自关联去掉组内重复数据

数据库环境:SQL SERVER 2005

  现有一个表的数据如下,id是主键,p1,p2是字符串类型,如果当前行的p1,p2字段的值分别等于其它行

的字段p2,p1的值,则视这2行记录为一组。比如,id=1和id=5就属于同一组数据。同一组数据只显示id最小

的那行记录,没有组的数据全部显示。

实现思路:

  将表进行自关联左联,假设表的别名是a,b,根据id进行关联,对关联后的结果集进行过滤。如果b.id是空的,则保留,

如果b.id不为空,则只保留a.id比b.id小的记录。

实现的SQL脚本:

/*1.数据准备*/
WITH    x0
          AS ( SELECT   1 AS id ,
                        'A' AS p1 ,
                        'B' AS p2
               /*UNION ALL
               SELECT   0 AS id ,
                        'A' AS p1 ,
                        'B' AS p2*/
               UNION ALL
               SELECT   2 AS id ,
                        'C' AS p1 ,
                        'D' AS p2
               UNION ALL
               SELECT   3 AS id ,
                        'E' AS p1 ,
                        'F' AS p2
               UNION ALL
               SELECT   4 AS id ,
                        'D' AS p1 ,
                        'C' AS p2
               UNION ALL
               SELECT   5 AS id ,
                        'B' AS p1 ,
                        'A' AS p2
               UNION ALL
               SELECT   6 AS id ,
                        'H' AS p1 ,
                        'J' AS p2
               UNION ALL
               SELECT   7 AS id ,
                        'T' AS p1 ,
                        'U' AS p2
               UNION ALL
               SELECT   8 AS id ,
                        'J' AS p1 ,
                        'H' AS p2
               /*UNION ALL
               SELECT   9 AS id ,
                        'I' AS p1 ,
                        'L' AS p2
               UNION ALL
               SELECT   10 AS id ,
                        'J' AS p1 ,
                        'K' AS p2*/
             ),/*2.去重*/
        x1
          AS ( SELECT   id ,
                        p1 ,
                        p2
               FROM     ( SELECT    id ,
                                    p1 ,
                                    p2 ,
                                    ROW_NUMBER() OVER ( PARTITION BY p1, p2 ORDER BY id ) AS rn
                          FROM      x0
                        ) t
               WHERE    rn = 1
             )
    /*3.求值*/
    SELECT  a.id ,
            a.p1 ,
            a.p2
    FROM    x1 a
            LEFT JOIN x1 b ON b.p1 = a.p2
                              AND b.p2 = a.p1
    WHERE   b.id IS NULL
            OR a.id < b.id
View Code

最终实现的效果如图:

也有网友提出通过ASCII来实现,他的实现SQL脚本如下:

WITH    c1
          AS ( SELECT   1 AS id ,
                        'A' AS p1 ,
                        'B' AS p2
               /*UNION ALL
               SELECT   0 AS id ,
                        'A' AS p1 ,
                        'B' AS p2*/
               UNION ALL
               SELECT   2 AS id ,
                        'C' AS p1 ,
                        'D' AS p2
               UNION ALL
               SELECT   3 AS id ,
                        'E' AS p1 ,
                        'F' AS p2
               UNION ALL
               SELECT   4 AS id ,
                        'D' AS p1 ,
                        'C' AS p2
               UNION ALL
               SELECT   5 AS id ,
                        'B' AS p1 ,
                        'A' AS p2
               UNION ALL
               SELECT   6 AS id ,
                        'H' AS p1 ,
                        'J' AS p2
               UNION ALL
               SELECT   7 AS id ,
                        'T' AS p1 ,
                        'U' AS p2
               UNION ALL
               SELECT   8 AS id ,
                        'J' AS p1 ,
                        'H' AS p2
               /*UNION ALL
               SELECT   9 AS id ,
                        'I' AS p1 ,
                        'L' AS p2
               UNION ALL
               SELECT   10 AS id ,
                        'J' AS p1 ,
                        'K' AS p2*/
             ),
        c2
          AS ( SELECT   MIN(id) AS min_id
               FROM     c1
               GROUP BY ASCII(p1) + ASCII(p2)
             )
    SELECT  c1.*
    FROM    c1
            JOIN c2 ON id = min_id
View Code

咋一看,似乎也可以实现同样的需求。实际上,这种写法存在2个问题:

  1.如果p1,p2是多个字符,ASCII的方式只会取第一个字符的ASCII

  2.ASCII('A')+ASCII('D')=ASCII('B')+ASCII('C'),对于这样的数据,用ASCII的方式无法区分

(本文完)

posted on 2015-08-12 18:34  ToBeHJH  阅读(697)  评论(0编辑  收藏  举报

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