Question

我有一个名为调查的表格，其中包含组列和主题列

CREATE TABLE survey (
  `group` INT NOT NULL,
  `subject` VARCHAR(16) NOT NULL,
  UNIQUE INDEX (`group`, `subject`)
);

INSERT INTO survey 
  VALUES
  (1, 'sports'),
  (1, 'history'),
  (2, 'art'),
  (2, 'music'),
  (3, 'math'),
  (3, 'sports'),
  (3, 'science')
;

我试图找出一个查询，它将返回不属于同一组的所有主题对。所以从上面的例子中，我希望看到这些对在表中返回：

science - history  
science - art  
science - music  
history - math  
sports  - art  
sports  - music  
history - art  
history - music

因此，查询不应该返回：

sports - history

作为一个例子，因为他们都在第1组。

非常感谢。

Answer 1

SELECT s1.subject,
       s2.subject
FROM   survey s1
       JOIN survey s2
         ON s1.subject < s2.subject
GROUP  BY s1.subject,
          s2.subject
HAVING COUNT(CASE
               WHEN s1.groupid = s2.groupid THEN 1
             END) = 0

Answer 2

样本表

create table Survey(groupid int, subject varchar(100))
insert into Survey select
1, 'sports' union all select
1, 'history' union all select
2, 'art' union all select
2, 'music' union all select
3, 'math' union all select
3, 'sports' union all select
3, 'science'

符合ANSI标准的查询，适用于所有主流RDBMS

select a.subject, b.subject
from (select distinct subject from Survey) A
inner join (select distinct subject from Survey) B on B.subject > A.subject
left join Survey C on C.subject = A.subject
left join Survey D on D.subject = B.subject and D.groupid = C.groupid
where D.groupid is null
order by a.subject, b.subject

Answer 3

这是一种略有不同的方法：

SELECT *
FROM (SELECT DISTINCT subject FROM yourtable) AS T1
JOIN (SELECT DISTINCT subject FROM yourtable) AS T2
ON T1.subject < T2.subject
WHERE NOT EXISTS
(
    SELECT *
    FROM yourtable T3
    JOIN yourtable T4
    ON T3.id = T4.id
    WHERE T1.subject = T3.subject
    AND T2.subject = T4.subject
)
ORDER BY t1.subject, t2.subject;

Answer 4

标准方法是使用MINUS来获取同一组中所有对的补充，但MySQL doesn't support MINUS。对于MySQL，您可以基于MINUS运算符和子查询将NOT IN转换为语句：

SELECT s1.subject, s2.subject
  FROM survey AS s1
    JOIN survey AS s2
  WHERE (s1.subject, s2.subject) NOT IN
  (
    SELECT s1.subject, s2.subject
      FROM survey AS s1
        JOIN survey AS s2
          ON s1.group = s2.group
  )
;

请注意，这可能会产生重复。如果您不想要它们，请使用SELECT DISTINCT。

使用索引和示例数据，扩展查询计划为：

+----+--------------------+-------+--------+---------------+-------+---------+--------------------+------+----------+---------------------------------------------+
| id | select_type        | table | type   | possible_keys | key   | key_len | ref                | rows | filtered | Extra                                       |
+----+--------------------+-------+--------+---------------+-------+---------+--------------------+------+----------+---------------------------------------------+
|  1 | PRIMARY            | s1    | index  | NULL          | group | 54      | NULL               |    7 |   100.00 | Using index; Using temporary                |
|  1 | PRIMARY            | s2    | index  | NULL          | group | 54      | NULL               |    7 |   100.00 | Using where; Using index; Using join buffer |
|  2 | DEPENDENT SUBQUERY | s1    | index  | group         | group | 54      | NULL               |    7 |    85.71 | Using where; Using index                    |
|  2 | DEPENDENT SUBQUERY | s2    | eq_ref | group         | group | 54      | test.s1.group,func |    1 |   100.00 | Using where; Using index                    |
+----+--------------------+-------+--------+---------------+-------+---------+--------------------+------+----------+---------------------------------------------+

Answer 5

Select S1.Subject As LeftSubject
    , S2.Subject As RightSubject
From SourceData As S1
    Join SourceData As S2
        On S2.subject > S1.subject
    Left Join   (
                Select S1.groupid
                    , S1.subject As LeftSubject
                    , S2.subject As RightSubject
                From SourceData As S1
                    Join SourceData As S2
                        On S2.groupid = S1.groupid
                            And S2.subject > S1.subject
                ) As Z
        On Z.groupid = S1.groupid
            And Z.LeftSubject = S1.subject
            And Z.RightSubject = S2.subject
Where Z.groupid is null

使用outis'元组格式的另一种变体：

Select S1.Subject As LeftSubject
    , S2.Subject As RightSubject
From SourceData As S1
    Join SourceData As S2
        On S2.subject > S1.subject
Where (S1.groupid, S1.subject, S2.subject) Not In (
                                                Select S1.groupid
                                                    , S1.subject
                                                    , S2.subject
                                                From SourceData As S1
                                                    Join SourceData As S2
                                                        On S2.groupid = S1.groupid
                                                Where S2.subject > S1.subject
                                                )

SQL查询仅适用于不同组的所有元素对

5 个答案: