Mysql sql重复查询结果

您好，我有如下表格：

Table: port 
tid    leftside    rightside 
1      2           3          
2      3           2
3      2           4
4      4           2
5      4           3
6      3           4
7      4           5
8      5           4
9      3           6
10     6           3

当我运行此查询时

SELECT * 
FROM port pt JOIN port p 
  ON (pt.leftside = p.rightside 
     and p.leftside <> 2) 
WHERE pt.rightside = 2

前两条线表示相同的关系，尽管它们位于相对的两侧。 如果存在重复记录，我希望查询只检索一行。 例如： 查询应返回第行下方

tid    leftside    rightside 
6      3           4

而不是

tid    leftside    rightside 
6      3           4 
5      4           3

---

这些行。

我在MS SQL中使用了它

with port as
(
select 1 as tid, 2 as leftside, 3 as rightside
UNION ALL
select 2, 3, 2
UNION ALL
select 3, 2, 4
UNION ALL
select 4, 4, 2
UNION ALL
select 5, 4, 3
UNION ALL
select 6, 3, 4
UNION ALL
select 7, 4, 5
UNION ALL
select 8, 5, 4
UNION ALL
select 9, 3, 6
UNION ALL
select 10, 6, 3
)
SELECT * 
FROM port pt JOIN port p 
  ON (pt.leftside = p.rightside 
     and p.leftside <> 2) 
WHERE pt.rightside = 2

通过向WHERE子句添加in子查询

SELECT *
FROM port pt JOIN port p 
  ON (pt.leftside = p.rightside 
    and p.leftside <> 2) 
WHERE (pt.rightside = 2)
   AND p.tid in (SELECT max(chk.tid) FROM port chk
                 GROUP BY Case when leftside>rightside
                             then rightside else leftside end,
                          Case when leftside<rightside
                             then rightside else leftside end
                )

它去掉了p.tid=5的行，因为它不是匹配对中较高的最大值

说明：这列出了所有10行，左侧和右侧按低/高顺序排列

SELECT tid,
  Case when leftside>rightside then rightside else leftside end as lower,
  Case when leftside<rightside then rightside else leftside end as higher
FROM port

---

因此，通过在lower&higher上分组并检索maxtid，我们可以得到匹配对的较高tid。

另一种方法是在leftside大于rightside时交换leftside和rightside。然后使用PARTITION BY和RANK删除重复项：

SELECT tid,
       leftside,
       rightside
FROM
(
  SELECT p.tid,
         CASE WHEN p.leftside <= p.rightside
              THEN p.leftside
              ELSE p.rightside
         END AS leftside,
         CASE WHEN p.leftside > p.rightside
              THEN p.leftside
              ELSE p.rightside
         END AS rightside,
         RANK() OVER
         (PARTITION BY CASE WHEN p.leftside <= p.rightside
                            THEN p.leftside
                            ELSE p.rightside
                       END,
                       CASE WHEN p.leftside > p.rightside
                            THEN p.leftside
                            ELSE p.rightside
                       END
          ORDER BY p.tid ASC) AS rank
  FROM port pt JOIN port p 
    ON (pt.leftside = p.rightside
        AND p.leftside <> 2)
  WHERE pt.rightside = 2  
) subquery
WHERE rank = 1;

---

请参见

pt.leftsidept.rights-如果你反向测试，你可以有0行或全部4行：你是对的-编辑我的答案以使用不同的方法，并在这次测试它！

SELECT tid,
  Case when leftside>rightside then rightside else leftside end as lower,
  Case when leftside<rightside then rightside else leftside end as higher
FROM port

SELECT tid,
       leftside,
       rightside
FROM
(
  SELECT p.tid,
         CASE WHEN p.leftside <= p.rightside
              THEN p.leftside
              ELSE p.rightside
         END AS leftside,
         CASE WHEN p.leftside > p.rightside
              THEN p.leftside
              ELSE p.rightside
         END AS rightside,
         RANK() OVER
         (PARTITION BY CASE WHEN p.leftside <= p.rightside
                            THEN p.leftside
                            ELSE p.rightside
                       END,
                       CASE WHEN p.leftside > p.rightside
                            THEN p.leftside
                            ELSE p.rightside
                       END
          ORDER BY p.tid ASC) AS rank
  FROM port pt JOIN port p 
    ON (pt.leftside = p.rightside
        AND p.leftside <> 2)
  WHERE pt.rightside = 2  
) subquery
WHERE rank = 1;