SQL255 给出employees表中排名为奇数行的first

题目来源：

给出employees表中排名为奇数行的first_name_牛客题霸_牛客网

描述

对于employees表中，输出first_name排名(按first_name升序排序)为奇数的first_name

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,PRIMARY KEY (`emp_no`));

如，输入为：

INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');

输出格式:

first

Georgi

Anneke

请你在不打乱原序列顺序的情况下，输出：按first_name排升序后，取奇数行的first_name。

如对以上示例数据的first_name排序后的序列为：Anneke、Bezalel、Georgi、Kyoichi。

则原序列中的Georgi排名为3，Anneke排名为1，所以按原序列顺序输出Georgi、Anneke。

drop table if exists  `employees` ; 
CREATE TABLE `employees` (`emp_no` int(11) NOT NULL,`birth_date` date NOT NULL,`first_name` varchar(14) NOT NULL,`last_name` varchar(16) NOT NULL,`gender` char(1) NOT NULL,`hire_date` date NOT NULL,PRIMARY KEY (`emp_no`));
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');

解决

解决方案一：

先用row_number()对first_name排序，再利用mod()函数取出奇数行，这样取出的行里面的first_name就是我们要的了，由于要返回的是初始的排序状态，所以我们再套一层select直接从employees表取first_name,只要这里的first_name在我们上面取出的first_name里面就好了：

select e.first_name from employees e
where e.first_name in
(select a.first_name as first_name from
(select first_name,ROW_NUMBER() over(order by first_name) as rk
from employees) a
where mod(a.rk,2)!=0);

解决方案二：

mysql> select m1.first_name from-> (select e1.first_name,count(*) as 'rowid' from-> employees e1-> left join employees e2-> on e1.first_name >= e2.first_name-> group by e1.first_name ) as m1-> where m1.rowid % 2 = 1;
+------------+
| first_name |
+------------+
| Georgi     |
| Anneke     |
+------------+
2 rows in set (0.01 sec)

以下是我解题过程：

在我的查询中，我使用了 JOIN 条件 e1.first_name >= e2.first_name，这意味着查询返回的结果中，e1 表中的每一行都会与 e2 表中所有满足条件的行进行连接，找出小于等于他名字的计数。连接的结果不再按照原表的顺序排列，而是根据连接条件的满足程度和其他优化因素来确定最终的顺序。（这是gpt说的）

mysql> select e1.first_name,count(*) from-> employees e1-> join employees e2-> on e1.first_name >= e2.first_name-> group by e1.first_name;
+------------+----------+
| first_name | count(*) |
+------------+----------+
| Kyoichi    |        4 |
| Georgi     |        3 |
| Bezalel    |        2 |
| Anneke     |        1 |
+------------+----------+
4 rows in set (0.00 sec)mysql>
mysql> select e1.first_name,count(e1.first_name) from-> employees e1-> join employees e2-> on e1.first_name >= e2.first_name-> group by e1.first_name;
+------------+----------------------+
| first_name | count(e1.first_name) |
+------------+----------------------+
| Kyoichi    |                    4 |
| Georgi     |                    3 |
| Bezalel    |                    2 |
| Anneke     |                    1 |
+------------+----------------------+
4 rows in set (0.00 sec)

但是这样查出来的数据改变了数据的顺序：

mysql> select first_name from employees; 
+------------+
| first_name |
+------------+
| Georgi     |
| Bezalel    |
| Kyoichi    |
| Anneke     |
+------------+
4 rows in set (0.00 sec)mysql> select e1.first_name from-> employees e1-> join employees e2-> on e1.first_name >= e2.first_name-> group by e1.first_name;
+------------+
| first_name |
+------------+
| Kyoichi    |
| Georgi     |
| Bezalel    |
| Anneke     |
+------------+
4 rows in set (0.00 sec)mysql> select e1.first_name from-> employees e1-> join employees e2-> on e1.first_name >= e2.first_name;
+------------+
| first_name |
+------------+
| Kyoichi    |
| Georgi     |
| Kyoichi    |
| Bezalel    |
| Georgi     |
| Kyoichi    |
| Anneke     |
| Kyoichi    |
| Bezalel    |
| Georgi     |
+------------+
10 rows in set (0.00 sec)

为了不改变表中的数据顺序，所以我想到了左连接，把e1表当作主表，外连接的功能就是把主表中的数据全部查出来，副表中没有一条数据与之匹配上的，就用null字段代替，且分组函数自动忽略null值：

mysql> select e1.first_name from-> employees e1-> left join employees e2-> on e1.first_name >= e2.first_name;
+------------+
| first_name |
+------------+
| Georgi     |
| Georgi     |
| Georgi     |
| Bezalel    |
| Bezalel    |
| Kyoichi    |
| Kyoichi    |
| Kyoichi    |
| Kyoichi    |
| Anneke     |
+------------+
10 rows in set (0.00 sec)


mysql> select * from-> employees e1-> left join employees e2-> on e1.first_name >= e2.first_name;
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date  | emp_no | birth_date | first_name | last_name | gender | hire_date  |
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
10 rows in set (0.00 sec)mysql> select * from-> employees e1-> left join employees e2-> on e1.first_name >  e2.first_name;
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date  | emp_no | birth_date | first_name | last_name | gender | hire_date  |
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |   NULL | NULL       | NULL       | NULL      | NULL   | NULL       |
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
7 rows in set (0.00 sec)

现在就跟这个顺序一样了：

最终代码：

mysql> select m1.first_name from-> (select e1.first_name,count(*) as 'rowid' from-> employees e1-> left join employees e2-> on e1.first_name >= e2.first_name-> group by e1.first_name ) as m1-> where m1.rowid % 2 = 1;
+------------+
| first_name |
+------------+
| Georgi     |
| Anneke     |
+------------+
2 rows in set (0.01 sec)

改进的方案二：

方案一虽然能通过，其实并不完美，因为方案一没有问题的前提是employees 中的first_name没有重名的，如果我加了一条重名字的数据进去，就出问题了：

mysql> INSERT INTO employees VALUES(10007,'1953-09-02','Georgi','Facello','M','1986-06-26');
Query OK, 1 row affected (0.01 sec)

没执行insert操作前：

如果你要像方案二一样通过join两张表来获得排名，那就必须去重，要不然Georgi 为8，意味着小于等于他名字的有八个，这不就摇身一变，变成了名字最大的了吗

执行了insert操作后：

mysql> select e1.first_name,count(*) as 'rowid' from-> employees e1-> left join employees e2-> on e1.first_name >= e2.first_name-> group by e1.first_name ;
+------------+-------+
| first_name | rowid |
+------------+-------+
| Georgi     |     8 |
| Bezalel    |     2 |
| Kyoichi    |     5 |
| Anneke     |     1 |
+------------+-------+
4 rows in set (0.00 sec)

原因如下：

mysql> select * from-> employees e1-> left join employees e2-> on e1.first_name >=    e2.first_name;
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
| emp_no | birth_date | first_name | last_name | gender | hire_date  | emp_no | birth_date | first_name | last_name | gender | hire_date  |
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10005 | 1955-01-21 | Kyoichi    | Maliniak  | M      | 1989-09-12 |  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
|  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10006 | 1953-04-20 | Anneke     | Preusig   | F      | 1989-06-02 |
|  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10002 | 1964-06-02 | Bezalel    | Simmel    | F      | 1985-11-21 |
|  10007 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |  10001 | 1953-09-02 | Georgi     | Facello   | M      | 1986-06-26 |
+--------+------------+------------+-----------+--------+------------+--------+------------+------------+-----------+--------+------------+
16 rows in set (0.00 sec)mysql>
mysql> select e1.first_name from-> employees e1-> left join employees e2-> on e1.first_name >= e2.first_name;
+------------+
| first_name |
+------------+
| Georgi     |
| Georgi     |
| Georgi     |
| Georgi     |
| Bezalel    |
| Bezalel    |
| Kyoichi    |
| Kyoichi    |
| Kyoichi    |
| Kyoichi    |
| Kyoichi    |
| Anneke     |
| Georgi     |
| Georgi     |
| Georgi     |
| Georgi     |
+------------+
16 rows in set (0.00 sec)

我们需要联合两个字段去重：

mysql> select  distinct e1.first_name ,e2.first_name #distinct去掉重名的-> from employees e1-> left join employees e2-> on e1.first_name >= e2.first_name;
+------------+------------+
| first_name | first_name |
+------------+------------+
| Georgi     | Georgi     |
| Georgi     | Anneke     |
| Georgi     | Bezalel    |
| Bezalel    | Anneke     |
| Bezalel    | Bezalel    |
| Kyoichi    | Georgi     |
| Kyoichi    | Anneke     |
| Kyoichi    | Kyoichi    |
| Kyoichi    | Bezalel    |
| Anneke     | Anneke     |
+------------+------------+
10 rows in set (0.00 sec)

mysql> select count(*) as rowid,m1.first_name from-> (->-> select  distinct e1.first_name  , e2.first_name  as name-> from employees e1-> left join employees e2-> on e1.first_name >= e2.first_name-> ) as m1 group by m1.first_name;
+-------+------------+
| rowid | first_name |
+-------+------------+
|     3 | Georgi     |
|     2 | Bezalel    |
|     4 | Kyoichi    |
|     1 | Anneke     |
+-------+------------+
4 rows in set (0.01 sec)

最终答案:

mysql> select m2.first_name from-> (select count(*) as rowid,m1.first_name from-> (->-> select  distinct e1.first_name  , e2.first_name  as name-> from employees e1-> left join employees e2-> on e1.first_name >= e2.first_name-> ) as m1 group by m1.first_name)as m2 where m2.rowid % 2 = 1;
+------------+
| first_name |
+------------+
| Georgi     |
| Anneke     |
+------------+
2 rows in set (0.00 sec)