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  • 利用SQL进行推理

    数据库环境:SQL SERVER 2008R2

    有如下需求:

    Baker, Cooper, Fletcher, Miller and Smith住在一座房子的不同楼层。
    Baker 不住顶层。Cooper不住底层。
    Fletcher 既不住顶层也不住底层。Miller住得比Cooper高。
    Smith住的楼层和Fletcher不相邻。
    Fletcher住的楼层和Cooper不相邻。
    用SQL写出来
     
    解题思路:
    先实现所有人住楼层的排列组合,然后把条件套进去即求得。如何实现排列组合,
    具体可以参考我前面的文章 http://www.cnblogs.com/boss-he/p/4534017.html
     
    1.基础数据准备
    --准备基础数据,用A、B、C、D、E分别表示Baker, Cooper, Fletcher, Miller and Smith
    CREATE TABLE ttb
        (
          subname VARCHAR(1) ,
          realname VARCHAR(10)
        )
    INSERT  INTO ttb
    VALUES  ( 'A', 'Baker' ),
            ( 'B', 'Cooper' ),
            ( 'C', 'Fletcher' ),
            ( 'D', 'Miller' ),
            ( 'E', 'Smith' )

    2.生成所有可能情况的排列组合

    --生成A、B、C、D、E所有的排列组合
    WITH    x0
              AS ( SELECT   CONVERT(VARCHAR(10), 'A') AS hid
                   UNION ALL
                   SELECT   CONVERT(VARCHAR(10), 'B') AS hid
                   UNION ALL
                   SELECT   CONVERT(VARCHAR(10), 'C') AS hid
                   UNION ALL
                   SELECT   CONVERT(VARCHAR(10), 'D') AS hid
                   UNION ALL
                   SELECT   CONVERT(VARCHAR(10), 'E') AS hid
                 ),
            x1
              AS ( SELECT   hid
                   FROM     x0
                   WHERE    LEN(hid) <= 5
                   UNION ALL
                   SELECT   CONVERT(VARCHAR(10), a.hid + b.hid) AS hid
                   FROM     x0 a
                            INNER JOIN x1 b ON CHARINDEX(a.hid, b.hid, 1) = 0
                 )
        SELECT  hid AS name
        INTO    #tt
        FROM    x1
        WHERE   LEN(hid) = 5
        ORDER BY hid

    3.加入条件,找出满足要求的楼层安排

    WITH    x2
              AS ( SELECT   name
                   FROM     #tt
                   WHERE    SUBSTRING(name, 5, 1) <> 'A'--Baker 不住顶层
                            AND SUBSTRING(name, 1, 1) <> 'B'--Cooper不住底层
                            AND ( SUBSTRING(name, 1, 1) <> 'C'
                                  AND SUBSTRING(name, 5, 1) <> 'C'--Fletcher 既不住顶层也不住底层
                                )
                            AND name LIKE '%B%D%'--Miller住得比Cooper高
                            AND name NOT LIKE '%CE%' AND name NOT LIKE '%EC%' --Smith住的楼层和Fletcher不相邻
                            AND name NOT LIKE '%BC%' AND name NOT LIKE '%CB%' --Fletcher住的楼层和Cooper不相邻
                 ),
            x3--生成楼层号
              AS ( SELECT   number AS id ,
                            SUBSTRING(x2.name, number, 1) AS name
                   FROM     master.dbo.spt_values
                            INNER JOIN x2 ON 1 = 1
                   WHERE    type = 'P'
                            AND number <= 5
                            AND number >= 1
                 )
        SELECT  a.id AS 楼层,
                b.realname AS 姓名
        FROM    x3 a
                INNER JOIN ttb b ON b.subname = a.name
        ORDER BY id

    楼层安排如下:

    (本文完)

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  • 原文地址:https://www.cnblogs.com/boss-he/p/4628305.html
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