MySQL唯一行

Question

我需要一排没有重复的独特行，我想找到相互间出现次数最多的一对（演员对）。 例如：

我的输出：

            +----------+-----------+
            |actor1_id | actor2_id | 
            +----------+-----------+
            | 27       |    60     | 
            | 60       |    27     | 
            +----------+-----------+

我需要：

            +----------+-----------+
            |actor1_id | actor2_id | 
            +----------+-----------+
            | 27       |    60     | 
            +----------+-----------+

这是代码：

SELECT T3.actor1_id ,T3.actor2_id 
FROM    actor,
        (SELECT T1.actor_id AS actor1_id , T2.actor_id AS actor2_id, COUNT(T1.film_id) AS mutual_films
        FROM(
             SELECT film_actor.film_id AS film_id , film_actor.actor_id AS actor_id
             FROM film_actor) AS T1 ,
             (
             SELECT film_actor.film_id AS film_id , film_actor.actor_id AS actor_id
             FROM film_actor) AS T2
        WHERE T1.film_id = T2.film_id AND T1.actor_id != T2.actor_id 
        GROUP BY T1.actor_id , T2.actor_id) AS T3

WHERE T3.mutual_films = 
                        (SELECT MAX(T3.mutual_films) AS max_mutual
                        FROM
                                (SELECT T1.actor_id AS actor1_id , T2.actor_id AS actor2_id, COUNT(T1.film_id) AS mutual_films
                                FROM(
                                     SELECT film_actor.film_id AS film_id , film_actor.actor_id AS actor_id
                                     FROM film_actor) AS T1 ,
                                     (
                                     SELECT film_actor.film_id AS film_id , film_actor.actor_id AS actor_id
                                     FROM film_actor) AS T2
                                WHERE T1.film_id = T2.film_id AND T1.actor_id != T2.actor_id 
                                GROUP BY T1.actor_id , T2.actor_id) AS T3)

AND T3.actor1_id != T3.actor2_id
GROUP BY T3.actor1_id , T3.actor2_id

Answer 1

您可以使用UNION语句将两列合并为一列，然后对它们进行DISTINCT SELECT。 就像是：

SELECT DISTINCT comb.actor_id FROM (
    SELECT s1.actor1_id AS actor_id FROM source_table AS s1
    UNION
    SELECT s2.actor2_id AS actor_id FROM source_table AS s2
) AS comb

会导致

+---------+
|actor_id |
+---------+
| 27      | 
| 60      | 
+---------+

现在，您可以在派生表上使用JOIN了。

Answer 2

只需使用CASE表达式在2列之间交换值即可。 然后找到不同的行。

询问

select t.`actor1_id_sub` as `actor1_id`,
t.`actor2_id_sub` as `actor2_id` from(
  select `actor1_id`, `actor2_id`,
  case when `actor1_id` < `actor2_id` then `actor1_id`
    else `actor2_id` end as `actor1_id_sub`,
  case when `actor1_id` < `actor2_id` then `actor2_id` 
    else `actor1_id` end as `actor2_id_sub`
  from `actors`
)t
group by t.`actor1_id_sub`, t.`actor2_id_sub`;

Answer 3

只需在WHERE子句中使用<而不是!= ：

T3.actor1_id < T3.actor2_id

Answer 4

这可以使用not exists来实现，如下所示-

SELECT *
  FROM film_actor t1
 WHERE NOT EXISTS (SELECT 1
          FROM film_actor t2
         WHERE t1.actor2_id = t2.actor1_id
           AND t1.actor1_id = t2.actor2_id
           AND t1.actor1_id > t1.actor2_id)

Answer 5

我只能想到的一种方法是：在表中创建另一个字段，然后将两个ID以相同的顺序（从小到大或从大到大）存储在该字段中。 所以你的桌子看起来像这样

+----------+-----------+-----------+
|actor1_id | actor2_id | actor_ids |
+----------+-----------+-----------+
| 27       |    60     |   27-60   |
| 60       |    27     |   27-60   |
+----------+-----------+-----------+

并使用actor_ids行创建SELECT DISTINCT语句。

Answer 6

请检查此内容，此查询可能会解决您的问题。

SELECT test_status, system_generated FROM test_details WHERE test_status <= system_generated 
UNION 
SELECT system_generated, test_status FROM test_details WHERE system_generated < test_status