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：

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