递归实现集合求解

2015-11-21 01:57:34 · 作者: · 浏览: 5
数据库环境:SQL SERVER 2005
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在群里看到一道题,截图如下:
可能有些朋友还看不懂题目的意思,我们现在就来分析一下。假设有3种产品A、产品B、产品C,产品A和产品B存在替换关系
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如果产品A和产品B中任何一种产品和产品C存在替换关系,则它们就是一组。
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实现思路:把存在替换关系的任意2种产品放到集合1中,然后遍历后续产品,如果满足关系的则加入集合1,否则,新建集合2,
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遍历符合集合2的产品并加入。
/*测试数据*/
WITH    tab
          AS ( SELECT   1 id ,
                        '121500125' fru ,
                        '121500132' sub_fru
               UNION ALL
               SELECT   2 id ,
                        '121500125' fru ,
                        '5B19A4567R' sub_fru
               UNION ALL
               SELECT   3 id ,
                        '121500132' fru ,
                        '121500125' sub_fru
               UNION ALL
               SELECT   4 id ,
                        '121500132' fru ,
                        '5B19A4567R' sub_fru
               UNION ALL
               SELECT   5 id ,
                        '121500178' fru ,
                        '121500180' sub_fru
               UNION ALL
               SELECT   6 id ,
                        '121500180' fru ,
                        '121500178' sub_fru
               UNION ALL
               SELECT   7 id ,
                        '121500180' fru ,
                        'SB19A46291' sub_fru
             ),/*递归,遍历数据加入关系集合*/
        t ( id, fru, sub_fru, fru_r, gp )
          AS ( SELECT   id ,
                        fru ,
                        sub_fru ,
                        CONVERT(VARCHAR(100), fru + ',' + sub_fru) AS fru_r ,--关系集合
                        1 AS gp--组号
               FROM     tab
               WHERE    id = 1
               UNION ALL
               SELECT   t1.id ,
                        t1.fru ,
                        t1.sub_fru ,
                        CONVERT(VARCHAR(100), CASE WHEN ( CHARINDEX(t1.fru,
                                                              t2.fru_r) >
0 OR CHARINDEX(t1.sub_fru, t2.fru_r) > 0 )--如果新产品和关系中的产品有关系,则加入 THEN CASE WHEN CHARINDEX(t1.fru, t2.fru_r) = 0 THEN t2.fru_r + ',' + t1.fru WHEN CHARINDEX(t1.sub_fru, t2.fru_r) = 0 THEN t2.fru_r + ',' + t1.sub_fru ELSE t2.fru_r END ELSE t1.fru + ',' + t1.sub_fru--新建另一组的关系集合 END) AS fru_r , CASE WHEN ( CHARINDEX(t1.fru, t2.fru_r) > 0 OR CHARINDEX(t1.sub_fru, t2.fru_r) > 0 ) THEN t2.gp ELSE t2.gp + 1 END AS gp FROM tab t1 , t t2 WHERE t1.id = t2.id + 1 ),/*取出每组集合中的最大子集*/ t1 AS ( SELECT gp AS id , MAX(fru_r) AS fru FROM t GROUP BY gp )/*按逗号分割关系产品*/ SELECT t1.id , col = CAST(SUBSTRING(fru, number, CHARINDEX(',', fru + ',', number) - number) AS VARCHAR(100)) FROM t1 CROSS APPLY master..spt_values WHERE type = 'P' AND number >= 1 AND number <= LEN(fru + 'a') AND CHARINDEX(',', ',' + fru, number) = number

当然,也可以在递归的时候就把每组的产品都罗列出来,就省去后面拆分字符串的步骤。不过,