no;
+---------+--------+--------+
| dept_no | emp_no | salary |
+---------+--------+--------+
| d001    |  10001 |  88958 |
| d004    |  10003 |  74057 |
| d003    |  10005 |  94692 |
| d002    |  10006 |  43311 |
| d005    |  10007 |  88070 |
| d006    |  10009 |  95409 |
+---------+--------+--------+
# 通过group by分组，得到的max(s.salary)与前两项不匹配，
# group by分组得到的dept_no和emp_no都没有变，只把max(salary)变了
# ！！！！对比一下这个和上一个！！！！

mysql> select d.dept_no,s.emp_no,max(s.salary)
    -> from dept_emp d, salaries s
    -> where d.emp_no=s.emp_no
    -> ;
+---------+--------+---------------+
| dept_no | emp_no | max(s.salary) |
+---------+--------+---------------+
| d001    |  10001 |         95409 |
+---------+--------+---------------+

发现max(salary)是独立的，并不会提取max(salary)相对应的哪一行。

解决办法：
使用partition by，这也是分组，但group by是聚合分组，即一组数据最后得出了一个值
而partition by可以将分组后的数据都显示出来
代码如下：

mysql> SELECT  dept_no,emp_no,salary
    -> FROM (
    ->     SELECT dept_no,salary,d.emp_no,
    ->             row_number() OVER (PARTITION BY dept_no ORDER BY salary DESC) AS a
    ->     FROM dept_emp d
    ->     INNer JOIN salaries s
    ->     ON d.emp_no= s.emp_no) b
    -> WHERE a =1;
+---------+--------+--------+
| dept_no | emp_no | salary |
+---------+--------+--------+
| d001    |  10001 |  88958 |
| d002    |  10006 |  43311 |
| d003    |  10005 |  94692 |
| d004    |  10004 |  74057 |
| d005    |  10007 |  88070 |
| d006    |  10009 |  95409 |
+---------+--------+--------+
6 rows in set (0.00 sec)

# 分解步骤：
mysql>     SELECT dept_no,salary,d.emp_no,
    ->             row_number() OVER (PARTITION BY dept_no ORDER BY salary DESC) AS a
    ->     FROM dept_emp d
    ->     INNer JOIN salaries s
    ->     ON d.emp_no= s.emp_no;
+---------+--------+--------+---+
| dept_no | salary | emp_no | a |
+---------+--------+--------+---+
| d001    |  88958 |  10001 | 1 |
| d001    |  72527 |  10002 | 2 |
| d002    |  43311 |  10006 | 1 |
| d003    |  94692 |  10005 | 1 |
| d004    |  74057 |  10004 | 1 |
| d004    |  43311 |  10003 | 2 |
| d005    |  88070 |  10007 | 1 |
| d006    |  95409 |  10009 | 1 |
| d006    |  94409 |  10010 | 2 |
+---------+--------+--------+---+


把row_number()替换成max()函数试试~

mysql>     SELECT dept_no,salary,d.emp_no,
    ->             max(salary) OVER (PARTITION BY dept_no ORDER BY salary DESC) AS a
    ->     FROM dept_emp d
    ->     INNer JOIN salaries s
    ->     ON d.emp_no= s.emp_no;
+---------+--------+--------+-------+
| dept_no | salary | emp_no | a     |
+---------+--------+--------+-------+
| d001    |  88958 |  10001 | 88958 |
| d001    |  72527 |  10002 | 88958 |
| d002    |  43311 |  10006 | 43311 |
| d003    |  94692 |  10005 | 94692 |
| d004    |  74057 |  10004 | 74057 |
| d004    |  43311 |  10003 | 74057 |
| d005    |  88070 |  10007 | 88070 |
| d006    |  95409 |  10009 | 95409 |
| d006    |  94409 |  10010 | 95409 |
+---------+--------+--------+-------+
9 rows in set (0.00 sec)

知识点
row_number(),dense_rank(),rank()都是排序，但是比如100，100，200排序
row_number()排序结果为1，2，3（正常排序）
dense_rank()排序结果为1，1，2（总量变）
rank() 排序结果为1，1，3（总量不变）