SQL 自连接不重复

Question

员工table：

+--------+----------+-----------+------------+----------+-----------------+---------+--------------------+--------------------+
| emp_id | fname    | lname     | start_date | end_date | superior_emp_id | dept_id | title              | assigned_branch_id |
+--------+----------+-----------+------------+----------+-----------------+---------+--------------------+--------------------+
|      1 | Michael  | Smith     | 2005-06-22 | NULL     |            NULL |       3 | President          |                  1 |
|      2 | Susan    | Barker    | 2006-09-12 | NULL     |               1 |       3 | Vice President     |                  1 |
|      3 | Robert   | Tyler     | 2005-02-09 | NULL     |               1 |       3 | Treasurer          |                  1 |
|      4 | Susan    | Hawthorne | 2006-04-24 | NULL     |               3 |       1 | Operations Manager |                  1 |
|      5 | John     | Gooding   | 2007-11-14 | NULL     |               4 |       2 | Loan Manager       |                  1 |
|      6 | Helen    | Fleming   | 2008-03-17 | NULL     |               4 |       1 | Head Teller        |                  1 |
|      7 | Chris    | Tucker    | 2008-09-15 | NULL     |               6 |       1 | Teller             |                  1 |
|      8 | Sarah    | Parker    | 2006-12-02 | NULL     |               6 |       1 | Teller             |                  1 |
|      9 | Jane     | Grossman  | 2006-05-03 | NULL     |               6 |       1 | Teller             |                  1 |
|     10 | Paula    | Roberts   | 2006-07-27 | NULL     |               4 |       1 | Head Teller        |                  2 |
|     11 | Thomas   | Ziegler   | 2004-10-23 | NULL     |              10 |       1 | Teller             |                  2 |
|     12 | Samantha | Jameson   | 2007-01-08 | NULL     |              10 |       1 | Teller             |                  2 |
|     13 | John     | Blake     | 2004-05-11 | NULL     |               4 |       1 | Head Teller        |                  3 |
|     14 | Cindy    | Mason     | 2006-08-09 | NULL     |              13 |       1 | Teller             |                  3 |
|     15 | Frank    | Portman   | 2007-04-01 | NULL     |              13 |       1 | Teller             |                  3 |
|     16 | Theresa  | Markham   | 2005-03-15 | NULL     |               4 |       1 | Head Teller        |                  4 |
|     17 | Beth     | Fowler    | 2006-06-29 | NULL     |              16 |       1 | Teller             |                  4 |
|     18 | Rick     | Tulman    | 2006-12-12 | NULL     |              16 |       1 | Teller             |                  4 |
+--------+----------+-----------+------------+----------+-----------------+---------+--------------------+--------------------+

查询：

SELECT e1.fname, e1.lname,
'VS' AS vs,
e2.fname, e2.lname
FROM employee e1
INNER JOIN employee e2
ON e1.emp_id < e2.emp_id
WHERE e1.title = "Teller" AND e2.title = "Teller";

结果：

+----------+----------+----+----------+----------+
| fname    | lname    | vs | fname    | lname    |
+----------+----------+----+----------+----------+
| Chris    | Tucker   | VS | Sarah    | Parker   |
| Chris    | Tucker   | VS | Jane     | Grossman |
| Chris    | Tucker   | VS | Thomas   | Ziegler  |
| Chris    | Tucker   | VS | Samantha | Jameson  |
| Chris    | Tucker   | VS | Cindy    | Mason    |
| Chris    | Tucker   | VS | Frank    | Portman  |
| Chris    | Tucker   | VS | Beth     | Fowler   |
| Chris    | Tucker   | VS | Rick     | Tulman   |
| Sarah    | Parker   | VS | Jane     | Grossman |
| Sarah    | Parker   | VS | Thomas   | Ziegler  |
| Sarah    | Parker   | VS | Samantha | Jameson  |
| Sarah    | Parker   | VS | Cindy    | Mason    |
| Sarah    | Parker   | VS | Frank    | Portman  |
| Sarah    | Parker   | VS | Beth     | Fowler   |
| Sarah    | Parker   | VS | Rick     | Tulman   |
| Jane     | Grossman | VS | Thomas   | Ziegler  |
| Jane     | Grossman | VS | Samantha | Jameson  |
| Jane     | Grossman | VS | Cindy    | Mason    |
| Jane     | Grossman | VS | Frank    | Portman  |
| Jane     | Grossman | VS | Beth     | Fowler   |
| Jane     | Grossman | VS | Rick     | Tulman   |
| Thomas   | Ziegler  | VS | Samantha | Jameson  |
| Thomas   | Ziegler  | VS | Cindy    | Mason    |
| Thomas   | Ziegler  | VS | Frank    | Portman  |
| Thomas   | Ziegler  | VS | Beth     | Fowler   |
| Thomas   | Ziegler  | VS | Rick     | Tulman   |
| Samantha | Jameson  | VS | Cindy    | Mason    |
| Samantha | Jameson  | VS | Frank    | Portman  |
| Samantha | Jameson  | VS | Beth     | Fowler   |
| Samantha | Jameson  | VS | Rick     | Tulman   |
| Cindy    | Mason    | VS | Frank    | Portman  |
| Cindy    | Mason    | VS | Beth     | Fowler   |
| Cindy    | Mason    | VS | Rick     | Tulman   |
| Frank    | Portman  | VS | Beth     | Fowler   |
| Frank    | Portman  | VS | Rick     | Tulman   |
| Beth     | Fowler   | VS | Rick     | Tulman   |
+----------+----------+----+----------+----------+

意向：

我想将人们配对在一起，就像员工在玩国际象棋游戏一样，在第一轮中，一名员工将只与一个人对战，而不是 3 或 4 个不同的人。例如，我不希望 Chris Tucker 在第一轮对阵 Sarah、Jane、Thomas 等人。我希望他只与一个对手配对。如果他或他的对手输了，他们就会出局。

我该怎么做？

Answer 1

我会通过随机枚举名称然后成对选择它们来解决这个问题：

select max(case when seqnum % 2 = 0 then fname end) as fname_1,
       max(case when seqnum % 2 = 0 then lname end) as lname_1,
       max(case when seqnum % 2 = 1 then fname end) as fname_2,
       max(case when seqnum % 2 = 1 then lname end) as lname_2       
from (select e.*, (@rn := @rn + 1) as seqnum
      from employee e cross join
           (select @rn := 0) params
      order by rand()
     ) e
group by floor((seqnum - 1) / 2);

这假设您有偶数的员工，这似乎被隐含地假设为问题的一部分。

SQL 自连接不重复

SQL Self Join with no repetition

mysql

self-join